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Oxana [17]
3 years ago
10

A 50-loop circular coil has a radius of 3 cm. It is oriented so that the field lines of a magnetic field are perpendicular to th

e coil. Suppose that the magnetic field is varied so that B increases from 0.10 T to 0.35 T in 2 ms. Find the induced emf in the coil.
Physics
1 answer:
nikitadnepr [17]3 years ago
6 0

Answer:

-17.8 V

Explanation:

The induced emf in a coil is given as:

E = \frac{-NdB\pi r^2}{dt}

where N = number of loops

dB = change in magnetic field

r = radius of coil

dt = elapsed time

From the question:

N = 50

dB = final magnetic field - initial magnetic field

dB = 0.35 - 0.10 = 0.25 T

r = 3 cm

dt = 2 ms = 0.002 secs

Therefore, the induced emf is:

E = \frac{-50 * 0.25 * \pi * 0.03^2}{0.002} \\E = -17.8 V

Note: The negative sign implies that the EMf acts in an opposite direction to the change in magnetic flux.

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Answer:

c. from south to north

Explanation:

since we know that:

F = qv×B

where F is the force and in this case is pointed upward

where v is the velocity due east

the field must be due north by the right hand rule

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3 years ago
A baseball is thrown at an angle of 25 degrees relative to the ground at a speed of 23 m/s. If the ball was caught 42 m from the
ExtremeBDS [4]

Answer:

a) v_{0y} = 9.72 m/s

b) v_{0x} = 20.85 m/s

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Explanation:

a) The initial vertical velocity is given by:

v_{0y} = v*sin(\theta)

Where:

θ: 25°

v: is the magnitude of the speed = 23 m/s

v_{0y} = 23 m/s*sin(25) = 9.72 m/s

b) The initial horizontal velocity can be calculated as follows:

v_{0x} = v*cos(\theta) = 23 m/s*cos(25) = 20.85 m/s

c) The flight time can be calculated using the following equation:

v_{0x} = \frac{x}{t}

Where:

x: is the total distance = 42 m

t = \frac{x}{v_{0x}} = \frac{42 m}{20.85 m/s} = 2.01 s

d) The maximum height is given by:

v_{fy}^{2} = v_{0y}^{2} - 2gh_{max}

Where:

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g: is the gravity = 9.81 m/s²

h_{max} = \frac{v_{0y}^{2}}{2g} = \frac{(9.72 m/s)^{2}}{2*9.81 m/s^{2}} = 4.82 m

I hope it helps you!                                  

7 0
3 years ago
The position of a particle moving along an x axis is given by x = 14.0t2 - 2.00t3, where x is in meters and t is in seconds. det
Mumz [18]

a) x = 14t^{2} -2t^{3}

at t = 5s

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a = 28 -12*5= 28-60= -32 m/s^2

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So 0= 28t - 6 t^2

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8 0
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