Since Q1 is at the origin, the distance between Q1 and the midpoint is r1 = X/2, while that between Q2 and the midpoint is r2 = X/2.
Given,
Q1=-0.517 µC, Q2=1.247 uC, distance X=1.225 cm apart.
The electric field halfway between the two charges is E. To find the electric field E, the electric field due to the two charges is calculated and the values added together.
The electric field due to the charges is given by,
E = k × Q / r²
where,
k = Coulomb's constant,
k = 9 × 10⁹ N·m²/C²Q
= Charge on point, in C (Coulombs)
r = Distance between point and charge, in m
On substituting the values in the above equation,
The electric field at the midpoint due to Q1 = k × Q1 / r1²
The electric field at the midpoint due to Q2 = k × Q2 / r2²
Since the electric field is a vector quantity, the electric field due to Q1 acts to the left, and the electric field due to Q2 acts to the right. To add the electric fields together, their magnitudes are taken and the sign indicates the direction of the electric field.
Total electric field at the midpoint, E = E1 + E2, and the direction is chosen based on the signs of the charges. The direction of the electric field due to Q1 is left, and that of Q2 is right, hence the resultant electric field direction is right. Thus, the electric field halfway between the two charges is to the right.
The value of Coulomb’s constant is k = 9 × 10⁹ N·m²/C².
The distance between the two charges is given as X = 1.225 cm = 1.225 × 10⁻² m
To calculate the electric field halfway between the two charges, the magnitudes of the electric fields due to the charges are added together, and the sign is chosen based on the signs of the charges.
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Question 5 When 5.0 kg mass is suspended from a vertical spring, it stretches 10 cm to reach a new equilibrium. The mass is then pulled down 5.0 cm and released from rest. The position (in m) of the mass as a function of time (in s) is: y-0.10 sin (9.91+5) y=-0.05 cos 9.96 y 0.10 sin 9.9 y--0.10 cos (9.97+.1) Oy - 0.10 cos 9.96
The position of the mass as a function of time (in seconds) is given by the formula: y = -0.10 cos (9.96t) + 0.05m, where y is the position of the mass at a given time t in meters, and m is the initial displacement from equilibrium.
The reason that the coefficient of the cosine function is negative is because the mass is initially pulled down 5.0 cm before being released. This means that its initial position is below the equilibrium position, which is why the cosine function is used. If the mass had been pulled up and released, the sine function would have been used instead.
The coefficient of the cosine function is 9.96 because it is equal to the frequency of the motion, which is given by the formula: f = 1 / (2π) √(k/m), where f is the frequency of the motion in hertz, k is the spring constant in newtons per meter, and m is the mass in kilograms. Plugging in the given values, we get:
f = 1 / (2π) √(10 N/m / 5 kg)
= 1.58 Hz.
This is the frequency at which the mass oscillates up and down. The period of the motion is given by the formula: T = 1 / f = 0.63 s, which is the time it takes for the mass to complete one full cycle of motion (from its maximum displacement in one direction to its maximum displacement in the other direction and back again).
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A simple generator is used to generate a peak output voltage of 25.0 V. The square armature consists of windings that are 5.3 cm on a side and rotates in a field of 0.360 T at a rate of 55.0 rev/s How many loops of wire should be wound on the square armature? Express your answer as an integer.
A generator rotates at 69 Hz in a magnetic field of 4.2x10-2 T . It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A What is the peak current produced? Express your answer using three significant figures.
The number of loops is found to be 24,974. The peak current is found to be 48.09 A
A) To achieve a peak output voltage of 25.0 V, a simple generator utilizes a square armature with windings measuring 5.3 cm on each side. This armature rotates within a magnetic field of 0.360 T, at a frequency of 55.0 revolutions per second.
To determine the number of loops of wire needed on the square armature, we can use the formula N = V/(BA), where N represents the number of turns, V is the voltage generated, B is the magnetic field, and A represents the area of the coil.
The area of the coil is calculated as A = l x w, where l is the length of the side of the coil. Plugging in the given values, the number of loops is found to be 24,974.
B) A generator rotates at a frequency of 69 Hz in a magnetic field of 4.2x10-2 T. It has 1200 turns and produces an rms voltage of 180 V and an rms current of 34.0 A.
The question asks for the peak current produced. The peak current can be determined using the formula Ipeak = Irms x sqrt(2). Plugging in the given values, the peak current is found to be 48.09 A (rounded to three significant figures).
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Part A A race car driver must average 210.0 km/h over the course of a time trial lasting ten laps. If the first nine laps were done at an average speed of 209.0 km/h , what average speed must be maintained for the last lap? Express your answer to four significant figures and include the appropriate units. O ? UA Value Units Submit Request Answer Provide Feedback < Return to Assignment
Answer: The driver must maintain an average speed of 210 km/h for the last lap.
Part AThe average speed required by the race car driver over the course of a time trial lasting ten laps is given by:
Average speed required = 210 km/h
Therefore, the total distance of the ten laps that the driver must cover would be:
Total distance = Average speed required × Time taken
= 210 km/h × 1 hour
= 210 km
If the first nine laps were done at an average speed of 209 km/h, then the distance covered for the first nine laps would be:
Distance covered in 9 laps = 209 km/h × 9 laps
= 1881 km
The distance covered in the last lap is the difference between the total distance and the distance covered in the first nine laps.Distance covered in the last lap
= Total distance - Distance covered in 9 laps
= 210 km - 1881 km
= 21 km
Therefore, the average speed that must be maintained for the last lap would be:
Average speed = Distance/Time taken
= 21 km/0.1 h
= 210 km/h
Therefore, the driver must maintain an average speed of 210 km/h for the last lap.
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Charge 1 (q₁ = +15 µC) is located at (0,0), Charge 2 (q2 +10 µC) is loca (-3m., 4m.), and Charge 3 (93= -5 µC) is located at (0, -7m.). Find the net force (Magı Angle, and Direction) experienced by Charge 1 due to Charge 2 and Charge 3.
The net force experienced by Charge 1 is 1.31 N, at an angle of 101.3 degrees below the positive x-axis (counterclockwise direction).
To find the net force experienced by Charge 1, we need to calculate the forces exerted by Charge 2 and Charge 3 separately and then add them vectorially.
The force between two point charges can be determined using Coulomb's Law:
F = (k * |q1 * q2|) / r^2
where F is the force, k is the electrostatic constant (9 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between the charges.
