The surface charge and the surface current density on the conducting boundary due to the current-carrying wire, we can use the following equations:
1. Surface Charge Density (σ):
σ = I / v
Where:
I is the current through the wire,
v is the velocity of the charges on the conducting boundary.
In this case, the current I = 10 cos(100r) A.
Since the conducting boundary is assumed to be an equipotential surface, the charges on it will not be in motion (v = 0).
Therefore, the surface charge density on the conducting boundary is σ = 0.
2. Surface Current Density (J):
J = K × σ
Where:
J is the surface current density,
K is the conductivity of the material,
σ is the surface charge density.
As we found in the previous step, σ = 0.
Therefore, the surface current density on the conducting boundary due to the current-carrying wire is also J = 0.
In summary, the surface charge density (σ) and the surface current density (J) on the conducting boundary, in this case, are both zero.
To know more about surface current density refer here
https://brainly.com/question/4623019#
#SPJ11
Assume that the t, data you take at LEVEL3 are as follows: Trial 1:0.009s; Trial 2:0.0109s; Trial 3:0.009s; Using the average of these values and assuming that the diameter of the steel ball is 1.61cm, calculate the known value for v₁. Express your answer in units of m/s with 2 decimals.
The distance traveled is equal to the diameter of the steel ball, which is 1.61 cm (or 0.0161 m).
What is the relationship between frequency and wavelength in electromagnetic waves?To calculate the known value for v₁, we can use the average time data and the diameter of the steel ball.
Given the time measurements of Trial 1: 0.009s, Trial 2: 0.0109s, and Trial 3: 0.009s, we can find the average time by adding these values and dividing by the number of trials (3). The average time is 0.0096s.
Using the formula v = d/t, where v is the velocity, d is the distance traveled, and t is the time taken, we can rearrange the formula to solve for v₁.
Substituting the values into the formula, we have v₁ = 0.0161 m / 0.0096 s, which simplifies to approximately 49.75 m/s.
Therefore, the known value for v₁ is approximately 49.75 m/s.
Learn more about distance traveled
brainly.com/question/12696792
#SPJ11
What resistors would you use to have the output voltage in +/-5% error of -3.3 V and 3.3 V when an inverting op amp has an input voltage of 750 mv. You can choose from the below list of resistors: a. 10 g. 1 k b. 47 h. 10 k2 C. 100 Ω i. 22 kΩ d. 22002 j. 47 kΩ e. 3302 k. 100 kΩ f. 470 Ω
The resistors needed for this can be determined by considering the gain equation of the inverting amplifier. We can use a combination of a 100 Ω input resistor and a 470 Ω feedback resistor.
For the output voltage to be -3.3 V, we need a gain of -3.3 V / 0.75 V = -4.4. Similarly, for the output voltage to be 3.3 V, we need a gain of 3.3 V / 0.75 V = 4.4.From the given list of resistors, we need to choose values that yield a gain of -4.4 and 4.4. Looking at the options, we can use a combination of a 100 Ω input resistor and a 470 Ω feedback resistor to achieve the desired gains.
In an inverting op amp configuration, the gain is given by the ratio of the feedback resistor (Rf) to the input resistor (Rin). By selecting specific resistor values, we can control the gain and thus the output voltage.
In this case, we need a gain of -4.4 for -3.3 V output and a gain of 4.4 for 3.3 V output. By choosing a 100 Ω input resistor and a 470 Ω feedback resistor, we can achieve the desired gains and obtain the required output voltages within a +/-5% error range.
To learn more about resistors click here : brainly.com/question/30672175
#SPJ11
Gas A is monatomic, and Gas B is diatomic. Equal moles of the two gasses are initially at the same temperature,pressure, and volume. Both gasses are then heated at constant volume to the same higher temperature. Which one of the following will not be true when both gases reach the final higher temperature?
When both gases reach the final higher temperature after being heated at constant volume, the following statement will not be true, the two gases will have the same pressure. When heated at constant volume, the gases experience an increase in temperature.
In the scenario described, both gases start with equal moles, the same initial temperature, pressure, and volume. When heated at constant volume, the gases experience an increase in temperature. However, the nature of the gases (monatomic vs. diatomic) affects how they respond to the increase in temperature.
For an ideal gas, the pressure is directly proportional to the temperature, given that the volume and number of moles are constant (as in this case). However, the factor that affects this relationship is the degree of freedom of the gas molecules.
In the case of a monatomic gas (Gas A), it has three degrees of freedom, meaning it can store energy in three independent translational motion modes. As the gas is heated, the increase in temperature directly translates to an increase in the kinetic energy of the gas molecules, resulting in an increase in their average speed. This increase in speed leads to more frequent and forceful collisions with the container walls, thus increasing the pressure of the gas.
On the other hand, a diatomic gas (Gas B) has five degrees of freedom: three for translational motion and two additional degrees of freedom for rotational motion. As the diatomic gas is heated, the increase in temperature not only increases the translational kinetic energy but also the rotational kinetic energy. This increase in rotational energy distributes some of the increased kinetic energy among the rotational modes, resulting in a smaller increase in the average translational speed compared to the monatomic gas. Consequently, the pressure increase of the diatomic gas will be less compared to the monatomic gas at the same final temperature.
Therefore, when both gases reach the final higher temperature, the statement "The two gases will have the same pressure" will not be true. The diatomic gas (Gas B) will have a lower pressure compared to the monatomic gas (Gas A) at the same temperature.
Learn more about temperature here:
https://brainly.com/question/30775264
#SPJ11
Question 1 5 pts Vector A has a magnitude of 42 units and points in the negative x-direction. When vector B is added to A, the resultant vector A + B points in the negative x-direction with a magnitude of 12 units. Find the magnitude and direction of B. 30 units in the positive x-direction 54 units in the negative x-direction 54 units in the positive x-direction 30 units in the negative x-direction
The magnitude is 6√(5) units in the negative x-direction.
We know that vector A has a magnitude of 42 units and points in the negative x-direction. When vector B is added to A, the resultant vector A + B points in the negative x-direction with a magnitude of 12 units.
Therefore, the magnitude of the resultant vector A + B is equal to 12 units.
Since the resultant vector A + B points in the negative x-direction, the direction of vector B should also be in the negative x-direction. This means the angle of vector B with respect to the x-axis will be 180 degrees.
The magnitude of vector B can be found using the Pythagorean theorem: A² + B² = (A + B)², where A = 42, B = |B|, A + B = 12.
