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vazorg [7]
2 years ago
10

A horizontal circular wire loop of radius 0.7 m lies in a plane perpendicular to a uniform magnetic field that is pointing down

from above into the plane of the loop, and has a constant magnitude of 0.44 T. If in 0.14 s the wire is reshaped from a circle into a square, but remains in the same plane, what is the magnitude of the average induced?
Physics
1 answer:
Maru [420]2 years ago
3 0

Answer:

\epsilon=1.10\ V

Explanation:

It is given that,

Radius of the circular loop, r = 0.7 m

Magnetic field, B = 0.44 T

In 0.14 s the wire is reshaped from a circle into a square, but remains in the same plane.

Area of the circular wire,

A_1=\pi r^2

A_1=\pi (0.7)^2=1.539\ m^2

For the area of square,

The circumference of wire, C=2\pi r=2\pi \times 0.7=4.39\ m

Side of square, l=\dfrac{4.39}{4}=1.09\ m

Area of square, A_2=1.09^2=1.188\ m^2

An emf is induced in the loop due to change in its area. The induced emf is given by :

\epsilon=-B\dfrac{dA}{dt}

\epsilon=-B\dfrac{A_2-A_1}{t}

\epsilon=-0.44\times \dfrac{1.188-1.539}{0.14}  

\epsilon=1.10\ V

So, the magnitude of the average induced emf is 1.10 volts. Hence, this is the required solution.

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at a certain moment, an object has an amount of 200 of motion energy and 400 of gravitational potential energyThe object is also
Radda [10]

Answer:

twice as much energy

Explanation:

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1 year ago
A space vehicle accelerates uniformly from 85 m/s at t = 0 to 164 m/s at t = 10.0 s .How far did it move between t = 2.0 s and t
creativ13 [48]

First, we have a change in the velocity from 85 to 164 m/s in 10 sec.

Then, we calculate the <u>acceleration </u>as:

a=\frac{v_{f}-v_{i} }{t} =\frac{164-85}{10}=7.9 m/s^2

Hence we need to calculate the velocity of the space vehicle at t = 2 sec using the first equation of motion:

v_{f}=v_{i}+at=85+7.9*2=100.8m/s

Then, using the second equation of motion to calculate the distance:

d=v_{i}  t+\frac{1}{2}at^2

d=100.8*2+\frac{1}{2}*7.9*(2)^2=217.4m

5 0
3 years ago
Read 2 more answers
platform diving in the olympic games takes place at two heights: 5 meters and 10 meters. What is the velocity of a diver enterin
posledela

1) Velocity: 9.9 m/s and 14 m/s

The motion of the diver is a free-fall motion, so it is a uniform accelerated motion. Choosing downward as positive direction, the final velocity can be found by using the following suvat equation:

v^2-u^2=2as

where

v is the final velocity

u = 0 is the initial velocity (the diver starts from rest)

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

s is the displacement

For the diver jumping from 5 m, s = 5 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(5)}=9.9 m/s

For the diver jumping from 10 m, s = 10 m, so

v=\sqrt{2as}=\sqrt{2(9.8)(10)}=14 m/s

2) Time: 1.01 s and 1.43 s

The time of flight of each diver can be found by using the other suvat equation

s=ut+\frac{1}{2}at^2

And since u = 0, it can be reduced to

s=\frac{1}{2}at^2

For the diver jumping from 5 m, s = 5 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(5)}{9.8}}=1.01 s

For the diver jumping from 10 m, s = 10 m, so we find

t=\sqrt{\frac{2s}{a}}=\sqrt{\frac{2(10)}{9.8}}=1.43 s

5 0
3 years ago
15) A 1720 kg car accelerates at a rate of 3.0 m/s2. How much force is the car's engine producing? (use the formula F=ma
Anastasy [175]

Answer:

<h3>The answer is 5160 N</h3>

Explanation:

To find the force acting on an object given it's mass and acceleration we use the formula

<h3>Force = mass × acceleration</h3>

From the question

mass = 1720 kg

acceleration = 3.0 m/s²

We have

Force = 1720 × 3

We have the final answer as

<h3>5160 N</h3>

Hope this helps you

4 0
3 years ago
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A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer

given,

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swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

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Work done by the gravity =- 3 \times 9.8 \times 4

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work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

7 0
3 years ago
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