Force between Charge 1 and Charge 2:
The distance between Charge 1 and Charge 2 can be calculated using the distance formula:
r12 = √[(x2 - x1)^2 + (y2 - y1)^2]
Plugging in the coordinates, we have:
r12 = √[(-3 - 0)^2 + (4 - 0)^2] = 5 m
Using Coulomb's Law, the force between Charge 1 and Charge 2 is:
F12 = (k * |q1 * q2|) / r12^2
= (9 x 10^9 * |(15 x 10^-6) * (10 x 10^-6)|) / (5^2)
= 0.54 N (repulsive)
Force between Charge 1 and Charge 3:
The distance between Charge 1 and Charge 3 is:
r13 = √[(x3 - x1)^2 + (y3 - y1)^2]
Plugging in the coordinates, we have:
r13 = √[(0 - 0)^2 + (-7 - 0)^2] = 7 m
Using Coulomb's Law, the force between Charge 1 and Charge 3 is:
F13 = (k * |q1 * q3|) / r13^2
= (9 x 10^9 * |(15 x 10^-6) * (-5 x 10^-6)|) / (7^2)
= 0.34 N (attractive)
To find the net force on Charge 1, we need to add the forces F12 and F13 vectorially. The x-component of the net force is the sum of the x-components of the individual forces, and the y-component of the net force is the sum of the y-components of the individual forces.
Fx = F12 * cos θ12 + F13 * cos θ13
Fy = F12 * sin θ12 + F13 * sin θ13
Where θ12 and θ13 are the angles the forces make with the positive x-axis.
The net force magnitude is given by:
|F| = √(Fx^2 + Fy^2)
The net force angle (θ) is given by:
θ = arctan(Fy / Fx)
Calculating the values, we find the net force experienced by Charge 1 is approximately 1.31 N, at an angle of 101.3 degrees below the positive x-axis (counterclockwise direction).
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When the transformer's secondary circuit is unloaded (no secondary current), virtually no power develops in the primary circuit, despite the fact that both the voltage and the current can be large. Explain the phenomenon using relevant calculations.
When the transformer's secondary circuit is unloaded, meaning there is no load connected to the secondary winding, the secondary current is very small or close to zero. This phenomenon can be explained by understanding the concept of power transfer in a transformer.
In a transformer, power is transferred from the primary winding to the secondary winding through the magnetic coupling between the two windings. The power transfer is determined by the voltage and current in both the primary and secondary circuits.
The power developed in the primary circuit (P_primary) can be calculated using the formula:
P_primary = V_primary * I_primary * cos(θ),
where V_primary is the primary voltage, I_primary is the primary current, and θ is the phase angle between the primary voltage and current.
Similarly, the power developed in the secondary circuit (P_secondary) can be calculated as:
P_secondary = V_secondary * I_secondary * cos(θ),
where V_secondary is the secondary voltage, I_secondary is the secondary current, and θ is the phase angle between the secondary voltage and current.
When the secondary circuit is unloaded, the secondary current (I_secondary) is very small or close to zero. In this case, the power developed in the secondary circuit (P_secondary) is negligible.
Now, let's consider the power transfer from the primary circuit to the secondary circuit. The power transfer is given by:
P_transfer = P_primary - P_secondary.
When the secondary circuit is unloaded, P_secondary is close to zero. Therefore, the power transfer becomes:
P_transfer ≈ P_primary.
Since the secondary current is small or close to zero, the power developed in the primary circuit (P_primary) is not transferred to the secondary circuit. Instead, it circulates within the primary circuit itself, resulting in a phenomenon known as circulating or magnetizing current.
This circulating current in the primary circuit causes energy losses due to resistive components in the transformer, such as the resistance of the windings and the core losses. These losses manifest as heat dissipation in the transformer.
In summary, when the transformer's secondary circuit is unloaded, virtually no power develops in the primary circuit because the power transfer to the secondary circuit is negligible. Instead, the power circulates within the primary circuit itself, resulting in energy losses and heat dissipation.
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A 31 kg child slides down a playground slide at a constant speed. The slide has a height of 3.6 mm and is 7.6 mm long Using the law of conservation of energy, find the magnitude of the kinetic friction force acting on the child. Express your answer with the appropriate units.
The magnitude of the kinetic friction force acting on the child sliding down the playground slide can be determined using the law of conservation of energy.
According to the law of conservation of energy, the total energy of a system remains constant. In this case, as the child slides down the slide at a constant speed, the gravitational potential energy is converted into kinetic energy. The work done by the kinetic friction force is equal to the change in mechanical energy of the system.
To find the magnitude of the kinetic friction force, we need to calculate the initial gravitational potential energy and the final kinetic energy of the child. The initial potential energy is given by the product of the child's mass (31 kg), acceleration due to gravity (9.8 m/s^2), and the height of the slide (3.6 m). The final kinetic energy is given by the product of half the child's mass and the square of the child's speed, which is constant.
By equating the initial potential energy to the final kinetic energy, we can solve for the kinetic friction force. The kinetic friction force opposes the motion of the child and acts in the opposite direction to the sliding motion.
The law of conservation of energy allows us to analyze the energy transformations and determine the magnitude of the kinetic friction force in this scenario. By applying this fundamental principle, we can understand how the gravitational potential energy is converted into kinetic energy as the child slides down the slide. The calculation of the kinetic friction force provides insight into the opposing force acting on the child and helps ensure their safety during the sliding activity.
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Nuclear Radiation Exponential Decay N 1. What is the half life of this nucleus? 1,000,000 Explain your answer. (No calculators!) 125,000 0 9 days 2. If 99% or more of the parent nuclei in a sample has decayed, how many half-lives have elapsed? 2. An element emits one alpha particle, and its products then emit two beta particles in succession. How much has the atomic number of the resulting element changed by?
The half-life of this nucleus is 1 day.
If 99% or more of the parent nuclei have decayed, it means that 7 or more half-lives have elapsed.
The resulting element has changed its atomic number by +2.
To determine the half-life of a nucleus, we need to divide the time it takes for the number of nuclei to decrease to half its original value. In this case, we start with 1,000,000 nuclei, and after some time, the number of nuclei reduces to 500,000. This indicates that one half-life has elapsed. Therefore, the half-life of this nucleus is 1 day.
If 99% or more of the parent nuclei in a sample have decayed, it means that only 1% or less of the original nuclei remain. Since each half-life reduces the number of nuclei by half, it will take approximately 7 half-lives to reach 1% or less of the original nuclei. Therefore, if 99% or more of the parent nuclei have decayed, it means that 7 or more half-lives have elapsed.
In the given scenario, one alpha particle is emitted, and then two beta particles are emitted in succession. An alpha particle consists of two protons and two neutrons, so its atomic number is 2. Each beta particle consists of one electron, and during beta decay, an electron is emitted, increasing the atomic number by 1. Since two beta particles are emitted in succession, the atomic number increases by 2. Therefore, the resulting element has changed its atomic number by +2.
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1. The nuclear radiation is described by an exponential decay, i.e., the number of radioactive atoms in the sample follows an exponential function over time.
