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3 years ago

A physics student skis down a hill, accelerating at a constant

Physics

1 answer:

ikadub [295]3 years ago

6 0

<h3>Answer:</h3>

225 meters

<h3>Explanation:</h3>

Acceleration is the rate of change in velocity of an object in motion.

In our case we are given;

Acceleration, a = 2.0 m/s²

Time, t = 15 s

We are required to find the length of the slope;

Assuming the student started at rest, then the initial velocity, V₀ is Zero.

<h3>Step 1: Calculate the final velocity, Vf</h3>

Using the equation of linear motion;

Vf = V₀ + at

Therefore;

Vf = 0 + (2 × 15)

= 30 m/s

Thus, the final velocity of the student is 30 m/s

<h3>Step 2: Calculate the length (displacement) of the slope </h3>

Using the other equation of linear motion;

S = 0.5 at + V₀t

We can calculate the length, S of the slope

That is;

S = (0.5 × 2 × 15² ) - (0 × 15)

= 225 m

Therefore, the length of the slope is 225 m

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Answer:

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(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

Explanation:

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Magnetic field = 0.45 T

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Using formula of flux

$\phi_{2}=BA\cos\theta$

Put the value into the formula

$\phi_{2}=0.45\times(\pi\times(3\times10^{-2})^2)\cos0$

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We need to calculate the magnitude of the average induced emf

Using formula of emf

$\epsilon=-N(\dfrac{\Delta \phi}{\Delta t})$

Put the value into t5he formula

$\epsilon=-1\times(\dfrac{0.00127-0.01021}{0.53})$

$\epsilon=0.016867\ V$

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(C). If the coil resistance is 2.5 Ω.

We need to calculate the induced current

Using formula of current

$I=\dfrac{\epsilon}{R}$

Put the value into the formula

$I=\dfrac{0.016867}{2.5}$

$I=0.00675\ A$

$I=6.75\ mA$

Hence, (A). The direction of the induced current will be clockwise.

(B). The magnitude of the average induced emf 16.87 mV.

(C). The induced current is 6.75 mA.

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