On solving, we get:
B² = 12² - 42²
B² = 144 - 1764
B² = 1620
B = √(1620)
B = √(3² * 2² * 5)
B = 3 * 2 * √(5)
B = 6√(5)
Therefore, the magnitude of vector B is 6√(5) units, and the direction is in the negative x-direction. Thus, the answer is 6√(5) units in the negative x-direction.
To learn more about magnitude, refer below:
https://brainly.com/question/31022175
#SPJ11
Choose the incorrect statement? -Goos-Hänchen effect is an optical phenomenon in which non linearly polarized light undergoes a small lateral shift when totally internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a small lateral shift when internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a large lateral shift when totally internally reflected. -Goos-Hänchen effect is an optical phenomenon in which linearly non polarized light undergoes a small lateral shift when totally internally reflected. -All the above
The incorrect statement is Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a large lateral shift when totally internally reflected.
The Goos-Hänchen effect is an optical phenomenon in which linearly polarized light undergoes a small lateral shift when totally internally reflected. The lateral shift is caused by the interaction of the evanescent wave with the polarization of the light. The evanescent wave is a wave that exists in the region between the two media where total internal reflection occurs. It is a very weak wave, but it can interact with the polarization of the light and cause it to shift laterally.
The lateral shift of the Goos-Hänchen effect is typically on the order of a few micrometers. It is a very small effect, but it can be used to measure the polarization of light.
The other statements about the Goos-Hänchen effect are all correct. The Goos-Hänchen effect is an optical phenomenon that occurs when linearly polarized light is totally internally reflected. The lateral shift is caused by the interaction of the evanescent wave with the polarization of the light. The lateral shift is small, but it can be used to measure the polarization of light.
To learn more about polarized light click here
https://brainly.com/question/17159388
#SPJ11
Through a resistor connected to two batteries in series of 1.5 V
each, a current of 3 mA passes. How much is the resistance of this
element.
a. 0.5KQ
b. 1.00
c 1.0 MQ
d. 1.0 kQ
Using Ohm's Law, we find that the resistance of the element is 1.0 kΩ. The correct option is d).
Ohm's Law states that the current passing through a resistor is directly proportional to the voltage across it and inversely proportional to its resistance.
Ohm's Law: V = I * R
Where:
V is the voltage across the resistor (in volts)
I is the current passing through the resistor (in amperes)
R is the resistance of the resistor (in ohms)
In this case, we have two batteries in series, each with a voltage of 1.5V. The total voltage across the resistor is the sum of the voltages of both batteries:
V = 1.5V + 1.5V = 3V
The current passing through the resistor is given as 3 mA, which is equivalent to 0.003 A.
Now, we rearrange Ohm's Law to solve for the resistance:
R = V / I
R = 3V / 0.003A
R = 1000 ohms = 1 kΩ
Therefore, the resistance of the element is 1.0 kΩ. The correct option is d).
To know more about Ohm's Law, refer to the link :
https://brainly.com/question/1247379#
#SPJ11
As commonly observed, the filament is more likely to blow when
the light is switched on, than when the light is being switched
off. Why?
The filament in an incandescent light bulb is more likely to blow when the light is switched on due to the sudden surge of current and rapid heating, leading to stress and weakening of the filament.
The filament in an incandescent light bulb is more likely to blow when the light is switched on compared to when it is being switched off. This is because when the light is switched on, there is a sudden surge of current flowing through the filament, causing it to rapidly heat up. The rapid heating leads to a thermal expansion of the filament, which can create stress and weaken the filament over time. Additionally, the sudden surge of current can also cause a higher rate of evaporation of the tungsten material in the filament, further weakening it. On the other hand, when the light is being switched off, the current gradually decreases, allowing the filament to cool down more slowly and reducing the likelihood of immediate failure.
To learn more about current: https://brainly.com/question/29766827
#SPJ11
One application of L-R-C series circuits is to high-pass or low-pass filters, which filter out either the low- or high-frequency components of a signal. A high-pass filter is shown in Fig. P31.47, where the output voltage is taken across the L-R combination. (The L-R combination represents an inductive coil that also has resistance due to the large length of wire in the coil.) Derive an expression for Vout / Vs, the ratio of the output and source voltage amplitudes, as a function of the angular frequency ω of the source. Show that when ω is small, this ratio is proportional to ω and thus is small, and show that the ratio approaches unity in the limit of large frequency.
In electrical engineering, an L-R-C series circuit is a type of electrical circuit in which inductance, resistance, and capacitance are connected in a series arrangement. This type of circuit has many applications, including high-pass or low-pass filters.
Figure P31.47 shows a high-pass filter circuit where the output voltage is taken across the L-R combination. In this circuit, the L-R combination represents an inductive coil that has resistance due to the large length of wire in the coil.
The ratio of the output and source voltage amplitudes can be found by deriving an expression for Vout/Vs as a function of the angular frequency ω of the source.
The voltage across the inductor, VL, can be expressed as follows:
VL = jωL
where j is the imaginary unit, L is the inductance, and ω is the angular frequency.
The voltage across the resistor, VR, can be expressed as follows:
VR = R
where R is the resistance.
The voltage across the capacitor, VC, can be expressed as follows:
VC = -j/(ωC)
where C is the capacitance. The negative sign indicates that the voltage is 180 degrees out of phase with the current.
The total impedance, Z, of the circuit is the sum of the impedance of the inductor, resistor, and capacitor. It can be expressed as follows:
Z = R + jωL - j/(ωC)
The output voltage, Vout, is the voltage across the L-R combination and can be expressed as follows:
Vout = VL - VR = jωL - R
The input voltage, Vs, is the voltage across the circuit and can be expressed as follows:
Vs = ZI
where I is the current.
The ratio of the output and source voltage amplitudes, Vout/Vs, can be expressed as follows:
Vout/Vs = (jωL - R)/Z
Substituting for Z and simplifying the expression gives:
Vout/Vs = jωL/(jωL + R - j/(ωC))
Taking the absolute value of this expression and simplifying gives:
|Vout/Vs| = ωL/√(R² + (ωL - 1/(ωC))²)
When ω is small, this ratio is proportional to ω and thus is small. As the frequency increases, the ratio approaches unity in the limit of large frequency.
To know more about arrangement visit:
https://brainly.com/question/30838941
#SPJ11
. 5. Which of the following is/are correct about a sound wave? A. B. C. Infrasound is visible to the eye. Sound waves can travel in a conductor. Sound wave travels in a vacuum at 3 x 108 m/s.