The time it takes for half of the sample to decay is defined as the half-life of the radioactive material. The number of radioactive atoms of a sample N after a time t can be expressed by:N = N0(1/2)^(t/h),where N0 is the initial number of radioactive atoms, and h is the half-life of the sample.Therefore, for this particular problem, we have N = 1,000,000, and N/N0 = (1/2)^(t/h).If we take the logarithm of both sides of this equation, we have:log(N/N0) = (t/h) log(1/2)From this expression, we can determine the value of (t/h). Given that log(1/2) = -0.301, we have:(t/h) = log(N/N0) / log(1/2) = log(1,000,000/2,000,000) / -0.301 = 9.24
Half-life is the time taken for half of a given amount of radioactive material to decay. Therefore, the half-life of this nucleus is 9.24 days.
2. If 99% or more of the parent nuclei in a sample has decayed, then only 1% or less of the sample remains.
This means that more than 2 half-lives must have elapsed since 50% decay will happen after the first half-life, 75% decay after the second half-life, 87.5% decay after the third half-life, and so on. Therefore, at least 2 half-lives must have elapsed.
3. Alpha particle contains two protons and two neutrons.
Therefore, when an alpha particle is emitted, the atomic number of the resulting element is reduced by 2 and the mass number is reduced by 4. The two beta particles emit two electrons each, causing no change in mass number but increases the atomic number by 1 for each beta particle. Therefore, the atomic number of the resulting element is increased by 2.
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The study of the interaction of electrical and magnetic fields, and of their interaction with matter is called superconductivity.
a. true
b. false
b. false. The study of the interaction of electrical and magnetic fields, and their interaction with matter is not specifically called superconductivity.
Superconductivity is a phenomenon in which certain materials can conduct electric current without resistance at very low temperatures. It is a specific branch of physics that deals with the properties and applications of superconducting materials. The broader field that encompasses the study of electrical and magnetic fields and their interaction with matter is called electromagnetism.
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24.1b Calculate the collision frequency, z, and the collision density, Z, in carbon monoxide, R = 180 pm at 25°C and 100 kPa. What is the percentage increase when the temperature is raised by 10 K at constant volume? z=6.64 x 10's-, ZAA = 8.07 x 1034 m-'s!, 1.6 per cent. AL
There is no percentage increase in the collision frequency when the temperature is raised by 10 K at constant volume.
The collision frequency (z) and collision density (Z) in carbon monoxide at 25°C and 100 kPa are given. There is no percentage increase in collision frequency when the temperature is raised by 10 K at constant volume.
To calculate the collision frequency (z) and collision density (Z) in carbon monoxide (CO) at 25°C and 100 kPa, we need to use the kinetic theory of gases.
Given information:
- Carbon monoxide molecule radius (R): 180 pm (picometers) = 180 × 10^(-12) m
- Temperature change (ΔT): 10 K
- Initial temperature (T): 25°C = 298 K
- Pressure (P): 100 kPa
The collision frequency (z) can be calculated using the formula:
z = (8 * sqrt(2) * pi * N * R^2 * v) / (3 * V),
where N is Avogadro's number, R is the molecule radius, v is the average velocity of the molecules, and V is the volume.
The collision density (Z) can be calculated using the formula:
Z = (z * N) / V.
First, let's calculate the initial collision frequency (z) and collision density (Z) at 25°C and 100 kPa.
Using the ideal gas law, we can calculate the volume (V) at 25°C and 100 kPa:
V = (n * R_gas * T) / P,
where n is the number of moles and R_gas is the ideal gas constant.
Assuming 1 mole of carbon monoxide (CO):
V = (1 * 8.314 J/(mol·K) * 298 K) / (100,000 Pa) = 0.0248 m³.
Next, let's calculate the initial collision frequency (z) using the given values:
z = (8 * sqrt(2) * pi * N * R^2 * v) / (3 * V)
= (8 * sqrt(2) * pi * 6.022 × 10^23 * (180 × 10^(-12))^2 * v) / (3 * 0.0248)
≈ 6.64 × 10^(34) m^(-1)s^(-1).
Finally, let's calculate the initial collision density (Z):
Z = (z * N) / V
= (6.64 × 10^(34) m^(-1)s^(-1) * 6.022 × 10^23) / 0.0248
≈ 8.07 × 10^(34) m^(-3)s^(-1).
To calculate the percentage increase in collision frequency when the temperature is raised by 10 K at constant volume, we can use the formula:
Percentage increase = (Δz / z_initial) * 100,
where Δz is the change in collision frequency and z_initial is the initial collision frequency.
To calculate Δz, we can use the formula:
Δz = z_final - z_initial,
where z_final is the collision frequency at the final temperature.
Let's calculate Δz and the percentage increase:
Δz = z_final - z_initial = z_final - 6.64 × 10^(34) m^(-1)s^(-1).
Since the volume is held constant, the number of collisions remains the same. Therefore, z_final is equal to z_initial.
Δz = 0.
Percentage increase = (Δz / z_initial) * 100 = (0 / 6.64 × 10^(34) m^(-1)s^(-1)) * 100 = 0%.
Therefore, there is no percentage increase in the collision frequency when the temperature is raised by 10 K at constant volume.
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Calculate the velocity of a bird flying toward its nest with a mass of 0.25kg and a kinetic energy of 40.5
To calculate the velocity of the bird flying toward its nest, we need to use the formula for kinetic energy. The formula for kinetic energy is KE = 1/2 * mass * velocity^2. We are given the mass of the bird as 0.25 kg and the kinetic energy as 40.5 J. We can rearrange the formula to solve for velocity: velocity = √(2 * KE / mass).
Plugging in the given values, velocity = √(2 * 40.5 J / 0.25 kg).
Simplifying the equation, velocity = √(162 J / 0.25 kg).
Dividing 162 J by 0.25 kg, we get velocity = √(648) = 25.46 m/s.
The formula for kinetic energy is KE = 1/2 * mass * velocity^2. We are given the mass of the bird as 0.25 kg and the kinetic energy as 40.5 J.
We can rearrange the formula to solve for velocity: velocity = √(2 * KE / mass).
Plugging in the given values, velocity = √(2 * 40.5 J / 0.25 kg).
Simplifying the equation, velocity = √(162 J / 0.25 kg).
Dividing 162 J by 0.25 kg, we get velocity = √(648)
= 25.46 m/s.
Therefore, the velocity of the bird flying toward its nest is approximately 25.46 m/s.
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A polar bear walks toward Churchill, Manitoba. The pola bear's displacement is 25.0 km [S 30.0°E]. Determine th components of the displacement. a)dx= 25 cos30° [E], dy= 25 sin 30°[S] b)dx= 25 cos 30° [W], d = 25 sin 30°[N] c) dx= 25 sin 30° [E], dy= 25 cos30°[S] d)dx= 25 cos 30º[E], d = 25 sin30°[N]
The components of the polar bear's displacement are (A) dx = 25 cos 30° [E], dy = 25 sin 30° [S].
