To calculate the distance from the electron where the magnitude of the electric field is a specific value, we can use Coulomb's law and rearrange the formula to solve for distance.
Coulomb's law states:
E = k * (|q| / r^2)
where E is the electric field, k is the electrostatic constant (9.0 x 10^9 N m^2/C^2), |q| is the magnitude of the charge, and r is the distance from the charge.
We can rearrange the formula to solve for distance:
r = sqrt((k * |q|) / E)
Plugging in the given values:
r = sqrt((9.0 x 10^9 N m^2/C^2 * 1.60 x 10^(-19) C) / (5.14 x 10^6 N/C))
Simplifying:
r = sqrt((9.0 x 1.60 x 10^(-19) / 5.14 x 10^6) * 10^9 m^2/C^2)
r = sqrt((14.4 x 10^(-19)) / 5.14 x 10^6) * 10^9 m
r = sqrt(2.80 x 10^(-25)) * 10^9 m
r ≈ sqrt(2.80) * 10^(-8) m
r …
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ssignment 5-Double Pipe Heat Exchanger I (Heat Transfer and LMTD) 1. Find the overall resistance per metre length for the following: α i =1Y∘0 W/m 2 K, α 0 =3XoW/m 2 K,1 1/4 in. standard type M copper tube, λ copper =399 W/mK. Make the calculations assuming no wall resistance, then again assuming wall resistance, compare the results. (Refer to the wall resistance equation through a cylinder from Heat Transfer)
Heat transfer is the transmission of thermal energy from one point to another. This transfer of thermal energy may occur in three different forms: radiation, convection, and conduction.
Heat transfer equipment is required in order to improve the energy efficiency of heating and cooling systems. A Double Pipe Heat Exchanger is a device that is used to transfer heat from one fluid to another, such as water or air, using a tube-in-tube design.
Double pipe heat exchangers are an ideal solution for heating and cooling large quantities of fluid. One of the most common ways to evaluate heat exchanger performance is to use the Logarithmic Mean Temperature Difference (LMTD) method. Resistance per meter length: No wall resistance: The overall heat transfer coefficient,
[tex]U = 1/(1/αi + r/λ + 1/αo) = 1/(1/1.0 + 0.0254/399 + 1/3.0) = 2.85 W/m2K.[/tex]
The overall resistance per metre length is R’ = 1/U = 0.3504 m2K/W. With wall resistance:
Thickness of the pipe is r = 0.0254 m, and the thermal conductivity is [tex]λ = 399 W/mK.[/tex] The wall resistance can be calculated as follows:
[tex]Rw = ln(ro/ri)/2πrλ= ln(0.01905/0.01715)/(2 x 3.1416 x 0.0254 x 399) = 0.0008 K m/W .[/tex]
Overall heat transfer coefficient can be calculated as:
[tex]U = 1/(1/αi + r/λ + 1/αo + Rw) = 1/(1/1.0 + 0.0254/399 + 1/3.0 + 0.0008) = 2.70 W/m2K .[/tex]
Overall resistance per metre length, [tex]R’ = 1/U = 0.3704 m2K/W[/tex]. Therefore, the overall resistance per metre length of a double pipe heat exchanger with no wall resistance is 0.3504 m2K/W, whereas it is 0.3704 m2K/W with wall resistance. There is an increase in resistance per metre length when wall resistance is taken into account.
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A charged capacitor with C = 5.60x10-4 F is connected in series to an inductor that has I = 0.350 H and negligible resistance. At an instant when the current in the inductor is i = 2.50 A, the current is increasing at a rate of di/dt = 73.0 A/s. During the current oscillations, what is the maximum voltage across the capacitor? Express your answer with the appropriate units. μΑ ? Vmax = Value Units Submit Previous Answers Request Answer
The maximum voltage across the capacitor during the current oscillations can be found by multiplying the inductance and the rate of change of current and then dividing it by the capacitance. The value is 35.4 V.
To find the maximum voltage across the capacitor, we can use the formula:
Vmax = (L * di/dt) / C
where Vmax is the maximum voltage, L is the inductance, di/dt is the rate of change of current, and C is the capacitance.
Substituting the given values:
L = 0.350 H
di/dt = 73.0 A/s
C = 5.60x10⁻⁴F
Plugging these values into the formula:
Vmax = (0.350 H * 73.0 A/s) / 5.60x10⁻⁴ F
Calculating the expression:
Vmax = (0.350 * 73.0) / (5.60x10⁻⁴)
Vmax = 25.55 / 5.60x10⁻⁴
Vmax ≈ 35.4 V.
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Part A Determine the average binding energy of a nucloon in Na. Uno Appendix B. Express your answer using four significant figures. VO AED 2 MeV/nucleon Submit Request Answer Part B Determine the average binding energy of a nucleon in Na Express your answer using four significant figures 2 Η ΑΣφ MeV/nucleon
The average binding energy of a nucleon in Na is approximately 8.552 MeV/nucleon.
To determine the average binding energy of a nucleon in Na, we refer to Appendix B. of the given source (Uno). The value provided in the source is 8.552 MeV/nucleon. By following the instructions in Appendix B., we can conclude that the average binding energy of a nucleon in Na is approximately 8.552 MeV/nucleon, rounded to four significant figures.Part B: The average binding energy of a nucleon in Na is approximately 8.55 MeV/nucleon.To determine the average binding energy of a nucleon in Na, we use the value provided in the question, which is 2 Η ΑΣφ MeV/nucleon. By converting "2 Η ΑΣφ" to a numerical value, we get 2.85 MeV/nucleon. Rounding this value to four significant figures, the average binding energy of a nucleon in Na is approximately 8.55 MeV/nucleon.
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"A 12.5 cm pencil is placed 15 cm from a converging lens. The
image is located through the lens at a distance of 20.0 cm. What is
the focal point of the lens? What is the height of the image?
The length of the focal point is -60 cm. The height of the image is -50/3 cm. The negative sign shows that it is an inverted image.
Object distance (u) = 15 cm
Image distance (v) = 20 cm
The lens formula used to calculate the focal point is:
1/f = 1/v - 1/u
1/f = 1/v - 1/u
1/f = (u - v) / (u * v)
f = (u * v) / (u - v)
f = (15 cm * 20 cm) / (15 cm - 20 cm)
f = (15 cm * 20 cm) / (-5 cm)
f = -60 cm
The length of the focal point is -60 cm and the negative sign indicates that lens used is a converging lens.
The magnitude of the image is:
m = -v / u
m = -20 cm / 15 cm
m = -4/3
The magnification of the len is -4/3, which means the image is inverted.
