Ask question

Ask question

All categories

3 years ago

15

The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional

area per turn of 0.018 m2 and contains 200 turns. Find (a) the frequency f of the generator in hertz, (b) the angular frequency in rad/s, and (c) the magnitude of the magnetic field.The drawing shows a plot of the output emf of a generator as a function of time t. The coil of this device has a cross-sectional area per turn of 0.018 m2 and contains 200 turns. Find (a) the frequency f of the generator in hertz, (b) the angular frequency in rad/s, and (c) the magnitude of the magnetic field.

1 answer:

adell [148]3 years ago

6 0

This question is incomplete, the missing image uploaded along this answer below;

Answer:

a) frequency is 2.381 Hz

b) the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field is 0.5199 T

Explanation:

Given that;

cross sectional Area A = 0.018 m²

Number of turns N = 200

from the diagram. maximum time T = 0.42 sec

a) the frequency f of the generator in hertz

frequency = 1 / T

we substitute

frequency = 1 / 0.42 = 2.381 Hz

Therefore, frequency is 2.381 Hz

b) the angular frequency in rad/s

angular speed ω = 2πf

we substitute

ω = 2π × 2.381

ω = 14.96 rad/s

Therefore, the angular frequency/speed is 14.96 rad/s

c) the magnitude of the magnetic field.

to determine the magnitude of the magnetic field, we use the following expression;

e = NBAω

from the diagram, e = 28.0 V

so we substitute

28.0 V = 200 × B × 0.018 × 14.96

28 = 53.856B

B = 28 / 53.856

B = 0.5199 T

Therefore, the magnitude of the magnetic field is 0.5199 T

You might be interested in

A unit mass of an ideal gas at temperature T undergoes a reversible isothermal process from pressure P1 to pressure P2 while los

mafiozo [28]

Answer:

Option B

Change in entropy of the process is $\Delta S= Rln(\frac{P_{1}}{P_{2}})$

Explanation:

The entropy of a system is a measure of the degree of disorderliness of the system.

The entropy of a system moving from process 1 to 2 is given as

$\Delta S = \int\limits^2_1 {\frac{\delta q}{T}}$

recall from first law, $\delta q =du +Pdv$

hence we have, $\Delta S = \int\limits^2_1 {\frac{du +Pdv}{T}}$

since the process is isothermal, $du= 0$

this gives us $\Delta S = \int\limits^2_1 {\frac{Pdv}{T}}$

integrating within the limits of 1 and 2, will give us

$\Delta S = R ln (\frac{V_{2}}{V_{1}})$

also from ideal gas laws,

$\frac{V_{2}}{V_{1}}=\frac{P_{1}}{P_{2}}$

hence we have $\Delta S = R ln (\frac{P_{1}}{P_{2}})$

This makes the correct option B

4 0

3 years ago

Read 2 more answers

While driving fast around a sharp right turn, you find yourself pressing against the car door. What is happening?

sergiy2304 [10]

Answer:

option C

Explanation:

The correct answer is option C

When the driver takes the sharp right turn the door will exert rightward pressure on the driver.

When the driver takes the sudden right turn the tendency of the body is to be in the straight line by the vehicle moves in the circular path so, as the vehicle turns it applies a rightward force on you.

The pushing of the door to you because of the centripetal force acting on the car due to sudden sharp turn.

3 0

3 years ago

If the beam carries 1015 electrons per second and is accelerated by a 350 kV source, find the current and power in the beam.

Y_Kistochka [10]

To solve this problem we will apply the concept of current defined as the electron charge flow by the number of electrons per second. That is,

I = q*N

Here q is Flow of electric charge in one second and N the number of electron flow per second.

A the same time the power is described as the applied voltage for the current.

P = VI

We know the charge of electron, $q = 1.602 * 10^{-19}$ Coulombs, then the current is

$I = (1.602*10^{-19})(10^{15})$

$I = 0.1602 mA$

And the power in the Beam is

$P = VI$

$P = (350*10^3)(0.1602)$

$P = 0.05607 Watts$

3 0

3 years ago

A football quarterback runs 15.0 m straight down the playing field in 3.00 s. He is then hit and pushed 3.00 m straight backward

algol [13]

Answer:

a) $v_{1}=14.29m/s\\v_{2}=9.25m/s\\v_{3}=6.36m/s$

b) $v=+9.97m/s$

Explanation:

From the exercise we know that

$x_{1} =15m, t_{1}=3s$

$x_{2} =-3m, t_{1}=1.74s$

$x_{3} =29m, t_{3}=5.20s$

From dynamics we know that the formula for average velocity is:

$v=\frac{x_{2}-x_{1} }{t_{2}-x_{1} }$

a) For the three intervals:

$v_{1}=\frac{x_{2}-x_{1} }{t_{2}-t_{1} }=\frac{(-3-15)m}{(1.74-3)s}=14.29m/s$

$v_{2}=\frac{x_{3}-x_{2} }{t_{3}-t_{2} }=\frac{(29-(-3))m}{(5.20-1.74)s}=9.25m/s$

$v_{3}=\frac{x_{3}-x_{1} }{t_{3}-t_{1} }=\frac{(29-15)m}{(5.20-3)s}=6.36m/s$

b) The average velocity for the entire motion can be calculate by the following formula:

$v=\frac{v_{1}+v_{2}+v_{3} }{n} =\frac{(14.29+9.25+6.36)m/s}{3}=+9.97m/s$

8 0

3 years ago

Comic-book superheroes are sometimes able to punch holes through steel walls. (a) If the ultimate shear strength of steel is tak

Katena32 [7]

Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.

shear stress is the measure of the material's resistance to cause "sliding" motion of its layers.
<span>Since the hero has to take a cylindrical piece of metal out, the area which will cause this force is 2*pi*r*l (put r and l in SI units to get answer in Newtons) </span>
<span>so the force required is shear strength into this area (2.60109 * 2 * pi * r * l) </span>

7 0

3 years ago