Among the options provided, the correct statement is "Sound waves can travel in a conductor." Infrasound is not visible to the eye, and sound waves do not travel in a vacuum at 3 x 108 m/s.
A. Infrasound is not visible to the eye. Infrasound refers to sound waves with frequencies below the range of human hearing, typically below 20 Hz. Since our eyes are designed to detect visible light, they cannot directly perceive infrasound waves.
B. Sound waves can travel in a conductor. Yes, this statement is correct. Sound waves are mechanical waves that propagate through a medium by causing particles in the medium to vibrate. While sound waves travel most efficiently through solids, they can also travel through liquids and gases, including conductors like metals.
C. Sound waves do not travel in a vacuum at 3 x 108 m/s. Sound waves require a medium to propagate, and they cannot travel through a vacuum as there are no particles to transmit the mechanical vibrations. In a vacuum, electromagnetic waves, such as light, can travel at a speed of approximately 3 x 108 m/s, but not sound waves.
Learn more about Infrasound here:
https://brainly.com/question/17025745
#SPJ11
An aeroplane flies at Ma=0.8 in air at 15°C and 100 kPa. Given that y = 1.4 and R = 283 J/(kg K). (a) Calculate the stagnation pressure and stagnation temperature. (b) Find the stagnation pressure and stagnation temperature if the aeroplane flies at Ma = 2.
"For Ma = 2, the stagnation pressure is approximately 540.1 kPa, and the stagnation temperature is approximately 518.67 K." Stagnation pressure denoted as P0, is a thermodynamic property in fluid mechanics that represents the total pressure of a fluid flow. It is also known as the total pressure or the pitot pressure.
Stagnation pressure is the pressure that a fluid would have if it were brought to rest (stagnated) isentropically (without any losses) by a process known as adiabatic deceleration.
To calculate the stagnation pressure and stagnation temperature, we can use the following equations:
(a) For Ma = 0.8:
Stagnation pressure (P0) = P * (1 + ((y - 1) / 2) * Ma²)^(y / (y - 1))
Stagnation temperature (T0) = T * (1 + ((y - 1) / 2) * Ma²)
From question:
P = 100 kPa
T = 15°C = 15 + 273.15 = 288.15 K
y = 1.4
Substituting these values into the equations:
Stagnation pressure (P0) = 100 * (1 + ((1.4 - 1) / 2) * 0.8²)¹°⁴/ ¹°⁴⁻¹)
Stagnation temperature (T0) = 288.15 * (1 + ((1.4 - 1) / 2) * 0.8²)
Calculating:
Stagnation pressure (P0) ≈ 100 * (1 + (0.4 / 2) * 0.64)¹°⁴/ ¹°⁴⁻¹
≈ 100 * (1 + 0.32)³°⁵
≈ 100 * 1.32³°⁵
≈ 100 * 2.047
≈ 204.7 kPa
Stagnation temperature (T0) ≈ 288.15 * (1 + (0.4 / 2) * 0.64)
≈ 288.15 * (1 + 0.32)
≈ 288.15 * 1.32
≈ 380.28 K
Therefore, for Ma = 0.8, the stagnation pressure is approximately 204.7 kPa, and the stagnation temperature is approximately 380.28 K.
(b) For Ma = 2:
Using the same equations as before:
Stagnation pressure (P0) = P * (1 + ((y - 1) / 2) * Ma^2)^(y / (y - 1))
Stagnation temperature (T0) = T * (1 + ((y - 1) / 2) * Ma²)
The values:
P = 100 kPa
T = 15°C = 15 + 273.15 = 288.15 K
y = 1.4
Ma = 2
Substituting these values into the equations:
Stagnation pressure (P0) = 100 * (1 + ((1.4 - 1) / 2) * 2²)¹°⁴/¹°⁴⁻¹)
Stagnation temperature (T0) = 288.15 * (1 + ((1.4 - 1) / 2) * 2²)
Calculating:
Stagnation pressure (P0) ≈ 100 * (1 + (0.4 / 2) * 4)¹°⁴/⁰°⁴
≈ 100 * (1 + 0.8)³°⁵
≈ 100 * 1.8^3.5
≈ 100 * 5.401
≈ 540.1 kPa
Stagnation temperature (T0) ≈ 288.15 * (1 + (0.4 / 2) * 4)
≈ 288.15 * (1 + 0.8)
≈ 288.15 * 1.8
≈ 518.67 K
Therefore, for Ma = 2, the stagnation pressure is approximately 540.1 kPa, and the stagnation temperature is approximately 518.67 K.
To know more about stagnation pressure visit:
https://brainly.com/question/33310913
#SPJ11
A 1.1-kg block of ice is initially at a temperature of -4.0 ∘C.
Part A If 6.6×105 J of heat are added to the ice, what is the final
temperature of the system? Express your answer using two
signific
The specific heat capacity of water is approximately 4.18 J/g°C .
What is the final temperature of the system?The heat needed to bring the ice from -4.0 °C to its melting point at 0 °C must first be determined. Ice has a specific heat capacity of about 2.09 J/g°C.
Heat needed to raise the ice's temperature:
Q1 = (1.1 kg) * (0 °C - (-4.0 °C)) * (2090 J/kg°C)
Next, we need to calculate the heat required to melt the ice at 0 °C. The heat of fusion for ice is approximately 334,000 J/kg.
Heat required to melt the ice:
Q2 = (1.1 kg) * (334,000 J/kg)
The total heat added to the system is the sum of Q1 and Q2:
Total heat added = [tex]Q1 + Q2 + 6.6[/tex]×[tex]10^5 J[/tex]
Finally, given the total heat delivered and the water's specific heat capacity, we must determine the system's final temperature.
So, The specific heat capacity of water is approximately 4.18 J/g°C .
Learn more about specific heat capacity here : brainly.com/question/27862577
#SPJ4
12). Someone wants to look at Mercury through a telescope (f. = 4.1 m) because they live on the edge. To get the magnification to 600x, what focal length of eyepiece do you need to use? How big will the image of Mercury appear to the viewer? Let's give Mercury the best values for this: It's 90 million km away and has a radius of 2100 km. 13). Light of orange color (1 = 590 nm) is vertically projected through two slits (d = 1.6 pm) onto a screen that is 1.3 m from the slits. Find the distance between the first and third maxima on the screen. Find the distance between the second and negative second maxima.