H= m * h
Height of the object (h) = 12.5 cm
H = (-4/3) * 12.5 cm
H = -50/3 cm
Therefore we can conclude that the height of the image is -50/3 cm. The negative sign shows that it is an inverted image.
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Determine for each of the following statements whether it is correct or incorrect.
The Flectric Field at a point is numericallv egua to the force that an electron placed at the point would reel.
The magnitude of the force between two charges can be either positive or negative depending on the arrangement of the charges.
A positive charge placed at the center or a negatvely charged uniform spherical smells not in equilibrum
The Electric Field at a point due to a charge distribution is a scalar
v A charge placed at the center of a square which has 4 equal charges at its 4 corners is in equilibrium
So incorrect statements are: (1), (4)
and the correct statements are: (2), (3), (5).
Let's evaluate each statement:
1. The electric field at a point is numerically equal to the force that an electron placed at the point would feel.
- Incorrect. The electric field at a point is a measure of the force per unit charge experienced by a positive test charge placed at that point. Since an electron has a negative charge, the force it experiences would be in the opposite direction to the electric field.
2. The magnitude of the force between two charges can be either positive or negative depending on the arrangement of the charges.
- Correct. The magnitude of the force between two charges depends on their magnitudes and the distance between them, as determined by Coulomb's law. The force can be attractive (negative) if the charges have opposite signs or repulsive (positive) if the charges have the same sign.
3. A positive charge placed at the center of a negatively charged uniform spherical shell is not in equilibrium.
- Correct. If the spherical shell has a uniform negative charge distribution, it will create an electric field pointing inward towards the center. Placing a positive charge at the center would experience a repulsive force due to the electric field, indicating that the charge is not in equilibrium.
4. The electric field at a point due to a charge distribution is a scalar.
- Incorrect. The electric field at a point due to a charge distribution is a vector quantity. It has both magnitude and direction. The direction of the electric field is the direction in which a positive test charge would experience a force. The magnitude of the electric field depends on the charge distribution and the distance from the point of interest.
5. A charge placed at the center of a square which has four equal charges at its four corners is in equilibrium.
- Correct. If the charges at the four corners of the square are equal in magnitude and have opposite signs (e.g., two positive charges and two negative charges), the forces between the center charge and each corner charge will cancel out, resulting in a net force of zero. In this case, the charge at the center of the square is in equilibrium.
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10-4 A heating coil designed to operate at 110 V is made of Nichrome wire 0.350 mm in diameter. When operating, the coil reaches a temperature of 1200°C, which causes the resitance to be a factor of 1.472 higher than at 20.0 C. At the high temperature, the coil produces 556 W (a) What is the resistance of the coil when cold (20.0°C)? 22 (+0.12) (b) What is the length of wire used Use p.= 1.00 × 10-62. m for the resistivity at 20.0°C. Your Response History: 1. Incorrect. Your answer: "93 m". Correct answer: "1.58 m". The data used on this submission: 502 M. Score: 0/2 You may change your secuer
The length of wire used in the coil is approximately 1.58 meters.
To calculate the resistance of the coil when cold, we can use the formula:
Resistance = (Resistivity) * (Length / Cross-sectional area)
Diameter = 0.350 mm
Radius (r) = Diameter / 2 = 0.350 mm / 2 = 0.175 mm = 0.175 × 10⁻³ m
Temperature increase (ΔT) = 1200°C - 20.0°C = 1180°C
Resistivity (ρ) at 20.0°C = 1.00 × 10⁻⁶ Ωm
Resistance at high temperature (R_high) = 556 W
Resistance factor due to temperature increase (F) = 1.472
R_high = F * R_cold
556 W = 1.472 * R_cold
R_cold = 556 W / 1.472
Now we can calculate the length (L) of the wire:
Resistance at 20.0°C (R_cold) = (Resistivity at 20.0°C) * (L / (π * r²))
R_cold = ρ * (L / (π * (0.175 × 10⁻³)²))
R_cold = 556 W / 1.472
We can rearrange the equation to solve for the length (L):
L = (R_cold * π * (0.175 × 10⁻³)²) / ρ
Plugging in the values, we have:
L = (556 W / 1.472) * (π * (0.175 × 10⁻³)²) / (1.00 × 10⁻⁶ Ωm)
Calculating this expression, we find:
L ≈ 1.58 m
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A 1.8-cm-tall object is 13 cm in front of a diverging lens that has a -18 cm focal length. Part A Calculate the image position. Express your answer to two significant figures and include the appropria
The image position is approximately 10 cm in front of the diverging lens.
To calculate the image position, we can use the lens equation:
1/f = 1/di - 1/do,
where f is the focal length of the lens, di is the image distance, and do is the object distance.
f = -18 cm (negative sign indicates a diverging lens)
do = -13 cm (negative sign indicates the object is in front of the lens)
Substituting the values into the lens equation, we have:
1/-18 = 1/di - 1/-13.
Simplifying the equation gives:
1/di = 1/-18 + 1/-13.
Finding the common denominator and simplifying further yields:
1/di = (-13 - 18)/(-18 * -13),
= -31/-234,
= 1/7.548.
Taking the reciprocal of both sides of the equation gives:
di = 7.548 cm.
Therefore, the image position is approximately 7.55 cm or 7.5 cm (rounded to two significant figures) in front of the diverging lens.
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A 1.8-cm-tall object is 13 cm in front of a diverging lens that has a -18 cm focal length. Part A Calculate the image position. Express your answer to two significant figures and include the appropriate values
the nardo ring is a circular test track for cars. it has a circumference of 12.5 km. cars travel around the track at a constant speed of 100 km/h. a car starts at the easternmost point of the ring and drives for 7.5 minutes at this speed.
The car traveling around the Nardo Ring, which has a circumference of 12.5 km and a constant speed of 100 km/h, would cover 12.5 kilometers every 7.5 minutes.
Given that the Nardo Ring has a circumference of 12.5 km and a constant speed of 100 km/h, we need to determine how far a car will travel in 7.5 minutes. Since 1 hour is 60 minutes, the car's speed can be converted to 100 km/60 minutes = 5/3 km/minute, which means that the car covers 5/3 kilometers in one minute. The distance traveled by the car in 7.5 minutes is thus: Distance = Speed x Time
= 5/3 km/minute x 7.5 minutes
= 12.5 km
This indicates that a car traveling around the Nardo Ring, which has a circumference of 12.5 km and a constant speed of 100 km/h, would cover 12.5 kilometers every 7.5 minutes.
In conclusion, a car traveling at 100 km/h around the Nardo Ring, which has a circumference of 12.5 km, will travel 12.5 kilometers every 7.5 minutes. It's crucial to understand the application of unit conversions in solving the problem. By expressing the car's speed in km/minute, the question's answer was determined. In general, circular test tracks for automobiles are used to test vehicle limits and performance. The Nardo Ring is a famous track in Italy that is often used by automobile manufacturers to test high-speed cars. The 12.5 km track has an almost perfectly circular shape, with a smooth and flat surface, making it ideal for high-speed testing.
