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Minchanka [31]
3 years ago
12

A circular loop of flexible iron wire has an initial circumference of 161 cm , but its circumference is decreasing at a constant

rate of 11.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.
Physics
1 answer:
miss Akunina [59]3 years ago
7 0

Complete question is;

A circular loop of flexible iron wire has an initial circumference of 161 cm , but its circumference is decreasing at a constant rate of 11.0 cm/s due to a tangential pull on the wire. The loop is in a constant uniform magnetic field of magnitude 0.800 T , which is oriented perpendicular to the plane of the loop. Assume that you are facing the loop and that the magnetic field points into the loop.

Find the emf induced in the loop, at the instant when 9.0s has passed.

Answer:

Magnitude of EMF induced = 0.008686 V

Explanation:

We are given;

Initial circumference; C = 161 cm

rate of decrease of circumference; dC/dt = 11 cm/s

magnetic field; B = 0.8 T

time; t = 9 sec

To find emf in loop, we'll use the formula;

EMF = -d(BAcosθ)/dt

Now, we are told magnetic field is uniform, thus angle is zero.

Thus;

EMF = -d(BAcos0)/dt

cos 0 = 1

Thus;

EMF = -B(dA/dt)

so area will be

Area of the loop is (πr²)

Thus;

dA/dt = d(πr²)/dt

Differentiating with respect to r, we have;

dA/dt = 2πr(dr/dt)

Formula for circumference is;

C = 2πr

Making r the subject, we have;

r = C/2π

Thus;

R = 161/2π

r = 25.6239 cm

c is circumference so from equation 2

From EMF = -B(dA/dt) , let's replace dA/dt with 2πr(dr/dt) since dA/dt = 2πr(dr/dt)

Thus;

EMF = -2πrB(dr/dt)

From C = 2πr, we can derive the rate of increase in radius.

Thus;

dC/dt = 2π(dr/dt)

Plugging in the relevant values, we have;

dr/dt = 11/2π = 1.75 cm/s

At the instance when 9 s has passed, radius of the could is now;

r1 = r - 9(dr/dt)

r1 = 25.6239 - 9(1.75)

r1 = 9.8739 cm

We will need to plug in 0.8 for B, 9.8739 cm for r, 1.75 cm/s for dr/dt into;

EMF = -2πrB(dr/dt)

But, we have to convert cm to m.

Thus: 9.8739 cm = 9.8739 × 10^(-2) m

1.75 cm = 0.0175 m

Thus;

EMF = -2π × 9.8739 × 10^(-2) × 0.8 × 0.0175

EMF = -0.008686 V

For magnitude of emf, we will take the absolute value.

Thus;

Magnitude of EMF = 0.008686 V

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The given parameters of the ball are;

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The location of the other tennis player when the ball is launched = 10 m from the ball

The time at which the other tennis player begins to run = 0.3 seconds after the ball is launched

The height at which the ball is hit back = 2.1 m above the height from which the ball is launched

The vertical position, 'y', at time, 't', of a projectile motion is given as follows;

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The horizontal distance, 'x', the ball travels at t ≈ 2.145 seconds is given as follows;

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Therefore, we have;

The time the other player has to reach the ball, t₂ =2.145 s - 0.3 s ≈ 1.845 s

The distance the other player has to run, d = 20.682 m - 10 m = 10.682 m

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Answer:

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Explanation:

When the gravitational field strength is g, the weight (\text{weight}) of an object of mass m would be m\, g.

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Assuming that g =9.806\; {\rm N \cdot kg^{-1}}. The mass of this 63.8\; {\rm N}-object would be:

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The mass of this object is 6.50\; {\rm kg}. Hence, the density of this object would be:

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The mass of that much oil would be:

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