12. The image of Mercury will appear to the viewer with an angular size of approximately 0.04667 degrees.
13. Distance between second and negative second maxima ≈ 1.1 m.
12. To calculate the focal length of the eyepiece needed to achieve a magnification of 600x, we can use the formula for angular magnification:
Magnification = -f_objective / f_eyepiece,
where f_objective is the focal length of the objective lens and f_eyepiece is the focal length of the eyepiece.
Given that the focal length of the telescope (objective lens) is f = 4.1 m and the desired magnification is 600x, we can rearrange the formula to solve for f_eyepiece:
f_eyepiece = -f_objective / Magnification,
f_eyepiece = -4.1 m / 600 = -0.00683 m.
The negative sign indicates that the eyepiece should be a diverging lens.
Regarding the size of the image of Mercury, we can calculate the angular size of the image using the formula:
Angular size = Actual size / Distance,
where the actual size of Mercury is its radius (r = 2100 km) and the distance is the distance from the viewer to Mercury (90 million km).
Converting the radius to meters and the distance to meters, we have:
Angular size = (2 * 2100 km) / (90 million km) = 0.04667 degrees.
So, the image of Mercury will appear to the viewer with an angular size of approximately 0.04667 degrees.
13. To find the distance between the first and third maxima on the screen, we can use the formula for the position of the mth maximum in the double-slit interference pattern:
Position of mth maximum = (m * λ * D) / d,
where λ is the wavelength of light, D is the distance between the slits and the screen, d is the slit separation, and m is the order of the maximum.
Given that the wavelength of orange light is λ = 590 nm = 590 × 10^(-9) m, the distance between the slits and the screen is D = 1.3 m, and the slit separation is d = 1.6 mm = 1.6 × 10^(-3) m, we can calculate the distances between the maxima:
Distance between first and third maxima = [(3 * λ * D) / d] - [(1 * λ * D) / d],
Distance between first and third maxima = [(3 * 590 × 10^(-9) m * 1.3 m) / (1.6 × 10^(-3) m)] - [(590 × 10^(-9) m * 1.3 m) / (1.6 × 10^(-3) m)].
Simplifying the expression, we get:
Distance between first and third maxima ≈ 1.3 m.
Similarly, we can find the distance between the second and negative second maxima:
Distance between second and negative second maxima = [(2 * λ * D) / d] - [(-2 * λ * D) / d],
Distance between second and negative second maxima = [(2 * 590 × 10^(-9) m * 1.3 m) / (1.6 × 10^(-3) m)] - [(-2 * 590 × 10^(-9) m * 1.3 m) / (1.6 × 10^(-3) m)].
Simplifying the expression, we get:
Distance between second and negative second maxima ≈ 1.1 m.
To learn more about wavelength
https://brainly.com/question/14316836
#SPJ11
The position of an object connected to a spring varies with time according to the expression x = (4.7 cm) sin(7.9nt). (a) Find the period of this motion. S (b) Find the frequency of the motion. Hz (c) Find the amplitude of the motion. cm (d) Find the first time after t = 0 that the object reaches the position x = 2.6 cm.
The period of oscillation is `0.796 n` and the frequency of the motion`1.26 Hz`.
Given that the position of an object connected to a spring varies with time according to the expression `x = (4.7 cm) sin(7.9nt)`.
Period of this motion
The general expression for the displacement of an object performing simple harmonic motion is given by:
x = A sin(ωt + φ)Where,
A = amplitude
ω = angular velocity
t = timeφ = phase constant
Comparing the given equation with the general expression we get,
A = 4.7 cm,
ω = 7.9 n
Thus, the period of oscillation
T = 2π/ω`= 2π/7.9n = 0.796 n`...(1)
Thus, the period of oscillation is `0.796 n`.
Frequency of the motion The frequency of oscillation is given as
f = 1/T
Thus, substituting the value of T in the above equation we get,
f = 1/0.796 n`= 1.26 n^-1 = 1.26 Hz`...(2)
Thus, the frequency of the motion is `1.26 Hz`.
Amplitude of the motion
The amplitude of oscillation is given as
A = 4.7 cm
Thus, the amplitude of oscillation is `4.7 cm`.
First time after
t = 0 that the object reaches the position
x = 2.6 cm.
The displacement equation of the object is given by
x = A sin(ωt + φ)
Comparing this with the given equation we get,
4.7 = A,
7.9n = ω
Thus, the equation of displacement becomes,
x = 4.7 sin (7.9nt)
Now, we need to find the time t when the object reaches a position of `2.6 cm`.
Thus, substituting this value in the above equation we get,
`2.6 = 4.7 sin (7.9nt)`Or,
`sin(7.9nt) = 2.6/4.7`
Solving this we get,
`7.9nt = sin^-1 (2.6/4.7)``7.9n
t = 0.6841`Or,
`t = 0.0867/n`
Thus, the first time after t=0 that the object reaches the position x=2.6 cm is `0.0867/n`
To know more about displacement visit :
brainly.com/question/11934397
#SPJ11
An object located 18 cm from a convex mirror produces a virtual image 9 cm from the mirror. What is the magnification of the image? Express your answer in 2 decimal places.
Answer: The magnification of the image is 0.50. This means the image is half the size of the object.
Explanation:
The magnification (m) of an image produced by a mirror is given by the ratio of the image distance (di) to the object distance (do). The formula is:
[tex]$$m = -\frac{di}{do}$$[/tex]
In this case, the object distance (do) is 18 cm and the image distance (di) is -9 cm (the negative sign indicates that the image is virtual and located behind the mirror). Substituting these values into the formula, we can calculate the magnification.
The magnification of the image is 0.50. This means the image is half the size of the object.
can
i please get the answer to this
Question 4 (1 point) The frequency at which a material vibrates most easily. Doppler shift Destructive interference Resonance Standing waves Resonant Frequency Constructive interference
The frequency at which a material vibrates most easily is called the resonant frequency. Resonance occurs when an external force or vibration matches the natural frequency of an object, causing it to vibrate with maximum amplitude.
Resonant frequency is an important concept in physics and engineering. When a system is subjected to an external force or vibration at its resonant frequency, the amplitude of the resulting vibration becomes significantly larger compared to other frequencies. This is because the energy transfer between the external source and the system is maximized when the frequencies match.
Resonance can occur in various systems, such as musical instruments, buildings, bridges, and electronic circuits. In each case, there is a specific resonant frequency associated with the system. By manipulating the frequency of the external source, one can identify and utilize the resonant frequency to achieve desired effects.
When resonance is achieved, it often leads to the formation of standing waves. These are stationary wave patterns that appear to "stand still" due to the constructive interference between waves traveling in opposite directions. Standing waves have specific nodes (points of no vibration) and antinodes (points of maximum vibration), which depend on the resonant frequency.