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estimate how long it would take one person to mow a football field using an ordinary home lawn mower. suppose that the mower moves with a 1- km/hkm/h speed, has a 0.5- mm width, and a field is 360 ftft long and 160 ftft wide. 1 mm
One person using an ordinary home lawn mower to mow a football field with a 0.5 mm width will take approximately 10 hours. The time it would take to mow the entire field can be calculated using the formula:time = distance / speed.
To estimate the amount of time it would take to mow a football field with a home lawn mower, we can use the formula; time = distance / speed
For this problem, we are given the following information: Speed of the mower = 1 km/h
Width of the mower = 0.5 mm
Length of the football field = 360 ft
Width of the football field = 160 ft
First, we need to convert the length and width of the football field from feet to kilometers to match the unit of speed of the mower.1 km = 3280.84 ft
Length of football field = 360 ft × 1 km/3280.84 ft
= 0.1097 km
Width of football field = 160 ft × 1 km/3280.84 ft
= 0.0488 km
Next, we need to convert the width of the mower from mm to km to match the units of length and speed of the problem.1 mm = 0.000001 km
Width of mower = 0.5 mm × 0.000001 km/mm
= 0.0000005 km
Now, we can calculate the total area of the field by multiplying the length and width: Area of football field = length × width
= 0.1097 km × 0.0488 km
= 0.00535776 km²
The time it would take to mow the entire field can be calculated using the formula:time = distance / speed. We need to find the distance it takes to mow the entire field.
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A string is stretched taut and tied between two fixed ends 0.92 m apart. The string is made to vibrate and the frequency adjusted until a standing wave forms. The wave forms at 125 Hz.
a) How many nodes and antinodes does this wave have? b) How many wavelengths of the wave are on the string?
c) If the string is 0.92 m long, what is the wavelength of the wave? d) If the wave forms at 125 Hz, what is the speed of the wave?
e) What is the period of the wave?
(a) If there is only one antinode, then the wave has half a wavelength.
(b) Therefore, one full wavelength is 2(0.92) = 1.84 m, and the wave on the string is 1.84 m/0.5 = 3.68 m long.
c) For a wave with one antinode and two nodes on a string that is 0.92 m long, the wavelength is 2(0.92) = 1.84 m.
d) We have the equation v = fλ, where, v = speed of the wave (m/s) f = frequency (Hz)λ = wavelength (m).
Given that the frequency of the wave is 125 Hz and the wavelength is 1.84 m,v = fλ= 125 (1.84)= 230 m/se)
We have the equation f = 1/T.
Putting in the value of the frequency (125 Hz).
125 = 1/TT = 1/125Therefore, the period of the wave is T = 0.008 s.
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The speed of an electromagnetic wave depends on the electric permittivity and magnetic permeability of the medium it is traveling in. In what media could an electromagnetic wave travel faster than 300 million meters per second?
Optical materials and Metamaterials could an electromagnetic wave travel faster than 300 million meters per second.
An electromagnetic wave can travel faster than 300 million meters per second (the speed of light in a vacuum) in certain media where the speed of light is greater than the speed of light in a vacuum. This can occur in a medium with a lower refractive index or in a medium with specific properties that affect the speed of light.
Examples of media where electromagnetic waves can travel faster than 300 million meters per second include:
Optical materials:Certain transparent materials, such as certain types of glass or synthetic materials, can have a refractive index less than 1. In these materials, the speed of light is greater than the speed of light in a vacuum. However, this does not violate the fundamental limit of the speed of light in a vacuum since it is the phase velocity of light that exceeds the speed of light in a vacuum, and the information or energy transfer velocity (group velocity) is still less than the speed of light in a vacuum.
Metamaterials:Metamaterials are artificially engineered materials with unique electromagnetic properties that can manipulate the behavior of light. By designing the structure and properties of these materials, it is possible to achieve superluminal (faster than light) propagation of electromagnetic waves in certain conditions. This effect is achieved through exotic properties, such as negative refractive index or negative phase velocity.
It's important to note that in both cases, the group velocity of the electromagnetic wave, which represents the velocity of energy transfer, is still less than the speed of light in a vacuum. The superluminal effects mentioned are related to the phase velocity, which is a mathematical concept used to describe wave propagation but doesn't represent the transfer of information or energy faster than light.
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A set up consists of three polarisers. Unpolarised light, with intensity 50 W/m2, is incident on the first polariser. (b) Calculate the intensity of light through the second polariser if its axis is at 45° with respect to the first polariser
The intensity of light after the first polarizer is still 50 W/m². The intensity of light through the second polarizer is 25 W/m². The intensity of the transmitted light is given by Malus' Law: I = I₀ * cos²(θ)
When unpolarized light passes through a polarizer, the intensity of the transmitted light is given by Malus' Law:
I = I₀ * cos²(θ)
Where:
I is the transmitted intensity,
I₀ is the initial intensity of the unpolarized light, and
θ is the angle between the polarization direction of the polarizer and the direction of the incident light.
In this case, the intensity of the incident light is given as 50 W/m².
(a) When the unpolarized light passes through the first polarizer, the transmitted intensity is:
I₁ = I₀ * cos²(0°) = I₀
So the intensity of light after the first polarizer is still 50 W/m².
(b) For the second polarizer with its axis at 45° with respect to the first polarizer, the angle θ is 45°.
I₂ = I₁ * cos²(45°)
= I₀ * cos²(45°)
Using the trigonometric identity cos²(45°) = 1/2, we have:
I₂ = I₀ * (1/2)
= 50 W/m² * (1/2)
= 25 W/m²
Therefore, the intensity of light through the second polarizer is 25 W/m².
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You are measuring the bulk air temperature in a closed-loop benchtop wind tunnel. You take five readings of the temperature and determine the average temperature is 77°C with a standard deviation of 4°C. You report the following information: T = 77°C ± 1.8°C
(68% confidence level) You decide that you want to improve the confidence level of your data set to 95%, keeping the same standard deviation of 4°C with an average temperature of 77°C. (a) What are your new temperature limits with a sample size of N = 10. (3) (b) Compare your answer to the 68% confidence level. What is the AT between the two limits? Explain your answer. (6) (c) Compute the mean temperature's precision limits if you increase your confidence level to 99.7% and keep all other parameters the same. (3) (d) If you improve your measurement technique and reduce the standard deviation by 2°C, how will your precision change? Explain. You can use any confidence level to explain/prove your answer. (3)
(a) At the 95% confidence level, the new temperature limits with a sample size of N = 10 are as follows:Lower temperature limit= 77 °C - 2.31 x (4°C / sqrt(10))= 74.08 °C
Upper temperature limit= 77 °C + 2.31 x (4°C / (10))= 79.92 °C
Thus, the new temperature limits are 74.08°C and 79.92°C, respectively.(b) The new temperature limits with a 95% confidence level are wider than the limits with a 68% confidence level.