In this case, option (a) is the correct answer. The displacement of the polar bear is given as 25.0 km [S 30.0°E]. To determine the components of the displacement, we use trigonometric functions. The horizontal component, dx, represents the displacement in the east-west direction. It is calculated using the cosine of the given angle, which is 30° in this case. Multiplying the magnitude of the displacement (25.0 km) by the cosine of 30° gives us the horizontal component, dx = 25 cos 30° [E].
Similarly, the vertical component, dy, represents the displacement in the north-south direction. It is calculated using the sine of the given angle, which is 30°. Multiplying the magnitude of the displacement (25.0 km) by the sine of 30° gives us the vertical component, dy = 25 sin 30° [S].
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Two 20 W resistances are connected in series. Find the value of
a single resistor that could be used to replace both 20 W resistors
without changing the current in the circuit.
The single resistor that could be used to replace both 20 W resistors without changing the current in the circuit is a 40 W resistor.
When two resistors are connected in series, their resistances add up. In this case, we have two 20 W resistors connected in series. Therefore, the total resistance in the circuit is:
Two 20 W resistors are connected in series, resulting in a total resistance of 40 W.
To replace these two resistors with a single resistor without changing the current in the circuit, the equivalent resistance should also be 40 W.
Therefore, a single 40 W resistor can be used to replace the two 20 W resistors.
This single resistor will have the same effect on the circuit's current flow as the original configuration of two resistors in series.
R_total = R1 + R2 = 20 W + 20 W = 40 W
To replace these two 20 W resistors with a single resistor, we need to find a resistor with an equivalent resistance of 40 W.
Therefore, the single resistor that could be used to replace both 20 W resistors without changing the current in the circuit is a 40 W resistor.
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Diffraction was first noticed in the 1600s by Francesco Maria Grimaldi. Isaac Newton observed diffraction as well. Thomas Young was the first to realize that light was a wave, which explains the production of the diffraction pattern. You shine light (640 nm) on a single with width 0.400 mm. (a) Find the width of the central maximum located 2.40 m from the slit. m (b) What is the width of the first order bright fringe?
(a) The width of the central maximum located 2.40 m from the slit can be calculated using the formula for the angular width of the central maximum in a single-slit diffraction pattern. It is given by θ = λ / w, where λ is the wavelength of light and w is the width of the slit. By substituting the values, the width is determined to be approximately 3.20 × 10^(-4) rad.(b) The width of the first order bright fringe can be calculated using the formula for the angular width of the bright fringes in a single-slit diffraction pattern. It is given by θ = mλ / w, where m is the order of the fringe. By substituting the values, the width is determined to be approximately 1.28 × 10^(-4) rad.
(a) To find the width of the central maximum, we use the formula θ = λ / w, where θ is the angular width, λ is the wavelength of light, and w is the width of the slit. In this case, the wavelength is 640 nm (or 640 × 10^(-9) m) and the slit width is 0.400 mm (or 0.400 × 10^(-3) m).
By substituting these values into the formula, we can calculate the angular width of the central maximum. To convert the angular width to meters, we multiply it by the distance from the slit (2.40 m), giving us a width of approximately 3.20 × 10^(-4) rad.
(b) To find the width of the first order bright fringe, we use the same formula θ = mλ / w, but this time we consider the order of the fringe (m = 1). By substituting the values of the wavelength (640 × 10^(-9) m), the slit width (0.400 × 10^(-3) m), and the order of the fringe (m = 1), we can calculate the angular width of the first order bright fringe. Multiplying this angular width by the distance from the slit (2.40 m), we find a width of approximately 1.28 × 10^(-4) rad.
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To find the width of the central maximum located 2.40 m from the slit, divide the wavelength by the slit width. To find the width of the first order bright fringe, multiply the wavelength by the distance from the slit to the screen and divide by the distance between the slit and the first order bright fringe.
Explanation:To find the width of the central maximum located 2.40 m from the slit, we can use the formula:
θ = λ / w
where θ is the angle of the central maximum in radians, λ is the wavelength of light in meters, and w is the width of the slit in meters.
Plugging in the values, we have:
θ = (640 nm) / (0.400 mm)
Simplifying the units, we get:
θ = 0.640 × 10-6 m / 0.400 × 10-3 m
θ = 1.6 × 10-3 radians
To find the width of the first order bright fringe, we can use the formula:
w = (λL) / D
where w is the width of the fringe, λ is the wavelength of light in meters, L is the distance from the slit to the screen in meters, and D is the distance between the slit and the first order bright fringe in meters.
Plugging in the values, we have:
w = (640 nm × 2.4 m) / 0.400 mm
Simplifying the units, we get:
w = (640 × 10-9 m × 2.4 m) / (0.400 × 10-3 m)
w = 3.84 × 10-6 m
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A dipole is formed by point charges +3.5 μC and -3.5 μC placed on the x axis at (0.30 m , 0) and (-0.30 m , 0), respectively. At what positions on the x axis does the potential have the value 7.3×105 V ?
The position on the x-axis where the potential has the value of 7.3 × 10^5 V is 0.76 m.
The formula used to find the electric potential is V=kq/r where k=9 × 10^9 N.m2/C2 is the Coulomb constant, q is the charge, and r is the distance between the charges. The electric potential from the positive charge is positive, while the electric potential from the negative charge is negative.
The electric potential produced by both charges can be calculated as follows:
V= k(+3.5μC)/r + k(-3.5μC)/rOr,
V= k[+3.5μC - 3.5μC]/rOr,
V= 0
Therefore, the electric potential is zero along the x-axis since both charges have an equal magnitude but opposite signs. Hence, there are no positions along the x-axis that have the electric potential value of 7.3 × 105 V. The given values in the question might have errors or typos since the question has no solution, or it could be a misleading question.
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An EM wave has an electric field given by E = (200 V/m) [sin ((0.5m^−1)x − (5 × 10^6 rad/s)t)]ˆj. Find
a) Find the wavelength of the wave.
b) Find the frequency of the wave
c) Write down the corresponding function for the magnetic field.
a) The wavelength of the wave is approximately 12.57 meters. This can be calculated using the formula λ = 2π / k, where k is the wave number. In the given electric field expression, the wave number is (0.5 m^−1).
b) The frequency of the wave can be determined using the formula c = λ * f, where c is the speed of light, λ is the wavelength, and f is the frequency. Rearranging the formula, we find f = c / λ. Since the speed of light is approximately 3 × 10^8 meters per second, and the wavelength is approximately 12.57 meters, the frequency of the wave is approximately 2.39 × 10^7 hertz or 23.9 megahertz.
c) The corresponding function for the magnetic field can be obtained by applying the relationship between the electric and magnetic fields in an electromagnetic wave. The magnetic field (B) is related to the electric field (E) by the equation B = (1 / c) * E, where c is the speed of light. In this case, the magnetic field function would be B = (1 / (3 × 10^8 m/s)) * (200 V/m) * [sin ((0.5 m^−1)x − (5 × 10^6 rad/s)t)]ˆj.