Understanding the resonant frequency of a material or system is crucial in various applications, such as designing musical instruments, optimizing structural integrity, or tuning electronic circuits for efficient performance.
To learn more about Resonance click here brainly.com/question/31781948
#SPJ11
will upvote if RIGHT && answered asap!! thsnk you so much
An 6 hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.600 cm diameter eardrum so exposed? Enter a number Additional Materials
[tex]91.3\times10^{6} J[/tex] of energy falls on a 0.600 cm diameter eardrum so exposed.
To calculate the energy falling on the eardrum, we need to convert the sound intensity level from decibels (dB) to watts per square meter (W/m²) and then calculate the total energy using the formula:
Energy = Intensity × Area × Time
First, let's convert the sound intensity level from dB to W/m²:
[tex]Intensity = 10^{(dB - 12) / 10)}[/tex]
Substituting the given intensity level:
[tex]Intensity = 10^{\frac{(90 - 12)}{ 10}}=10^{7.8}[/tex]
Next, let's calculate the area of the eardrum:
[tex]Radius = \frac{0.800 cm }{2} = 0.004 m[/tex]
[tex]Area = \pi \times (radius)^2[/tex]
Now, we can calculate the energy:
Energy = Intensity × Area × Time
Substituting the values:
[tex]Energy = Intensity \times \pi \times (0.004)^2 \times (8 hours \times 3600 seconds/hour)[/tex]
[tex]Energy = 10^{7.8}\times\pi\times(0.004)^2\times8\times3600\\Energy = 91.3 \times 10^{6} J[/tex]
Thus, [tex]91.3\times10^{6}J[/tex] energy falls on a 0.600 cm diameter eardrum so exposed.
Learn more about energy here: brainly.com/question/8101588
#SPJ11
COMPLETE QUESTION
An 8-hour exposure to a sound intensity level of 90.0 dB may cause hearing damage. What energy in joules falls on a 0.800-cm-diameter eardrum so exposed?
: An airplane whose airspeed is 620 km/h is supposed to fly in a straight path 35.0 North of East. But a steady 95 km/h wind blows from the North. In what direction should the plaire N head ?
The plane should head approximately 10.7° north of east. To find the direction, we have to break down the airspeed vector into its east and north components.
Firstly, we need to break down the airspeed vector into its east and north components.
The angle between the airplane's direction and due east is (90° - 35°) = 55°.
Therefore,
The eastward component of the airplane's airspeed is: (620 km/h) cos 55° = 620 × 0.5736
≈ 355 km/h.
The northward component of the airplane's airspeed is: (620 km/h) sin 55° = 620 × 0.8192
≈ 507 km/h.
Now consider the velocity of the airplane relative to the ground. The plane's velocity relative to the ground is the vector sum of the airplane's airspeed velocity and the velocity of the wind.
Therefore, We have, tan θ = (95 km/h) / (507 km/h)θ
= tan⁻¹ (95/507)θ
≈ 10.7°.T
This is the direction that the plane must head, which is approximately 10.7° north of east.
Therefore, the plane should head approximately 10.7° north of east.
To know more about airspeed , refer
https://brainly.com/question/30529519
#SPJ11
In the following three scenarios, an object is located on one side of a converging lens. In each case, you must determine if the lens forms an image of this object. If it does, you also must determine the following.whether the image is real or virtual
whether the image is upright or inverted
the image's location, q
the image's magnification, M
The focal length is
f = 60.0 cm
for this lens.
Set both q and M to zero if no image exists.
Note: If q appears to be infinite, the image does not exist (but nevertheless set q to 0 when entering your answers to that particular scenario).
(a)
The object lies at position 60.0 cm. (Enter the value for q in cm.)
q= cmM=
Select all that apply to part (a).
realvirtualuprightinvertedno image
(b)
The object lies at position 7.06 cm. (Enter the value for q in cm.)
q= cmM=
Select all that apply to part (b).
realvirtualuprightinvertedno image
(c)
The object lies at position 300 cm. (Enter the value for q in cm.)
q= cmM=
Select all that apply to part (c).
realvirtualuprightinvertedno image
The image is real, it is inverted. Here's how you can determine whether a lens forms an image of an object, whether the image is real or virtual, upright or inverted, the image's location (q), and the image's magnification (M).
In the following scenarios, an object is placed on one side of a converging lens. Here are the solutions:
(a) The object is located at a distance of 60.0 cm from the lens. Given that f = 60.0 cm, the lens's focal length is equal to the distance between the lens and the object. As a result, the image's location (q) is equal to 60.0 cm. The magnification (M) is determined by the following formula:
M = - q / p
= f / (p - f)
In this case, p = 60.0 cm, so:
M = - 60.0 / 60.0 = -1
Thus, the image is real, inverted, and the same size as the object. So the answers for part (a) are:q = -60.0 cmM = -1real, inverted
.(b) The object is located 7.06 cm away from the lens. For a converging lens, the distance between the lens and the object must be greater than the focal length for a real image to be created. As a result, a virtual image is created in this scenario. Using the lens equation, we can calculate the image's location and magnification.
q = - f . p / (p - f)
q = - (60 . 7.06) / (7.06 - 60)
q = 4.03cm
The magnification is calculated as:
M = - q / p
= f / (p - f)
M = - 4.03 / 7.06 - 60
= 0.422
As the image is upright and magnified, it is virtual. Thus, the answers for part (b) are:
q = 4.03 cm
M = 0.422 virtual, upright.
(c) The object is located at a distance of 300 cm from the lens. Since the object is farther away than the focal length, a real image is formed. Using the lens equation, we can calculate the image's location and magnification.
q = - f . p / (p - f)
q = - (60 . 300) / (300 - 60)
q = - 50 cm
The magnification is calculated as:
M = - q / p
= f / (p - f)M
= - (-50) / 300 - 60
= 0.714
As the image is real, it is inverted. Thus, the answers for part (c) are:
q = -50 cmM = 0.714real, inverted.
To know more about lens visit:
https://brainly.com/question/29834071
#SPJ11
Consider a particle in the delta-function barrier V (x)= Bδ(x-2), where B is a positive constant.
1. How many bound states are there? Find their energies.
2. Show that the scattering states have a transmission coefficient
The delta-function barrier potential V(x) = Bδ(x-2) has one bound state with energy E = -B²/2, and scattering states exhibit a transmission coefficient.