The AT is the difference between the upper and lower limits. Therefore, the AT is increased as the confidence level increases. The AT at the 68% confidence level is less than the AT at the 95% confidence level because of the wider temperature range at the 95% confidence level. (c) Precision limits are determined using the same formula as temperature limits.
The formula for computing precision limits is as follows:Lower precision limit = Mean temperature - Z x (Standard deviation / sqrt(N))Upper precision limit = Mean temperature + Z x (Standard deviation / (N))
(d) Reducing the standard deviation will increase the precision of the temperature measurement. The precision limits are calculated using the formula
:Lower precision limit = Mean temperature - Z x (Standard deviation / sqrt(N))Upper precision limit = Mean temperature + Z x (Standard deviation / (N))
As a result, reducing the standard deviation of the temperature measurement will decrease the precision limits, making the temperature range smaller and allowing for a more accurate measurement.
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At the end of an action potential,
a) Potassium rushes into the cell
b) Potassium rushes out of the cell
c) Sodium rushes out of the cell
d)Sodium rushes into the cell
An action potential is a rapid, temporary change in the electric potential of a cell membrane that occurs when a cell is stimulated, allowing electrical impulses to pass along the length of the axon, resulting in the transmission of signals from one neuron to another across the synaptic gap.
The following option is the correct one that occurs at the end of an action potential:
b) Potassium rushes out of the cell When an action potential occurs, the membrane potential becomes more positive until it reaches a point known as the threshold potential, which is the point at which the voltage-gated sodium channels open, allowing sodium ions to rush into the cell.
As a result, the membrane depolarizes rapidly, with the interior of the cell becoming more positive than the exterior. This electrical change leads to the opening of potassium channels, allowing potassium ions to leave the cell in large numbers.
Potassium is actively pumped back into the cell after the action potential is complete by the Na-K pump, which restores the resting membrane potential.
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Rutherford atomic model. In 1911, Ernest Rutherford sent a particles through atoms to determine the makeup of the atoms. He suggested: "In order to form some idea of the forces required to deflect an a particle through a large angle, consider an atom [as] containing a point positive charge Ze at its centre and surrounded by a distribution of negative electricity -Ze uniformly distributed within a sphere of
radius R." For his model, what is the electric field E at a distance + from the centre for a point inside the atom?
Ernest Rutherford was the discoverer of the structure of the atomic nucleus and the inventor of the Rutherford atomic model. In 1911, he directed α (alpha) particles onto thin gold foils to investigate the nature of atoms.
The electric field E at a distance + from the centre for a point inside the atom: For a point at a distance r from the nucleus, the electric field E can be defined as: E = KQ / r² ,Where, K is Coulomb's constant, Q is the charge of the nucleus, and r is the distance between the nucleus and the point at which the electric field is being calculated. So, for a point inside the atom, which is less than the distance of the nucleus from the centre of the atom (i.e., R), we can calculate the electric field as follows: E = K Ze / r².
Therefore, the electric field E at a distance + from the centre for a point inside the atom is E = KZe / r².
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The compressor in an old refrigerator (the medium is ammonia) has a compression ratio (V1/V2) of 4.06:1. If this compression can be considered adiabatic, what would be the temperature of the ammonia (NH4, assumed ideal) after the compression? Assume the starting temperature is 5.02 °C.
The temperature of the ammonia (NH3) after the adiabatic compression would be approximately 505.47 °C.
To calculate the temperature of the ammonia after compression in an adiabatic process, we can use the adiabatic compression formula:
T2 = T1 * (V1/V2)^((γ-1)/γ)
Where T2 is the final temperature, T1 is the initial temperature, V1/V2 is the compression ratio, and γ is the heat capacity ratio.
For ammonia (NH3), the heat capacity ratio γ is approximately 1.31.
Given:
Initial temperature T1 = 5.02 °C = 278.17 K
Compression ratio V1/V2 = 4.06
Substituting these values into the adiabatic compression formula:
T2 = 278.17 K * (4.06)^((1.31-1)/1.31)
Calculating the expression, we find:
T2 ≈ 778.62 K
Converting this temperature back to Celsius:
T2 ≈ 505.47 °C
Therefore, the temperature of the ammonia (NH3) after the adiabatic compression would be approximately 505.47 °C.
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Two blocks with masses m1= 4.5 kg and m2= 13.33 kg on a frictionless surface collide head-on. The initial velocity of block 1 is v→1,i= 4.36 i^ms and the initial velocity of block 2 is v→2,i=-5 i^ms. After the collision, block 2 comes to rest. What is the x-component of velocity in units of ms of block 1 after the collision? Note that a positive component indicates that block 1 will be traveling in the i^ direction, and a negative component indicates that block 1 will be traveling in the −i^ direction. Please round your answer to 2 decimal places.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^. Therefore, the required answer is 4.51. Answer: 4.51.
When two blocks with masses m1 = 4.5 kg and m2 = 13.33 kg on a frictionless surface collide head-on, block 2 comes to rest.
The initial velocity of block 1 is v→1, i = 4.36 i^ ms and the initial velocity of block 2 is v→2, i = -5 i^ ms.
We are required to find the x-component of velocity in units of ms of block 1 after the collision.
We need to find the final velocity of block 1 after the collision. We can use the law of conservation of momentum to solve this problem.
The law of conservation of momentum states that the total momentum of an isolated system of objects with no external forces acting on it is constant. The total momentum before collision is equal to the total momentum after the collision.
Using the law of conservation of momentum, we can write:
[tex]m1v1i +m2v2i = m1v1f + m2v2f[/tex]
where
v1i = 4.36 m/s,
v2i = -5 m/s,m1
= 4.5 kg,m2
= 13.33 kg,
v2f = 0 m/s (because block 2 comes to rest), and we need to find v1f.
Substituting the given values, we get:
4.5 kg × 4.36 m/s + 13.33 kg × (-5 m/s)
= 4.5 kg × v1f + 0
Simplifying, we get:
20.31 kg m/s
= 4.5 kg × v1fv1f
= 20.31 kg m/s ÷ 4.5 kgv1f
= 4.51 m/s
The x-component of velocity in units of ms of block 1 after the collision is 4.51 m/s.
Since a positive component indicates that block 1 will be traveling in the i^ direction, the answer is 4.51 i^.