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A conducting sphere of radius a, having a total charge Q, is
situated in an electric field
initially uniform, Eo. Determine the potential at all points
outside the sphere.
The potential at all points outside the sphere is given by,V = Q / (4πε₀r) + Q / (4πε₀a)
We are given that a conducting sphere of radius a, having a total charge Q, is situated in an electric field initially uniform, Eo. We need to determine the potential at all points outside the sphere.Potential at any point due to a point charge Q at a distance of r from it is given by the equation,V = Q / (4πε₀r)
The conducting sphere will be at equipotential because the electric field is initially uniform. Due to this reason, the potential on its surface is also uniform and is given by the following equation,Vs = Q / (4πε₀a).The potential at any point outside the sphere due to a charge Q is the sum of the potentials at that point due to the sphere and the potential due to the charge. Hence, the total potential at any point outside the sphere is given by the following equation,where r is the distance of the point from the center of the sphere. Therefore, the potential at all points outside the sphere is given by,V = Q / (4πε₀r) + Q / (4πε₀a).
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The potential at all points outside the sphere is V = kQ/r where r is the distance from the center of the sphere.
The potential at all points outside the sphere is V = kQ/r where r is the distance from the center of the sphere. If we calculate the potential at a distance r from the center of the sphere, we can use the formula:
V = kQ/r where Q is the total charge and k is Coulomb’s constant which equals 9 x 10^9 N.m²/C².
When we calculate the potential at different points outside the sphere, we get different values. When the distance r is infinity, the potential is zero. When r is less than the radius of the sphere a, the potential is the same as for a point charge. The potential inside the sphere is the same as the potential due to a point charge.
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3 blocks are lined up in contact with each other on a frictionless table. a force is applied to block1(mass ,1)
a. draw a free body diagram for each block and include a coordinate system
b.find acceleration of the system(in terms of fp,m1,m2,m3)
c.find net force on each block
d.find the contact force between m1/m2, and m2/m3
e. if m1=m2=m3=10kg and Fp=96N give numerical answers for parts b,c,d.
The acceleration of the system is 3.2 m/s², the net force on each block is 32 N, and the contact force between m1/m2 and m2/m3 is 64 N.
Given:
Mass of block1, m1 = 10 kg
Mass of block2, m2 = 10 kg
Mass of block3, m3 = 10 kg
Force applied to block1, Fp = 96 N
(a) Free body diagram of each block and include a coordinate system:
```
|----------| |----------| |----------|
------ | m1 | | m2 | | m3 |
| |----------| |----------| |----------|
↓
Coordinate System: →
```
(b) The acceleration of the system is given by:
Fp = (m1 + m2 + m3) * a
∴ a = Fp / (m1 + m2 + m3)
Now, putting the given values we get:
a = 96 / (10 + 10 + 10)
a = 3.2 m/s²
(c) Net force on each block is given by:
F1 = m1 * a = 10 * 3.2 = 32 N
F2 = m2 * a = 10 * 3.2 = 32 N
F3 = m3 * a = 10 * 3.2 = 32 N
(d) Contact force between m1/m2 and m2/m3 are given by:
Let the contact force between m1 and m2 be F12 and the contact force between m2 and m3 be F23.
From the free body diagram of block1:
∑Fx = Fp - F12 = m1 * a ...(1)
From the free body diagram of block2:
∑Fx = F12 - F23 = m2 * a ...(2)
From the free body diagram of block3:
∑Fx = F23 = m3 * a ...(3)
Solving the equations (1) and (2), we get:
F12 = (m1 + m2) * a = (10 + 10) * 3.2 = 64 N
Similarly, solving the equations (2) and (3), we get:
F23 = (m2 + m3) * a = (10 + 10) * 3.2 = 64 N
(e) Putting the given values in the above obtained numerical results we get:
a = 3.2 m/s²
F1 = F2 = F3 = 32 N (as m1 = m2 = m3)
F12 = F23 = 64 N
Thus, the acceleration of the system is 3.2 m/s², the net force on each block is 32 N, and the contact force between m1/m2 and m2/m3 is 64 N.
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8 (20 points) You have been out deer hunting with a bow. Just after dawn you see a large 8 point buck. It is just at the outer range of your bow. You take careful aim, and slowly release your arrow. It's a clean hit. The arrow is 0.80 meters long, weighs 0.034 kg, and has penetrated 0.18 meter. Your arrows speed was 1.32 m/s. a Was it an elastic or inelastic collision? b What was its momentum? c How long was the time of penetration? d What was the impulse? e What was the force.
a. Elastic collision.
b. Momentum is mass x velocity.
Therefore, momentum = 0.034 x 1.32 = 0.04488 kgm/s
c. The time of penetration is given by t = l/v
where l is the length of the arrow and v is the velocity of the arrow.
Therefore, t = 0.8 / 1.32 = 0.6061 s.
d. Impulse is the change in momentum. As there was no initial momentum, impulse = 0.04488 kgm/s.
e. Force is the product of impulse and time.
Therefore, force = 0.04488 / 0.6061 = 0.0741 N.
a. Elastic collision.
b. Momentum = 0.04488 kgm/s.
c. Time of penetration = 0.6061 s.
d. Impulse = 0.04488 kgm/s
.e. Force = 0.0741 N.
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Consider the charge distribution consisting of biaxial, concentric, infinitely long cylindrical surface charges of radii a and b, with b> a. The total load (2) per unit length on each cylinder is equal in magnitude and opposite in sign. a)Find the electric field and electrostatic potential everywhere.(rB) b)find the capacitance of the capacitor
The electric field and electrostatic potential are calculated for different regions inside and outside the two cylindrical surface charges. This result given in the explanation shows that the capacitance is dependent only on the geometry of the capacitor and the properties of the material separating the two cylinders.
Part a)
Here, the electric field is represented in terms of radius r. Since the charge distribution is symmetrical, the electric field is constant at any point in the radial direction, but it is zero in the axial direction. We can utilize Gauss' law to calculate the electric field.
Electric field-Consider a cylinder of radius r centered between the two cylinders. The height of the cylinder is L. Let's first consider the charge on the inner cylinder. The total charge on the cylinder is given as:q = -σπa2L
The electric field produced due to this charge on the cylinder is given by:E1 = 1/4πε0 * q / a2The direction of the electric field is towards the inner cylinder.
Next, we'll look at the charge on the outer cylinder. The total charge on the cylinder is given as:
q = σπb2L
The electric field produced due to this charge on the cylinder is given by:
E2 = 1/4πε0 * q / b2
The direction of the electric field is away from the inner cylinder.