1. To determine the number of bound states and their energies, we solve the time-independent Schrödinger equation for the given potential. In this case, the Schrödinger equation is:
[-(ħ²/2m) * d²ψ/dx² + Bδ(x-2)ψ] = Eψ,
where ħ is the reduced Planck's constant, m is the mass of the particle, ψ is the wavefunction, and E is the energy.
Since the potential is localized at x = 2, we can solve the Schrödinger equation separately on both sides of x = 2. The wavefunction should be continuous, but its derivative can have a jump at x = 2.
By solving the Schrödinger equation, it can be shown that there is one bound state with energy E = -B²/2.
2. Scattering states can be represented by plane waves on both sides of the potential barrier. We can calculate the transmission coefficient (T) to determine the probability of the particle passing through the barrier. The transmission coefficient is given by:
T = |(4k₁k₂)/(k₁ + k₂)²|,
where k₁ and k₂ are the wave numbers of the incident and transmitted waves, respectively.
For a delta-function barrier, the transmission coefficient can be derived by matching the wavefunctions and their derivatives at x = 2. By calculating the transmission coefficient, we can determine the probability of the particle transmitting through the barrier.
It is important to note that the detailed calculations and solutions depend on the specific form of the wavefunction and the potential. These equations provide a general framework for understanding the behavior of the particle in the given potential.
Learn more about Schrödinger here:
https://brainly.com/question/27850342
#SPJ11
At what angle is the first-order maximum for 440-nm wavelength blue light falling on double slits separated by 0.05 mm? Hint First-order maximum is at degrees from the central maximum.
The first-order maximum for the blue light with a wavelength of 440 nm occurs at an angle of approximately 0.505 degrees from the central maximum.
To find the angle at which the first-order maximum occurs, we can use the formula for the location of the maxima in a double-slit interference pattern:
dsinθ = mλ
where d is the slit separation, θ is the angle from the central maximum, m is the order of the maximum, and λ is the wavelength of light.
In this case, we are given a blue light with a wavelength of 440 nm (or 440 × 10^-9 m) and a slit separation of 0.05 mm (or 0.05 × 10^-3 m). We want to find the angle at which the first-order maximum occurs (m = 1).
Substituting the given values into the formula:
0.05 × 10^-3 × sinθ = (1) × (440 × 10^-9)
Simplifying the equation, we get:
sinθ = (440 × 10^-9) / (0.05 × 10^-3)
sinθ = 0.0088
To find the angle θ, we take the inverse sine (or arcsine) of 0.0088:
θ = arcsin(0.0088)
Using a calculator, we find:
θ ≈ 0.505 degrees
Therefore, the first-order maximum for the blue light with a wavelength of 440 nm occurs at an angle of approximately 0.505 degrees from the central maximum.
Learn more about wavelength at: https://brainly.com/question/10750459
#SPJ11
12. (II) (a) Show that the nucleus Be (mass = 8.005308 u) is unstable to decay into two a particles. (b) Is 'C stable against decay into three a particles? Show why or why not. tum what off b SECTIONS
(a) To determine the stability of the Be nucleus against decay into two alpha particles, we must compute the mass of the products (2 alpha particles) and compare it to the mass of the Be nucleus. Two alpha particles are equivalent to a helium nucleus. The mass of the helium nucleus is 4.001506 u. Therefore, the mass of two alpha particles is 8.003012 u.
The difference between the mass of the Be nucleus and the mass of two alpha particles is:Δm = M(Be) - M(2α) = 8.005308 u - 8.003012 u= 0.002296 u The decay into two alpha particles can proceed if the Q-value of the reaction is positive. The Q-value of the reaction is: Q = Δm c² = 0.002296 u x (1.6606 x 10-27 kg/u) x (2.998 x 108 m/s)²Q = 4.13 x 10-12 J This is a small amount of energy.
Therefore, the Be nucleus is unstable against decay into two alpha particles.(b) The carbon-12 nucleus is stable against decay into three alpha particles. To show why, we must compute the Q-value of the reaction. Three alpha particles are equivalent to a helium nucleus. The mass of the helium nucleus is 4.001506 u.
Therefore, the mass of three alpha particles is 12.004518 u. The difference between the mass of the C nucleus and the mass of three alpha particles is: Δm = M(C) - M(3α) = 12.000 u - 12.004518 u= -0.004518 u The decay into three alpha particles can proceed if the Q-value of the reaction is positive. The Q-value of the reaction is:
To know more about nucleus visit:
https://brainly.com/question/23366064
#SPJ11
1. In 2019, Sammy Miller drove a rocket powered dragster from rest to 402m (1/4 mile) in a
record 3.22s. What acceleration did he experience?
Show all steps
Sammy Miller experienced an acceleration of approximately 124.6 m/s².
To find the acceleration experienced by Sammy Miller, we can use the formula:
acceleration = (final velocity - initial velocity) / time
Given:
- The distance covered, d = 402 m
- The time taken, t = 3.22 s
First, let's calculate the final velocity. We know that the distance covered is equal to the average velocity multiplied by time:
d = (initial velocity + final velocity) / 2 * t
Substituting the values:
402 = (0 + final velocity) / 2 * 3.22
Simplifying the equation:
402 = (0.5 * final velocity) * 3.22
402 = 1.61 * final velocity
Dividing both sides by 1.61:
final velocity = 402 / 1.61
final velocity = 249.07 m/s
Now we can calculate the acceleration using the formula mentioned earlier:
acceleration = (final velocity - initial velocity) / time
Since Sammy Miller started from rest (initial velocity, u = 0), the equation simplifies to:
acceleration = final velocity / time
Substituting the values:
acceleration = 249.07 / 3.22
acceleration ≈ 77.29 m/s²
Therefore, Sammy Miller experienced an acceleration of approximately 124.6 m/s².
For more such questions on acceleration, click on:
https://brainly.com/question/24965358
#SPJ8
1. What is the gravitational energy (relative to the unstretched surface of the trampoline) of the 20 kg ball at its apex 2 m above the trampoline?
E= mgh = 20(10)(2) =400 J Therefore, the gravitational energy is 400 J.
2. What is the kinetic energy of the ball just before impacting the trampoline?
The kinetic energy is 400 J because energy can not be created or destroyed.
3. At maximum stretch at the bottom of the motion, what is the sum of the elastic and gravitational energy of the ball?
I need help with question 3
use g= 10 N/kg
At maximum stretch at the bottom of the motion, the sum of the elastic and gravitational energy of the ball is 800 J.
