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notsponge [240]

3 years ago

5

What is the peak emf generated (in V) by rotating a 1000 turn, 42.0 cm diameter coil in the Earth's 5.00 ✕ 10−5 T magnetic field

, given the plane of the coil is originally perpendicular to the Earth's field and is rotated to be parallel to the field in 12.0 ms?

1 answer:

ELEN [110]3 years ago

3 0

Answer:

5.77 Volt

Explanation:

N = 1000, Diameter = 42 cm = 0.42 m, t = 12 ms = 12 x 10^-3 s

Change in magnetic field, B = 5 x 10^-5 T

The peak value of emf is given by

e = N x dФ / dt

e = N x A x dB/dt

e = (1000 x 3.14 x 0.21 x 0.21 x 5 x 10^-5) / (1.2 x 10^-3)

e = 5.77 Volt

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The first step that Enrique must take in order to calculate the tangential speed of the satellite is to convert the period from days to seconds.

We know that the SI unit of speed is meter per second and now, we with to obtain the tangential speed of the satellite.

Since the period is given in days, the first step is to convert the period from days to seconds.

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2 years ago

A 1 kg mass is attached to a spring with spring constant 7 Nt/m. What is the frequency of the simple harmonic motion? What is th

Scorpion4ik [409]

1. 0.42 Hz

The frequency of a simple harmonic motion for a spring is given by:

$f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}$

where

k = 7 N/m is the spring constant

m = 1 kg is the mass attached to the spring

Substituting these numbers into the formula, we find

$f=\frac{1}{2\pi}\sqrt{\frac{7 N/m}{1 kg}}=0.42 Hz$

2. 2.38 s

The period of the harmonic motion is equal to the reciprocal of the frequency:

$T=\frac{1}{f}$

where f = 0.42 Hz is the frequency. Substituting into the formula, we find

$T=\frac{1}{0.42 Hz}=2.38 s$

3. 0.4 m

The amplitude in a simple harmonic motion corresponds to the maximum displacement of the mass-spring system. In this case, the mass is initially displaced by 0.4 m: this means that during its oscillation later, the displacement cannot be larger than this value (otherwise energy conservation would be violated). Therefore, this represents the maximum displacement of the mass-spring system, so it corresponds to the amplitude.

4. 0.19 m

We can solve this part of the problem by using the law of conservation of energy. In fact:

- When the mass is released from equilibrium position, the compression/stretching of the spring is zero: $x=0$ , so the elastic potential energy is zero, and all the mechanical energy of the system is just equal to the kinetic energy of the mass:

$E=K=\frac{1}{2}mv^2$

where m = 1 kg and v = 0.5 m/s is the initial velocity of the mass

- When the spring reaches the maximum compression/stretching (x=A=amplitude), the velocity of the system is zero, so the kinetic energy is zero, and all the mechanical energy is just elastic potential energy:

$E=U=\frac{1}{2}kA^2$

Since the total energy must be conserved, we have:

$\frac{1}{2}mv^2 = \frac{1}{2}kA^2\\A=\sqrt{\frac{m}{k}}v=\sqrt{\frac{1 kg}{7 N/m}}(0.5 m/s)=0.19 m$

5. Amplitude of the motion: 0.44 m

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}kA^2$ is the mechanical energy of the system when x=A (maximum displacement)

Equalizing the two expressions, we can solve to find A, the amplitude:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}kA^2\\A=\sqrt{x_0^2+\frac{m}{k}v_0^2}=\sqrt{(0.4 m)^2+\frac{1 kg}{7 N/m}(0.5 m/s)^2}=0.44 m$

6. Maximum velocity: 1.17 m/s

We can use again the law of conservation of energy.

- $E_i = \frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2$ is the initial mechanical energy of the system, with $x_0=0.4 m$ being the initial displacement of the mass and $v_0=0.5 m/s$ being the initial velocity

$- E_f = \frac{1}{2}mv_{max}^2$ is the mechanical energy of the system when x=0, which is when the system has maximum velocity, $v_{max}$

Equalizing the two expressions, we can solve to find $v_{max}$ , the maximum velocity:

$\frac{1}{2}kx_0^2 + \frac{1}{2}mv_0^2=\frac{1}{2}mv_{max}^2\\v_{max}=\sqrt{\frac{k}{m}x_0^2+v_0^2}=\sqrt{\frac{7 N/m}{1 kg}(0.4 m)^2+(0.5 m/s)^2}=1.17 m/s m$

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3 years ago

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26. A 40 kg boy jumps from a height of 4m onto a plate-form mounted on springs. As the

denpristay [2]

Answer:

c. 1600J

Explanation:

The loss in potential energy of the boy is given by:

$U=mg \Delta h$

where

m = 40 kg is the mass of the boy

g = 9.8 m/s^2 is the acceleration of gravity

$\Delta h = 4 m + 0.02 m = 4.02 m$ is the total change in the height of the boy (4 metres + 2 cm due to the compression of the spring)

Substituting, we find

$\Delta U = (40 kg)(9.8 m/s^2)(4.02 m) = 1577 J \sim 1600 J$

4 0

2 years ago

What is the net force acting upon this object? *

AnnZ [28]

2 N Right

.. .. .. .. .. .. .. ..

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3 years ago

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Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .

vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

$Ke=\frac{1}{2}Mass*Velocity^2$

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

$Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules$

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

7 0

3 years ago

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