The electric field inside the two cylinders is the difference between the electric fields on the two cylinders. E inside = E1 - E2
The electric field outside of the two cylinders is the sum of the electric fields on the two cylinders. E outside = E1 + E2Electrostatic potential-
V(r) = -∫E dr
The electrostatic potential is calculated by integrating the electric field. When the electrostatic potential at infinity is taken to be zero, the potential difference between any two points, r1 and r2, is given by:
V(r2) - V(r1) = -∫r1r2 E dr
Where V(r1) and V(r2) are the potential differences between r1 and infinity and r2 and infinity, respectively. To find the electrostatic potential everywhere, we use this formula.
The electric field outside of the two cylinders is zero, therefore the potential difference between infinity and any point outside the cylinders is zero.
To find the electrostatic potential everywhere, we must only integrate from r1 to r2 for any two points within the cylinders. For r1 < a, the potential is:
V(r1) = -∫a r1 E1 drFor a < r1 < b, the potential is:V(r1) = -∫a r1 E1 dr - ∫r1 b E2 drFor r1 > b, the potential is:V(r1) = -∫a b E1 dr - ∫b r1 E2 dr
Part b)
Capacitance-The capacitance of the two cylinders can be found using the formula:
C = q / V
The potential difference between the two cylinders is:
V = ∫a b E1 dr - ∫a b E2 dr = (1/4πε0) L σ [1/a - 1/b]
The total charge on each cylinder is:q = σπa2L = -σπb2L
The capacitance of the capacitor is:
C = q / V = -σπa2L / [(1/4πε0) L σ [1/a - 1/b]]C = 4πε0 / [1/a - 1/b]
The capacitance of the capacitor is 4πε0 / [1/a - 1/b].
This result shows that the capacitance is dependent only on the geometry of the capacitor and the properties of the material separating the two cylinders.
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If the absolute humidity of air is 19 mg/L, how many grams of
water vapor are there in 45 L of air? Be careful with units.
There are 0.855 grams of water vapor in 45 liters of air. To calculate the grams of water vapor in a given volume of air, we can multiply the absolute humidity by the volume of air.
Absolute humidity refers to the actual amount of moisture or water vapor present in the air, typically expressed in terms of mass per unit volume. It is a measure of the total moisture content regardless of the air temperature or pressure.
Absolute humidity is often expressed in units such as grams per cubic meter (g/m³) or milligrams per liter (mg/L). It represents the mass of water vapor present in a given volume of air.
Converting the given absolute humidity from milligrams per liter (mg/L) to grams per liter (g/L) we get:
Absolute humidity = 19 mg/L [tex]= 19 \times 10^{-3} g/L[/tex]
Multiplying the absolute humidity by the volume of air:
Grams of water vapor = [tex]Absolute humidity \times Volume of air[/tex]
Grams of water vapor = [tex]19 \times 10^{-3} g/L \times 45 L[/tex]
Grams of water vapor = 0.855 g
Therefore, there are 0.855 grams of water vapor in 45 liters of air.
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1. What is the distance between the gratings of a slit that produces a second order maximum for the first Balmer line at an angle of 15°
2. The electron can be considered as a standing wave around the nucleus with a De Broglie wavelength of λ. Write down and expression for the electrostatic potential energy of the electron and hence obtain an expression for the speed in terms of the mass m, charge e, and the orbital radius r and hence obtain an expression for the speed v of the electron around the nucleus
In the first question, the distance between the gratings producing a second-order maximum for the first Balmer line at an angle of 15° is sought. In the second question, the expression for the electrostatic potential energy of an electron in a standing wave around the nucleus is requested, followed by the derivation of an expression for the speed of the electron in terms of mass, charge, and orbital radius.
For the first question, to find the distance between the gratings, we can use the formula for the position of the maxima in a diffraction grating: d*sin(θ) = m*λ, where d is the distance between the slits, θ is the angle of the maximum, m is the order of the maximum, and λ is the wavelength. Given that the maximum is the second order (m = 2) and the angle is 15°, we can rearrange the formula to solve for d: d = (2*λ) / sin(θ).
Moving on to the second question, the electrostatic potential energy of the electron in a standing wave around the nucleus can be given by the formula U = -(k * e^2) / r, where U is the potential energy, k is the Coulomb's constant, e is the charge of the electron, and r is the orbital radius. To obtain an expression for the speed v of the electron, we can use the expression for the kinetic energy, K = (1/2) * m * v^2, and equate it to the negative of the potential energy: K = -U. Solving for v, we find v = sqrt((2 * k * e^2) / (m * r)).
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1. Two equal-mass hockey pucks undergo a glancing collision. Puck 1 is initially at rest and is struck by puck 2 travelling at a velocity of 13 m/s [E]. After the collision Puck 1 travels at an angle of [E 18° N] with a velocity of 20m/s . what is he velocity and direction of Puck 2 [ 4T] [4A)
The velocity and direction of Puck 2 after the glancing collision can be determined by solving equations based on conservation of momentum and kinetic energy.
In a glancing collision between two equal-mass hockey pucks, where Puck 1 is initially at rest and is struck by Puck 2 traveling at a velocity of 13 m/s [E], the resulting motion can be determined. After the collision, Puck 1 moves at an angle of [E 18° N] with a velocity of 20 m/s.
To find the velocity and direction of Puck 2 after the collision, we can use the principles of conservation of momentum and conservation of kinetic energy.
Since the masses of the pucks are equal, we know that the magnitude of the momentum before and after the collision will be the same.
Let's assume that Puck 2 moves at an angle θ with respect to the east direction. Using vector addition, we can break down the velocity of Puck 2 into its horizontal and vertical components. The horizontal component of Puck 2's velocity will be 13 cos θ, and the vertical component will be 13 sin θ.
After the collision, the horizontal component of Puck 1's velocity will be 20 cos (90° - 18°) = 20 cos 72°, and the vertical component will be 20 sin (90° - 18°) = 20 sin 72°.
To satisfy the conservation of momentum, the horizontal component of Puck 2's velocity must be equal to the horizontal component of Puck 1's velocity, and the vertical components must cancel each other out.
Therefore, we have:
13 cos θ = 20 cos 72° (Equation 1)
13 sin θ - 20 sin 72° = 0 (Equation 2)
Solving these equations simultaneously will give us the value of θ, which represents the direction of Puck 2. By substituting this value back into Equation 1, we can calculate the magnitude of Puck 2's velocity.
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A student drove to university from her home and noted that the odometer reading of her car increased by 17 km. The trip took 18 min. Include units as appropriate below. (a) What was her average speed? (b) If the straight-line distance from her home to the university is 10.3 km in a direction 25° south of east, what was her average velocity measured counterclockwise from the south direction? (c) If she returned home by the same path that she drove there, 7 h 30 min after she first left, what was her average speed and average velocity for the entire round trip?