To calculate the elastic energy, we need to consider the potential energy stored in the trampoline when it is stretched. When the ball reaches the bottom of its motion, it comes to a momentary rest before bouncing back up. At this point, the potential energy due to the stretched trampoline is at its maximum, and it is equal to the elastic potential energy stored in the trampoline.
The elastic potential energy (PEe) can be calculated using Hooke's Law, which states that the force exerted by a spring is proportional to its displacement. The formula for elastic potential energy is given as:
PEe = (1/2)k[tex]x^2[/tex]
Where k is the spring constant and x is the displacement from the equilibrium position. In this case, the trampoline acts like a spring, and the displacement (x) is equal to the maximum stretch of the trampoline caused by the ball's impact.
Since the values of the spring constant and maximum stretch are not given, we cannot calculate the exact elastic potential energy. However, we can still determine the sum of the elastic and gravitational energy by adding the previously calculated gravitational energy of 400 J to the kinetic energy just before impacting the trampoline, which is also 400 J.
Therefore, at maximum stretch at the bottom of the motion, the sum of the elastic and gravitational energy of the ball is 800 J (400 J from gravitational energy + 400 J from kinetic energy).
To know more about gravitational energy here https://brainly.com/question/15896499
#SPJ4
To stretch a certain spring by 2.80 cm from its equilibrium position requires 9.50 J of work.
What is the force constant of this spring?
What was the maximum force required to stretch it by that distance?
To determine the force constant of the spring, we can use Hooke's Law. The force constant of this spring is approximately 4,061.22 and the maximum force is approximately 113.89 N.
Mathematically, it can be expressed as F = -kx, where F is the force applied to the spring, k is the force constant, and x is the displacement from the equilibrium position.
k = 2 * 9.50 J / (0.028 m)^2
k = 2 * 9.50 J / (0.028^2 m^2)
k ≈ 4,061.22 N/m
Therefore, the force constant of this spring is approximately 4,061.22 N/m.
To find the maximum force required to stretch the spring by 2.80 cm, we can use Hooke's Law, F = -kx.
F = -4,061.22 N/m * 0.028 m
F ≈ -113.89 N
The negative sign indicates that the force is in the opposite direction of the displacement. Thus, the maximum force required to stretch the spring by 2.80 cm is approximately 113.89 N.
Learn more about force here : brainly.com/question/30507236
#SPJ11
Pollen particles are mixed in water and allowed to settle. If the water depth is 2 cm, what is the diameter of the largest particles that can remain in suspension after 1 h? The density of pollen is 1.8 g/cm3.
The diameter of the largest particles that can remain in suspension after 1 hour is approximately 34.18 micrometers.
To determine the diameter of the largest particles that can remain in suspension after 1 hour, we need to consider the settling velocity and the conditions required for suspension.
The settling velocity of a particle in a fluid can be determined using Stokes' Law, which states:
v = (2 * g * (ρp - ρf) * r²) / (9 * η)
where v is the settling velocity, g is the acceleration due to gravity (approximately 9.8 m/s²), ρp is the density of the particle (1.8 g/cm³),
ρf is the density of the fluid (assumed to be the density of water, which is approximately 1 g/cm³), r is the radius of the particle, and η is the dynamic viscosity of the fluid (approximately 1.002 × 10⁻³ Pa·s for water at 20°C).
For the particle to remain in suspension, the settling velocity must be equal to or less than the upward velocity of the fluid caused by turbulence.
Given that the water depth is 2 cm, we can calculate the upward velocity of the fluid using the equation:
u = d / t
where u is the upward velocity, d is the water depth (2 cm = 0.02 m), and t is the time (1 hour = 3600 seconds).
Now we can set the settling velocity equal to the upward velocity and solve for the radius of the largest particle that can remain in suspension:
v = u
(2 * g * (ρp - ρf) * r²) / (9 * η) = d / t
Substituting the values and solving for r:
r = √((d * η) / (18 * g * (ρp - ρf)))
r = √((0.02 * 1.002 × 10⁻³) / (18 * 9.8 * (1.8 - 1)))
Now we can calculate the diameter of the largest particle using the equation:
diameter = 2 * r
Substituting the value of r and calculating:
diameter = 2 * √((0.02 * 1.002 × 10⁻³) / (18 * 9.8 * (1.8 - 1)))
After performing the calculations, the diameter of the largest particles that can remain in suspension after 1 hour is approximately 34.18 micrometers.
Learn more about density at: https://brainly.com/question/1354972
#SPJ11
The main water line enters a house on the first floor. The line has a gauge pressure of 285 x 10% Pa(a) A faucet on the second floor, 4.10 m above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it even if the faucet were open? (a) Number 1 Units (b) Number Units A water tower is a familiar sight in many towns. The purpose of such a tower is to provide storage capacity and to provide sufficient pressure in the pipes that deliver the water to customers. The drawing shows a spherical reservoir that contains 3.09 x 105 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet in (a) house A and (b) house B. Ignore the diameter of the delivery pipes. Vent 150 m Facet 12.30 m Faucet (a) Number i Units (b) Number Units
The gauge pressure at the faucet is [tex]325\times10^{3} Pa[/tex] and the maximum height is 29.169 m.
(a) To find the gauge pressure at the faucet on the second floor, we can use the equation for pressure due to the height difference:
Pressure = gauge pressure + (density of water) x (acceleration due to gravity) x (height difference).
Given the gauge pressure at the main water line and the height difference between the first and second floors, we can calculate the gauge pressure at the faucet on the second floor. So,
Pressure =[tex]2.85\times 10^{5}+(997)\times(9.8)\times(4.10) =325\times10^{3} Pa.[/tex]
Thus, the gauge pressure at the faucet on the second floor is [tex]325\times10^{3} Pa.[/tex]
(b) The maximum height at which water can be delivered from a faucet depends on the pressure needed to push the water up against the force of gravity. This pressure is related to the maximum height by the equation:
Pressure = (density of water) * (acceleration due to gravity) * (height).
By rearranging the equation, we can solve for the maximum height.
Maximum height = [tex]\frac{pressure}{density of water \times acceleration of gravity}\\=\frac{2.85 \times10^{5}}{997\times 9.8} \\=29.169 m[/tex]
Therefore, the gauge pressure at the faucet is [tex]325\times10^{3} Pa[/tex] and the maximum height is 29.169 m.
Learn more about pressure here: brainly.com/question/28012687
#SPJ11
CORRECT QUESTION
The main water line enters a house on the first floor. The line has a gauge pressure of [tex]2.85\times10^{5}[/tex] Pa. (a) A faucet on the second floor, 4.10 m above the first floor, is turned off. What is the gauge pressure at this faucet? (b) How high could a faucet be before no water would flow from it even if the faucet were open?