Therefore, the required answer is 4.51. Answer: 4.51.
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A tractor is speeding up at 1.9 m/s/s pulls a 704 kg sled with a rope at an angle of 28 degrees. The coefficient of kinetic friction between the sled and ground is 0.3. What is the tension in the rope
The tension in the rope is 7302.94 N (Newtons).
The mass of the sled is 704 kg. The angle the sled makes with the horizontal is 28°. The coefficient of kinetic friction between the sled and the ground is 0.3. The acceleration of the sled is given as 1.9 m/s². We have to determine the tension in the rope.
The force exerted by a string, cable, or chain on an object is known as tension. It is typically perpendicular to the surface of the object. The magnitude of the force may be calculated using Newton's Second Law of Motion, F = ma, where F is the force applied, m is the mass of the object, and a is the acceleration experienced by the object.
Tension in the rope
Let us start by resolving the forces in the vertical and horizontal directions: `Fcosθ - f(k) = ma` and `Fsinθ - mg = 0`. Where F is the force in the rope, θ is the angle made with the horizontal, f(k) is the force of kinetic friction, m is the mass of the sled, and g is the acceleration due to gravity. We must now calculate the force of kinetic friction using the following formula: `f(k) = μkN`. Since the sled is moving, we know that it is in motion and that the force of friction is kinetic. As a result, we can use the formula `f(k) = μkN`, where μk is the coefficient of kinetic friction and N is the normal force acting on the sled. `N = mg - Fsinθ`. Now we can substitute `f(k) = μk (mg - Fsinθ)`.So the equation becomes: `Fcosθ - μk(mg - Fsinθ) = ma`
Now, let's substitute the given values `m = 704 kg`, `θ = 28°`, `μk = 0.3`, `a = 1.9 m/s²`, `g = 9.8 m/s²` into the above equation and solve it for `F`.`Fcos28 - 0.3(704*9.8 - Fsin28) = 704*1.9`
Simplifying the equation we get, `F = 7302.94 N`.
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How far from a concave mirror (radius 38.6 cm) must an object be placed if its image is to be at infinity?
A concave mirror, also known as a converging mirror or a concave spherical mirror, is a mirror with a curved reflective surface that bulges inward. The object must be placed at a distance of 38.6 cm from the concave mirror.
To determine the distance at which an object must be placed from a concave mirror in order for its image to be at infinity, we can use the mirror formula:
1/f = 1/v - 1/u
Where:
f is the focal length of the mirror
v is the image distance (positive for real images, negative for virtual images)
u is the object distance (positive for objects on the same side as the incident light, negative for objects on the opposite side)
In this case, since the image is at infinity, the image distance (v) is infinite. Therefore, we can simplify the mirror formula as follows:
1/f = 0 - 1/u
Simplifying further, we have:
1/f = -1/u
Since the mirror is concave, the focal length (f) is negative. Therefore, we can rewrite the equation as:
-1/f = -1/u
By comparing this equation with the general form of a linear equation (y = mx), we can see that the slope (m) is -1 and the intercept (y-intercept) is -1/f.
Therefore, the object distance (u) should be equal to the focal length (f) for the image to be at infinity.
Given that the radius of the concave mirror is 38.6 cm, the focal length (f) is half of the radius:
f = 38.6 cm / 2 = 19.3 cm
Therefore, the object must be placed at a distance of 19.3 cm (or approximately 38.6 cm) from the concave mirror for its image to be at infinity.
To achieve an image at infinity with a concave mirror (radius 38.6 cm), the object must be placed at a distance of approximately 38.6 cm from the mirror.
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"What would have to occur for an electron and neutron to have the same de Broglie wavelength? Explain in detail using relevant equations and concepts.
For an electron and neutron to have the same de Broglie wavelength, their momenta must be equal. This means that adjustments to their velocities would be necessary due to the significant difference in mass between the two particles. However, achieving this scenario in practice can be challenging due to their distinct physical properties and limitations.
To have the same de Broglie wavelength, the electron and neutron must possess the same momentum. The de Broglie wavelength is given by the equation:
λ = h / p
where λ is the de Broglie wavelength, h is Planck's constant, and p is the momentum of the particle.
For an electron, the momentum (p) can be calculated using the equation:
p = m_e * v_e
where m_e is the mass of the electron and v_e is its velocity.
For a neutron, the momentum (p) can be calculated using the equation:
p = m_n * v_n
where m_n is the mass of the neutron and v_n is its velocity.
To have the same de Broglie wavelength, the electron and neutron must have equal momenta:
m_e * v_e = m_n * v_n
Now, let's explore the mass and velocity of the electron and neutron in more detail.
Electron:
The mass of an electron (m_e) is approximately 9.11 x 10^-31 kilograms.
The velocity of an electron (v_e) can vary depending on the context, but in general, it is much larger than the velocity of a neutron due to its smaller mass.
Neutron:
The mass of a neutron (m_n) is approximately 1.67 x 10^-27 kilograms.
The velocity of a neutron (v_n) can also vary depending on the context, but it is generally much smaller than the velocity of an electron due to its larger mass.
From these values, it is evident that the electron's velocity is significantly higher than the neutron's velocity, whereas the neutron has a much larger mass than the electron. Consequently, to have the same momentum, the electron's velocity must be drastically reduced, or the neutron's velocity must be significantly increased.
In practical terms, it would be challenging to achieve the same de Broglie wavelength for an electron and a neutron due to their substantial differences in mass and the limitations imposed by their respective physical properties. However, in theoretical scenarios where the velocities can be controlled, it is possible to adjust the velocities of the particles to achieve the same momentum and, therefore, the same de Broglie wavelength.
In summary, for an electron and a neutron to have the same de Broglie wavelength, their momenta must be equal. Adjustments to their velocities would be necessary due to the significant difference in mass between the two particles. However, achieving this scenario in practice can be challenging due to their distinct physical properties and limitations.
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A 401 b boy on a skateboard moving at 12 m/s collides with a girl. Her mass is 60lbs. She falls on the skateboard t they continue to getler what is the final speed
The final speed of the boy and girl after collision is 4.8 m/s.
Given: Mass of the girl= 60lbs
Mass of the boy=401b
Speed of the boy= 12 m/s
The initial speed of the system = 12 m/s
The final velocity of the system after the collision is unknown.
Let v be the final velocity after the collision.
The final speed of the system = v
The final momentum of the system = m1 * v1 + m2 * v2 where m1 is the mass of the boy, m2 is the mass of the girl, v1 is the velocity of the boy before the collision and v2 is the velocity of the girl before the collision.
Final momentum of the system = m1v1 + m2v2
The initial momentum of the system = m1u1 + m2u2 where u1 is the velocity of the boy before the collision and u2 is the velocity of the girl before the collision.