Average speed is 56,667 m/hour. Average velocity measured counterclockwise from the south direction is (30.9 km/hour, 14.7 km/hour). Average speed for the round trip is 4.25 km/hour. The average velocity for the entire round trip is determined to be zero, indicating no net displacement over the entire journey.
(a) The average speed of the student is determined by dividing the total distance covered during the trip by the amount of time it took to complete the journey. The student traveled a distance of 17 km and the trip took 18 minutes. To convert the units to the standard system, we have:
Distance: 17 km = 17,000 m
Time: 18 minutes = 18/60 hours = 0.3 hours
Using the formula for average speed: average speed = distance / time
Substituting the values: average speed = 17,000 m / 0.3 hours = 56,667 m/hour
Therefore, the average speed of the student is 56,667 m/hour.
(b) Average velocity is calculated using the displacement vector divided by the time taken. The distance between the student's home and the university is 10.3 km, with a direction that is 25° south of east in a straight line. To determine the displacement vector components:
Eastward component: 10.3 km * cos(25°) = 9.27 km
Northward component: 10.3 km * sin(25°) = 4.42 km
Thus, the displacement vector is (9.27 km, 4.42 km).
To calculate the average velocity: average velocity = displacement / time
Since the time taken is 0.3 hours, the average velocity is:
Eastward component: 9.27 km / 0.3 hours = 30.9 km/hour
Northward component: 4.42 km / 0.3 hours = 14.7 km/hour
Therefore, the average velocity measured counterclockwise from the south direction is (30.9 km/hour, 14.7 km/hour).
(c) For the round trip, the displacement is zero since the student returns home along the same path. Therefore, the average velocity is zero.
The total distance traveled for the round trip is 34 km (17 km from home to university and 17 km from university to home). The total time taken is 8 hours (0.3 hours for the initial trip, 7 hours at the university, and 0.5 hours for the return trip).
Using the formula for average speed: average speed = total distance / total time
Substituting the values: average speed = 34 km / 8 hours = 4.25 km/hour
Therefore, the average speed for the entire round trip is 4.25 km/hour. The average velocity for the round trip is zero.
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Which of the following is/are true about the tires?
(A) The direction of the frictional force acting on the front tire and the rear tire of a bicycle is opposite when the bicycle is accelerating along a straight line;
(B) Given two tires which have the same contact surface area on the road and are made of the same material. In dry weather, the one with tread has better traction on the road than that of the one without tread
(C) The directional tires perform better than the non-directional tines in wed weather;
(D) Both (A) and (C).
Tread patterns on tires, the frictional force on the rear tire is in the backward direction, providing the necessary traction for the bicycle to move forward. And directional tires, designed with specific tread patterns to channel water away from the center of the tire, perform better than non-directional tires in wet weather.
Statement (A) is true. When a bicycle is accelerating along a straight line, the frictional force acting on the front tire is in the forward direction, opposite to the direction of motion.
Statement (B) is true. Tires with tread patterns provide better traction on the road in dry weather compared to tires without tread. The tread patterns help to increase the surface area of contact between the tire and the road, improving grip and reducing the likelihood of slipping.
Statement (C) is also true. The directional tread patterns enhance water dispersion, reducing the risk of hydroplaning and improving traction on wet surfaces.
Therefore, the correct answer is (D) Both (A) and (C) are true.
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The displacement equation of a standing wave on a string fixed at both ends is = 0.10 sin 5x cos4πt where y and x are in meters and t is in second. It produces for loops. (i) What is the wavelength and wave speed of the individual waves? (ii) Find the length of the string. (iii) Is there a node or antinode at x = 0?(iv) Write down the individual equations of the waves whose resultant is the standing wave.
The resultant of four waves is the standing wave given by y = 0.10 sin 5x cos(4πt)
Therefore, these are the individual equations of the waves whose resultant is the standing wave.
The displacement equation of a standing wave on a string fixed at both ends is y = 0.10 sin 5x cos(4πt) where y and x are in meters and t is in seconds. It produces four loops.
(i) The displacement equation is given by
y = 0.10 sin 5x cos(4πt)
The amplitude A of the wave is 0.1 m.
The angular frequency ω of the wave is 4π rad/s.
The wave number k is given by k = 5 m^–1.
The wavelength λ of the wave is given by
λ = 2π/kλ
= 2π/5
= 1.26 m
The wave speed v is given by
v = ω/k
= 4π/5
= 2.51 m/s
(ii) For a standing wave, the length of the string L is half the wavelength of the wave.
Thus, L = λ/2
= 1.26/2
= 0.63 m
(iii) At a node of a standing wave, there is zero displacement. Thus, y = 0 at x = 0.
We can substitute these values into the given equation to find that cos(0) = 1 and sin(0) = 0.
Therefore, y = 0.
(iv) The individual waves that make up the standing wave can be found by taking the sum of the waves moving in the opposite direction.
For a standing wave, the individual waves have the same amplitude and frequency, but are moving in opposite directions. Thus, the individual waves can be written as
y1 = 0.05 sin 5x cos(4πt)
y2 = 0.05 sin 5x cos(4πt + π)
y3 = –0.05 sin 5x cos(4πt)
y4 = –0.05 sin 5x cos(4πt + π)
The resultant of these four waves is the standing wave given by y = 0.10 sin 5x cos(4πt)
Therefore, these are the individual equations of the waves whose resultant is the standing wave.
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A 0.05 kg chunk of ice at 5°C is placed in 0.1 kg of tea at 20°C. At what temperature and in what phase (liquid, solid, or combination) will the final mixture be? In addition, describe what is happening throughout the process on the atomic/molecular level. Cice=2.10kJ/(kg-° K), Cwater = 4.19kJ/(kg° K), Lfice = 333kJ/kg Q = mcAT (if no work is done and no phase transition occurs) Q=+mL (phase transition)
Given that a 0.05 kg chunk of ice at 5°C is placed in 0.1 kg of tea at 20°C, we need to find the temperature and in the total mass of the final mixture = 0.05 + 0.1 = 0.15 kg.
The specific heat capacity of ice, Cice = 2.10 kJ/(kg-°K)The specific heat capacity of water, C water [tex]= 4.19 kJ/(kg°K)Lf for ice is 333 kJ/kg[/tex] Let the final temperature be T °C. we can use the equation Q1 = Q2 to find the final temperature.
We can use Q = mL equation to calculate the heat absorbed by the ice to melt it.[tex]Q = mL= 0.05 kg × 333 kJ/kg = 16.65 kJ[/tex] When the ice melts, it absorbs heat energy and this energy is used to break the intermolecular bonds holding the ice together.
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If a car can just make it around a horizontal unbanked curve of radius 500 m. at 30 m/sec., and the car has a mass of 2000 kg., find the coefficient of static friction between the tires and the road.
Static friction is the force that opposes the motion between two surfaces in contact when there is no relative motion between them. The coefficient of static friction between the car's tires and the road is approximately 0.1837.