Buttercup is sliding on frictionless ice with a speed of 2.5 m/s when she runs into a large massless spring with a spring constant of 272 N/m. Buttercup has a mass of 31.5 kg. a) What is the amplitude
(a)The amplitude of the spring oscillations is 0.29 m.
In a scenario where Buttercup is sliding on a frictionless ice with a speed of 2.5 m/s and runs into a large massless spring with a spring constant of 272 N/m, her mass of 31.5 kg makes it possible to calculate the amplitude of the spring oscillations using the given formula.
Amplitude is defined as the magnitude of the maximum displacement of the oscillating object from its equilibrium position. It represents the maximum value of an oscillation or wave from its equilibrium or average value.
Spring constant (k) is defined as the ratio of the applied force to the deformation caused by that force. It is the amount of force required per unit deformation or lengthening of a spring.
The formula for the amplitude of the spring oscillations, A= (m × v) / k where A is the amplitude, m is the mass of the object (Buttercup) that collided with the spring, v is the velocity of the object before the collision, and k is the spring constant of the massless spring. Substituting the given values into the formula: A = (m × v) / k = (31.5 kg × 2.5 m/s) / 272 N/mA = 0.29 m.
Therefore, the amplitude of the spring oscillations is 0.29 m.
let's learn more about Spring constant:
https://brainly.com/question/22712638
#SPJ11
A copper block is removed from a 370 °C oven and dropped into 1.10 kg of water at 28.0 °C. The water quickly reaches 37.0 °C and then remains at that temperature. What is the mass of the copper block?
The mass of the copper block is approximately 400.2 grams.
We can solve this problem by applying the principle of energy conservation. According to this principle, the heat lost by the copper block is equal to the heat gained by the water.
To calculate the heat gained by the water, we can use the formula: Q = mcΔT, where Q represents the heat gained by the water, m is the mass of the water, c is the specific heat capacity of water, and ΔT is the change in temperature of the water.
Mass of water (m) = 1.10 kg
Specific heat capacity of water (c) = 4.18 J/g°C
Initial temperature of water (T1) = 28.0 °C
Final temperature of water (T2) = 37.0 °C
Calculating the heat gained by the water:
Q = (1.10 kg) * (4.18 J/g°C) * (37.0 °C - 28.0 °C)
Q = 51.47 kJ
Since the heat lost by the copper block is equal to the heat gained by the water, the heat lost by the copper block is also 51.47 kJ.
To find the mass of the copper block, we can use the equation:
Q = mcΔT
Specific heat capacity of copper (c') = 0.385 J/g°C
Initial temperature of copper (T1') = 370 °C
Final temperature of copper (T2') = 37.0 °C
Calculating the mass of the copper block:
51.47 kJ = m * (0.385 J/g°C) * (37.0 °C - 370 °C)
51.47 kJ = m * (0.385 J/g°C) * (-333 °C)
m = 51.47 kJ / [(0.385 J/g°C) * (-333 °C)]
m ≈ 400.2 g
Therefore, the mass of the copper block is approximately 400.2 grams.
Learn more about mass at: https://brainly.com/question/86444
#SPJ11
During a particular thunderstorm, the electric potential difference between a cloud and the ground is Vcloud-Vground= 1.8 x10^8 Volts. What is the change in potential energy of an electron as it moves from the cloud to the ground?
The change in potential energy of an electron as it moves from the cloud to the ground is -2.88 x 10^-11 Joules. The negative sign indicates a decrease in potential energy, as the electron moves from a higher potential (cloud) to a lower potential (ground). The change in potential energy of an electron as it moves from the cloud to the ground can be calculated using the formula:
ΔPE = q * ΔV,
where ΔPE is the change in potential energy, q is the charge of the electron, and ΔV is the potential difference between the cloud and the ground.
The charge of an electron is -1.6 x 10^-19 Coulombs (C).
Substituting the values into the formula, we have:
ΔPE = (-1.6 x 10^-19 C) * (1.8 x 10^8 V).
Calculating the value, we get:
ΔPE = -2.88 x 10^-11 Joules.
Therefore, the change in potential energy of an electron as it moves from the cloud to the ground is -2.88 x 10^-11 Joules. The negative sign indicates a decrease in potential energy, as the electron moves from a higher potential (cloud) to a lower potential (ground).
Learn more about potential energy here:
https://brainly.com/question/24284560
#SPJ11
quick answer
please
A 1.00-mm-radius, cylindrical copper wire carries a current of 8.00 A. If each copper atom in the wire contributes one free conduction electron to the current, what is the drift velocity of the electr
The drift velocity of the electrons in the wire is approximately 0.0000235 cm/s
The drift velocity of the electrons in the wire can be calculated using the formula
I = n×A×q×v
where:
I = current
n = number of free electrons per unit volume
A = cross-sectional area of the wire
q = charge of an electron
v = drift velocity
Given :
Current = 8.00 A
Density of copper = 8.96 g/cm³
1 cm³ = 1 mL
Molar mass of copper = 63.546 g/mole
Number of moles of copper in 1 mL = Density of copper / molar mass of copper
= (8.96 g/mL) / (63.546 g/mole)
= 0.141 moles/mL.
Avogadro’s number = (6.02 x 10²³)
Number of free atoms per unit volume = Number of moles of copper in 1 mL × Avogadro’s number
= (0.141 moles/mL) × (6.02 x 10²³ atoms/mole)
= 8.48 x 10²² atoms/mL
Each copper atom contributes one free electron,
n = 8.48 x 10²² electrons/cm³
The cross-sectional area of the wire
A = πr²
where
r = radius of the wire
substuting the r value in the equation we get:
A = π(0.1 cm)²
= 0.0314 cm²
The charge of an electron = q = 1.6 x 10⁻¹⁹ C/electron.
Substuting the values in the formula for current, we get:
I = n × A × q × v
8A = (8.48 x 10²² electrons/cm³) × (0.0314 cm²) × (1.6 x 10⁻¹⁹ C/electron) × v
v = (8 A) / ((8.48 x 10²² electrons/cm³)(0.0314 cm²)(1.6 x 10⁻¹⁹ C/electron))
= 0.0000235 cm/s
Therefore, the drift velocity of the electrons in the wire is 0.0000235 cm/s
To learn more about drift velocity:
https://brainly.com/question/28498217
#SPJ4