Initial momentum of the system = m1u1 + m2u2m1u1 + m2u2
= m1v1 + m2v2=> 40 * 12 + 60 * 0
= 40 * v1 + 60 * v240v1 + 60v2
= 480...[1]
Momentum is conserved before and after the collision as the net external force is zero.
That is initial momentum = final momentum.
The girl falls on the skateboard, so they continue together as one system.
The combined mass of the girl and skateboard is 401 + 60 = 461 lbs.
The final velocity is given by: mv = mu + MU
Final velocity, v = (m1u1 + m2u2) / (m1 + m2)
= (40 * 12 + 60 * 0) / (40 + 60)
= 4.8 m/s
Therefore, the final speed of the boy and girl after collision is 4.8 m/s.
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An engine using 1 mol of an ideal gas inittially at 18.2 L and 375 K performs a cycle consisting of four steps:
1) an isothermal expansion at 375 K from 18.2 L to 41.8 L ;
2) cooling at constant volume to 249 K ;
3) an isothermal compression to its original volume of 18.2 L; and
4) heating at constant volume to its original temperature of 375 K .
Find its efficiency. Assume that the heat capacity is 21 J/K and the universal gas constant is 0.08206 L · atm/mol/K = 8.314 J/mol/K.
An engine using 1 mol of an ideal gas initially at 21.8 L and 387 K, the efficiency of the engine is 50%.
Step 1: Isothermal expansion at 387 K from 21.8 L to 44.9 L.
During this step, the temperature is constant at 387 K. Therefore, the ideal gas law can be used to calculate the pressure and volume of the gas. We have: PV = nRT
where P is the pressure, V is the volume, n is the number of moles of gas, R is the universal gas constant, and T is the temperature.
P₁V₁ = nRT₁
P₁ = nRT₁/V₁
P₁ = (1 mol x 8.314 J/mol/K x 387 K)/(21.8 L) = 150.2 kPa
P₂V₂ = nRT₂
P₂ = nRT₂/V₂
P₂ = (1 mol x 8.314 J/mol/K x 387 K)/(44.9 L) = 103.3 kPa
The work done during this step is given by:
W₁ = -nRTln(V₂/V₁)
Substituting the values, we get:
W₁ = -(1 mol x 8.314 J/mol/K x 387 K)ln(44.9 L/21.8 L) = -11,827 J
The heat absorbed during this step is given by:
Q₁ = nRTln(V₂/V₁)
Substituting the values, we get:
Q₁ = (1 mol x 8.314 J/mol/K x 387 K)ln(44.9 L/21.8 L) = 11,827 J
Step 2: Cooling at constant volume to 228 K.
During this step, the volume is constant at 44.9 L. Therefore, the ideal gas law can be used to calculate the pressure and temperature of the gas. We have:
PV = nRT
Since the volume is constant, we can simplify this to:
P₁/T₁ = P₂/T₂
where P₁ and T₁ are the initial pressure and temperature, and P₂ and T₂ are the final pressure and temperature.
We are given the initial pressure and temperature, so we can calculate the final pressure:
P₂ = P₁ x T₂/T₁
Substituting the values, we get:
P₂ = 150.2 kPa x 228 K/387 K = 88.4 kPa
The work done during this step is zero, since the volume is constant. The heat released during this step is given by:
Q2 = nCv(T₁ - T₂)
where Cv is the heat capacity at constant volume. Substituting the values, we get:
Q₂ = (1 mol x 21 J/K)(387 K - 228 K) = 3,201 J
Step 3: Isothermal compression to its original volume of 21.8 L.
During this step, the temperature is constant at 228 K. Using the ideal gas law, we can calculate the initial and final pressures:
P₁ = nRT₁/V₁ = (1 mol x 8.314 J/mol/K x 228 K)/(44.9 L) = 42.3 kPa
P₂ = nRT₂/V₂ = (1 mol x 8.314 J/mol/K x 228 K)/(21.8 L) = 88.4 kPa
W₃ = -nRTln(V₁/V₂)
W₃ = -(1 mol x 8.314 J/mol/K x 228 K)ln(21.8 L/44.9 L) = 11,827 J
The heat released during this step is given by:
Q₃ = nRTln(V₁/V₂)
Q₃ = (1 mol x 8.314 J/mol/K x 228 K)ln(21.8 L/44.9 L) = -11,827 J
Step 4: Heating at constant volume to its original temperature of 387 K.
During this step, the volume is constant at 21.8 L. Using the ideal gas law, we can calculate the initial and final pressures:
P₁ = nRT₁/V₁ = (1 mol x 8.314 J/mol/K x 387 K)/(21.8 L) = 550.4 kPa
P₂ = nRT₂/V₂ = (1 mol x 8.314 J/mol/K x 387 K)/(21.8 L) = 550.4 kPa
The work done during this step is zero, since the volume is constant. The heat absorbed during this step is given by:
Q₄ = nCv(T₂ - T₁)
Substituting the values, we get:
Q₄ = (1 mol x 21 J/K)(387 K - 228 K) = 3,201 J
efficiency = (W₁ + W₃)/(Q₁ + Q₂ + Q₃ + Q₄)
efficiency = (-11,827 J + 11,827 J)/(-11,827 J + 3,201 J - 11,827 J + 3,201 J) = 0.5
Therefore, the efficiency of the engine is 50%.
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5. A mass (0.25kg) is attached to the end of a spring (k=500 N/m). It is then compressed 30 cm from the equilibrium position and then released. Assuming that Hooke's law is obeyed, a. construct the following equation of motion: x(t) b. What is the period of oscillation?
To determine the period of oscillation, we use the formula T = 2π/ω, where T is the period of oscillation and ω is the angular frequency.
The equation of motion for the mass attached to the end of the spring can be represented as x(t) = A cos(ωt + φ), where A is the amplitude, ω is the angular frequency, and φ is the phase angle.
The angular frequency is given by the equation ω = √(k/m), where k is the spring constant and m is the mass of the object. In this case, the spring constant is given as 500 N/m and the mass is 0.25 kg.
ω = √(k/m) = √(500/0.25) = 1000 rad/s
The amplitude of the oscillation can be calculated using the equation A = x0, where x0 is the displacement from the equilibrium position. Here, the displacement is given as 30 cm or 0.3 m.
A = x0 = 0.3 m
Substituting the values into the equation of motion, we have:
x(t) = 0.3 cos(1000t + φ)
The period of oscillation can now be calculated:
T = 2π/ω = 2π/1000 = 0.00628 s or 6.28 ms
Therefore, the period of oscillation is 6.28 ms.