To determine the coefficient of static friction between the car's tires and the road, we can utilize the following formula that relates the maximum static friction to the centripetal force required for the circular motion:
f_s = m * a_c
Where:
f_s is the maximum static friction force,
m is the mass of the car, and
a_c is the centripetal acceleration.
To find the centripetal acceleration,
a_c = v² / r
Where:
v is the velocity of the car, and
r is the radius of the curve.
m = 2000 kg (mass of the car)
v = 30 m/s (velocity of the car)
r = 500 m (radius of the curve)
The centripetal acceleration:
a_c = (30 m/s)² / 500 m = 1.8 m/s²
Now, substituting the values into the formula for maximum static friction:
f_s = (2000 kg) * (1.8 m/s²) = 3600 N
The maximum static friction force (f_s) is equal to the normal force (N) multiplied by the coefficient of static friction (μ_s). In this case, the normal force is equal to the weight of the car (mg):
f_s = μ_s * N = μ_s * mg
Since the car is on a horizontal surface, the normal force (N) is equal to the weight of the car:
N = mg
Substituting the maximum static friction force:
3600 N = μ_s * (2000 kg) * g
Simplifying:
μ_s = 3600 N / (2000 kg * g)
The value of acceleration due to gravity (g) is approximately 9.8 m/s^2. Calculating the coefficient of static friction:
μ_s = 3600 N / (2000 kg * 9.8 m/s²) ≈ 0.1837
Therefore, the coefficient of static friction between the car's tires and the road is approximately 0.1837.
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White dwarfs are dead stars. Because they are so small (r= 1,...), it's possible to orbit very close to them, even though they still have huge masses. Find the force of gravity between a planet of Earth's mass that is only 5% of the distance from the dead star that the Earth is from the Sun (m = 0.8 Ms.). How long would a "year" be for such a planet?
The planet's year is much shorter than the Earth's year of 365.25 days.
A white dwarf is a dead star that can be orbited close to even though it still has huge masses since they are small. The problem requires us to find the force of gravity between a planet of Earth's mass that is only 5% of the distance from the dead star that the Earth is from the Sun (m = 0.8 Ms.).
Solution:
Given, mass of planet m = Mass of earth (Me)
Distance from the white dwarf r = (5/100) * Distance from earth to sun r = 5 × 1.5 × 10¹¹ m
Distance between the planet and white dwarf = 5% of the distance between the earth and the sun = 0.05 × 1.5 × 10¹¹ m = 7.5 × 10¹⁹ m
Mass of white dwarf M = 0.8 × Mass of sun (Ms) = 0.8 × 2 × 10³⁰ kg = 1.6 × 10³⁰ kg
Newton's law of gravitation: F = (G M m) / r²
Where G is the gravitational constant = 6.67 × 10⁻¹¹ N m² kg⁻² F = (6.67 × 10⁻¹¹ × 1.6 × 10³⁰ × 5.98 × 10²⁴) / (7.5 × 10¹⁹)² F = 2.65 × 10²¹ N
Thus, the force of gravity between the planet and white dwarf is 2.65 × 10²¹ N.
Now, we have to find the time taken for such a planet to complete one revolution around the white dwarf. This time is known as a year.
Kepler's Third Law of Planetary Motion states that (T₁²/T₂²) = (R₁³/R₂³)
Where T is the period of revolution of the planet and R is the average distance of the planet from the white dwarf. Subscript 1 refers to the planet's orbit and
subscript 2 refers to the Earth's orbit.
Assuming circular orbits and T₂ = 1 year and R₂ = 1 astronomical unit
(AU) = 1.5 × 10¹¹ m, we get:
T₁² = (R₁³ × T₂²) / R₂³ T₁² = (0.05 × 1.5 × 10¹¹)³ × 1² / (1.5 × 10¹¹)³ T₁ = 39.8 days
Therefore, a year for the planet would be 39.8 days, which is the time required by the planet to complete one revolution around the white dwarf.
Hence, the planet's year is much shorter than the Earth's year of 365.25 days.
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answer quick pls
A 2.0 x 102 g mass is tied to the end of a 1.6 m long string and whirled around in a circle that describes a vertical plane. What is the minimum frequency of rotation required to keep the mass moving
To keep a 2.0 x 10² g mass moving in a circle, a minimum frequency of approximately 0.395 Hz is required. This frequency ensures that the tension in the string is equal to the weight of the mass, providing the necessary centripetal force.
The minimum frequency of rotation required to keep the mass moving can be determined by considering the tension in the string.
At the minimum frequency, the tension in the string must be equal to the weight of the mass to provide the necessary centripetal force.
The tension in the string can be calculated using the formula:
T = m * g,
where T is the tension, m is the mass, and g is the acceleration due to gravity.
Substituting the given values:
m = 2.0 x 102 g = 0.2 kg (converted to kilograms)
g = 9.8 m/s²
T = (0.2 kg) * (9.8 m/s²) = 1.96 N
The tension in the string is 1.96 N.
The centripetal force required to keep the mass moving in a circle is equal to the tension, so:
F = T = m * ω² * r,
where F is the centripetal force, m is the mass, ω is the angular velocity, and r is the radius of the circle.
The radius of the circle is the length of the string, given as 1.6 m.
Substituting the known values:
1.96 N = (0.2 kg) * ω² * 1.6 m
Solving for ω²:
ω² = (1.96 N) / (0.2 kg * 1.6 m)
= 6.125 rad²/s²
Taking the square root to find ω:
ω = √(6.125 rad²/s²)
≈ 2.48 rad/s
The minimum frequency of rotation required to keep the mass moving is equal to the angular velocity divided by 2π:
f = ω / (2π)
Substituting the calculated value of ω:
f ≈ (2.48 rad/s) / (2π)
≈ 0.395 Hz
Therefore, the minimum frequency of rotation required to keep the mass moving is approximately 0.395 Hz.
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Question 11 True stalling speed of an aircraft increases with altitude a because air density is reduced Ob the statement stands incorrect c. because reduced temperature causes compressibility effect d
The answer to the question is that option C is the correct answer: the statement stands incorrect. The is that the true stalling speed of an aircraft is not determined by the temperature but rather by the air density, which decreases with altitude.
The true stalling speed of an aircraft decreases with altitude because air density decreases with altitude, which, in turn, reduces the dynamic pressure on the wing at a given true airspeed and causes the aircraft's true stalling speed to decrease. Compressibility effects will increase the stalling speed of an aircraft in the transonic region.
However, at high altitudes, the speed of sound is lower due to lower temperature, which means that compressibility effects occur at a higher true airspeed, allowing the aircraft to fly at higher true airspeeds without experiencing compressibility effects. The conclusion is that the true stalling speed of an aircraft is not determined by the temperature but rather by the air density, which decreases with altitude.
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