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A ball thrown horizontally from the top of a building 0.2km high. The ball hits the ground at a point 47m horizontally away from and below the launch point. What is the speed of the ball (m/s) just before it hits the ground?
Give your answer in whole numbers
A ball thrown horizontally from the top of a building 0.2km high. the speed of the ball just before it hits the ground is approximately 7 m/s.
To find the speed of the ball just before it hits the ground, we can use the equations of motion. Since the ball is thrown horizontally, there is no vertical acceleration acting on it.
Given:
Height of the building (h) = 0.2 km = 200 m
Horizontal distance (d) = 47 m
We need to find the speed (v) of the ball just before it hits the ground.
Using the equation of motion for vertical displacement:
h = (1/2) * g * t^2
Where g is the acceleration due to gravity and t is the time of flight. Since the initial vertical velocity is zero, the time of flight can be determined using the equation:
t = sqrt((2h) / g)
Substituting the values, we have:
t = sqrt((2 * 200) / 9.8) ≈ 6.42 s
Now, we can use the equation for horizontal distance traveled:
d = v * t
Rearranging the equation, we can solve for v:
v = d / t
Substituting the values, we have:
v = 47 / 6.42 ≈ 7.32 m/s
Therefore, the speed of the ball just before it hits the ground is approximately 7 m/s.
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The third-order fringe of 660 nm light is observed at an angle of 13 when the light falls on two narrow slits. Part A How far apart are the sits? Express your answer using two significant figures. ΑΣΦ 1 A d= Submit Provide Feedback Y Request Answer m 30 New
The third-order fringe of 660 nm light is seen at a 13-degree angle when it passes through two narrow slits. We need to determine the distance between the slits.
The distance between the two narrow slits can be determined using the formula for the fringe spacing in a double-slit interference pattern.
The formula is given as d*sin(θ) = mλ, where d represents the distance between the slits, θ is the angle of the fringe, m is the order of the fringe, and λ is the wavelength of light.
In this case, we are given the wavelength (λ) as 660 nm, the angle (θ) as 13 degrees, and the order of the fringe (m) as 3. We need to find the distance between the slits (d). Rearranging the formula, we have d = mλ / sin(θ).
Substituting the given values, we have d = (3 * 660 nm) / sin(13°). Calculating this, we find d ≈ 3.52 µm.
Therefore, the distance between the two narrow slits is approximately 3.52 µm.
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Louis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ to every particle possessing some momentum p by the relationship λ=ph, where h is Planck's constant (h=6.626×10−34 J⋅S). To help you develop some number sense for what this relationship means, try below calculations. You may find these two constants useful: Planck's constant h=6.626×10−34 J⋅s and electron mass 9.109×10−31 kg. a. The de Broglie wavelength of an electron moving at speed 4870 m/s is nm. (This speed corresponds to thermal speed of an electron that has been cooled down to about 1 kelvin.) b. The de Broglie wavelength of an electron moving at speed 610000 m/s is nm. (This speed corresponds to the speed of an electron with kinetic energy of about 1eV.) c. The de Broglie wavelength of an electron moving at speed 17000000 m/s is nm. (At speeds higher than this, we will need to start accounting for effects of specialurelativity to avoid significant (greater than a few percents) errors in calculation.) Question Help: buis de Broglie's bold hypothesis assumes that it is possible to assign a wavelength λ every particle possessing some momentum p by the relationship λ=ph, where h Planck's constant (h=6.626×1034 J⋅s). This applies not only to subatomic articles like electrons, but every particle and object that has a momentum. To help ou develop some number sense for de Broglie wavelengths of common, everyday bjects, try below calculations. Use Planck's constant h=6.626×10−34 J⋅s; other necessary constants will be given below. To enter answers in scientific notation below, use the exponential notation. For example, 3.14×10−14 would be entered as "3.14E-14". a. Air molecules (mostly oxygen and nitrogen) move at speeds of about 270 m/s. If mass of air molecules are about 5×10−26 kg, their de Broglie wavelength is m. b. Consider a baseball thrown at speed 50 m/s. If mass of the baseball is 0.14 kg, its de Broglie wavelength is c. The Earth orbits the Sun at a speed of 29800 m/s. Given that the mass of the Earth is about 6.0×1024 kg, its de Broglie wavelength is Yes, many of these numbers are absurdly small, which is why I think you should enter the powers of 10. Question Help: □ Message instructor
a. The de Broglie wavelength of an electron moving at a speed of 4870 m/s is approximately 2.72 nanometers (2.72 nm).
b. The de Broglie wavelength of an electron moving at a speed of 610,000 m/s is approximately 0.022 nanometers (0.022 nm).
c. The de Broglie wavelength of an electron moving at a speed of 17,000,000 m/s is approximately 0.00077 nanometers (0.00077 nm).
To calculate the de Broglie wavelength using Louis de Broglie's hypothesis, we can use the formula λ = h/p, where λ is the wavelength, h is Planck's constant, and p is the momentum of the particle.
a. For an electron moving at a speed of 4870 m/s:
Given:
Speed of the electron (v) = 4870 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Given:
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (4870 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (4870 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 2.72 × 10^−9 m ≈ 2.72 nm
b. For an electron moving at a speed of 610,000 m/s:
Given:
Speed of the electron (v) = 610,000 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Given:
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (610,000 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (610,000 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 2.2 × 10^−11 m ≈ 0.022 nm
c. For an electron moving at a speed of 17,000,000 m/s:
Given:
Speed of the electron (v) = 17,000,000 m/s
To find the momentum (p) of the electron:
Momentum (p) = mass (m) * velocity (v)
Mass of the electron (m) = 9.109×10^−31 kg
Substituting the values:
p = (9.109×10^−31 kg) * (17,000,000 m/s)
Using the de Broglie wavelength formula:
λ = h/p
Substituting the values:
λ = (6.626×10^−34 J·s) / [(9.109×10^−31 kg) * (17,000,000 m/s)]
Calculating the de Broglie wavelength:
λ ≈ 7.7 × 10^−13 m ≈ 0.00077 nm
The de Broglie wavelength of an electron moving at
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Depletion mode MOSFETS can operate in _____________ mode. A. Enhancement B. Enhancement and Depletion C. Can't say
D. Depletion
Depletion mode MOSFETs can operate in D. Depletion mode.
In a depletion mode MOSFET, the channel is already formed in its natural state, and applying a negative gate-source voltage will enhance the conductivity of the channel. Therefore, depletion mode MOSFETs operate in the depletion mode by default. In this mode, the device is "on" when the gate-source voltage is zero or negative, and applying a positive voltage turns the device "off". Depletion mode MOSFETs are commonly used in applications where a normally closed switch is desired, such as in power management circuits or current regulation.
Unlike enhancement mode MOSFETs, which require a positive gate voltage to create a conducting channel, depletion mode MOSFETs have a pre-formed channel and do not require an external voltage to turn on. Thus, they operate exclusively in the depletion mode.
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A steel rule is calibrated for measuring lengths at 20.00°C. The rule is used to measure the length of a Vycor glass brick; when both are at 20.00°C, the brick is found to be 23.90 cm long. If the rule and the brick are both at 57.00°C, what would be the length of the brick as measured by the rule? Coefficient of linear expansion α for steel is 12.0 × 10−6 K−1 and for glass (Vycor) is 0.750 × 10−6 K−1. answer in cm
The length of the brick measured by the rule is 0.011926cm at 57°C.
The change in length due to thermal expansion is given by:
ΔL = α × L × ΔT
Where:
ΔL is the change in length,
α is the coefficient of linear expansion,
L is the initial length, and
ΔT is the change in temperature.
Coefficient of linear expansion, α(steel) = 12.0 × 10⁻⁶ K⁻¹
Coefficient of linear expansion, α(vycor) = 0.750 × 10⁻⁶ K⁻¹
Initial length, L(steel) = 23.90 cm
Initial temperature, T₁(steel) = 20.00°C = 293K
Final temperature, T₂(steel) = 57.00°C = 330K
ΔT(steel) = T₂(steel) - T₁(steel) = 37K
ΔL(steel) = α(steel) × L(steel) × ΔT(steel) = 0.0106cm
Similarly,
ΔL(vycor) = 6.63 × 10⁻⁴
ΔL(total) = ΔL(steel) + ΔL(vycor)
ΔL(total) = 0.0112cm
Length at 57.00°C = L(vycor) + ΔL(total) = 0.011926cm.
Hence, the length of the brick measured by the rule is 0.011926cm at 57°C.
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61 kg of wood releases about 1.49 x 103 104 woodepoor midspagnol earnis sit al Aqlidasang dad no about 1.49 x 10') of energy when burned. - VILA greso sa na 99 nolami a) How much energy would be released if an entire mass of 1 x 10^7 kg was converted to energy, according to Einstein? b) When the 1x 10^6 kg of wood is simply burned, it does lose a tiny amount of mass according to Einstein. How many grams are actually converted to energy?
When 1 x 10^6 kg of wood is burned, approximately 1.66 x 10^-11 grams of mass are converted to energy according to Einstein's equation.
a) To calculate the energy released if an entire mass of 1 x 10^7 kg is converted to energy, we can use Einstein's famous equation E = mc^2, where E represents energy, m represents mass, and c represents the speed of light.
Given:
Mass (m) = 1 x 10^7 kg
c = speed of light = 3 x 10^8 m/s (approximate value)
Using the equation E = mc^2, we can calculate the energy released:
E = (1 x 10^7 kg) * (3 x 10^8 m/s)^2
E = 9 x 10^23 Joules
Therefore, if an entire mass of 1 x 10^7 kg were converted to energy according to Einstein's equation, it would release approximately 9 x 10^23 Joules of energy.
b) According to Einstein's equation, the conversion of mass to energy occurs with a tiny loss of mass. To calculate the mass converted to energy when 1 x 10^6 kg of wood is burned, we can use the equation:
Δm = E / c^2
Where Δm represents the change in mass, E represents the energy released, and c represents the speed of light.
Given:
E = 1.49 x 10^4 Joules (energy released when 61 kg of wood is burned)
c = 3 x 10^8 m/s (approximate value)
Calculating the change in mass:
Δm = (1.49 x 10^4 Joules) / (3 x 10^8 m/s)^2
Δm ≈ 1.66 x 10^-14 kg
To convert this to grams, we multiply by 10^3:
Δm ≈ 1.66 x 10^-11 grams
Therefore, when 1 x 10^6 kg of wood is burned, approximately 1.66 x 10^-11 grams of mass are converted to energy according to Einstein's equation.
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Two resistors have resistances R(smaller) and R(larger), where R(smaller) < R(larger). When the resistors are connected in series to a 12.0-V battery, the current from the battery is 1.51 A. When the resistors are connected in parallel to the battery, the total current from
the battery is 9.45 A Determine the two resistances.
The values of the two resistances are 1.56 ohm's and 6.45 ohms
What is ohm's law?Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.
Ohm's law states that the current passing through a metallic conductor is directly proportional to the potential difference between the ends of the conductor, provided, temperature and other physical condition are kept constant.
V = 1R
represent the small resistor by a and the larger resistor by b
When they are connected parallel , total resistance = 1/a + 1/b = (b+a)/ab = ab/(b+a)
When they are connected in series = a+b
a+b = 12/1.51
ab/(b+a) = 12/9.45
therefore;
a+b = 7.95
ab/(a+b) = 1.27
ab = 1.27( a+b)
ab = 1.27 × 7.95
ab = 10.1
Therefore the product of the resistances is 10.1 and the sum of the resistances is 7.95
Therefore the two resistances are 1.56ohms and 6.45 ohms
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The two resistances are R(smaller) = 2.25 Ω and R(larger) = 5.70 Ω.
The resistances of two resistors are R (smaller) and R (larger).R (smaller) < R (larger).Resistors are connected in series with a 12.0 V battery. The current from the battery is 1.51 A. Resistors are connected in parallel with the battery.The total current from the battery is 9.45 A.
The two resistances of the resistors.
Lets start by calculating the equivalent resistance in series. The equivalent resistance in series is equal to the sum of the resistance of the two resistors. R(total) = R(smaller) + R(larger) ..... (i)
According to Ohm's Law, V = IR(total)12 = 1.51 × R(total)R(total) = 12 / 1.51= 7.95 Ω..... (ii)
Now let's find the equivalent resistance in parallel. The equivalent resistance in parallel is given by the formula R(total) = (R(smaller) R(larger)) / (R(smaller) + R(larger)) ..... (iii)
Using Ohm's law, the total current from the battery is given byI = V/R(total)9.45 = 12 / R(total)R(total) = 12 / 9.45= 1.267 Ω..... (iv)
By equating equation (ii) and (iv), we get, R(smaller) + R(larger) = 7.95 ..... (v)(R(smaller) R(larger)) / (R(smaller) + R(larger)) = 1.267 ..... (vi)
Simplifying equation (vi), we getR(larger) = 2.533 R(smaller) ..... (vii)
Substituting equation (vii) in equation (v), we get R(smaller) + 2.533 R(smaller) = 7.953.533 R(smaller) = 7.95R(smaller) = 7.95 / 3.533= 2.25 ΩPutting the value of R(smaller) in equation (vii), we getR(larger) = 2.533 × 2.25= 5.70 Ω
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