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3 years ago

9

sledge (including load) weighs 5000 N. It is pulled on level snow by a dog team exerting ahorizontal force on it. The coeﬃcient

of kinetic friction between sledge and snow is 0.05. Howmuch work is done by the dog team pulling the sledge 1000 m at constant speed?

1 answer:

lara31 [8.8K]3 years ago

6 0

Answer:

2.5 x 10^{5} J

Explanation:

weight = 5,000 N

coefficient of friction = 0.05

distance = 1000 m

how much work is done by the dogs pulling the sledge

work done = force x coefficient of friction x distance

work done = 5000 x 0.05 x 1000 = 2.5 x 10^{5} J

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Aloop of wire of area 71 cm^2 is placed with its plane parallel to a 16 mt magnetic field. the loop is then rotated so that its

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Answer:

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Explanation:

The average EMF in the coil is equal to

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Why does this formula work?

By Faraday's Law of Induction, the EMF $\epsilon$ induced in a coil (one loop) is equal to the rate of change in the magnetic flux $\Phi$ through the coil.

$\displaystyle \epsilon(t) = \frac{d}{dt}(\Phi(t))$ .

Finding the average EMF in the coil is similar to finding the average velocity.

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt$ .

However, by the Fundamental Theorem of Calculus, integration reverts the action of differentiation. That is:

$\displaystyle \int_0^{t} \epsilon(t)\cdot dt = \int_0^{t} \frac{d}{dt}\Phi(t)\cdot dt = \Phi(t) - \Phi(0)$ .

Hence the equation

$\displaystyle \text{Average}\; \epsilon = \frac{1}{t}\int_0^t \epsilon(t)\cdot dt = \frac{\Phi(t)- \Phi(0)}{t}$ .

Note that information about the constant term in the original function will be lost. However, since this integral is a definite one, the constant term in $\Phi(t)$ won't matter.

Apply this formula to this question. Note that $\Phi$ , the magnetic flux through the coil, can be calculated with the equation

$\Phi = B \cdot A \cdot N \; \sin{\theta}$ .

For this question,

$B = \rm 16\; mT = 16\times 10^{-3}\; T$ is the strength of the magnetic field.
$A = \rm 71\; cm^{2} = 71\times \left(10^{-2}\right)^2 \; m^{2}$ is the area of the coil.
$N = 1$ is the number of loops in the coil.
$\theta$ is the angle between the field lines and the coil.
At $\rm 0\;s$ , the field lines are parallel to the coil, $\theta = 0^{\circ}$ .
At $\rm 0.7\; s$ , the field lines are perpendicular to the coil, $\displaystyle \theta = 90^{\circ}$ .

Initial flux: $\Phi(0)= 0$ .

Final flux: $\Phi(0.7) = \rm 1.1136\times 10^{-4}\; Wb$ .

Average EMF, which is the same as the average rate of change in flux:

$\displaystyle \frac{\Phi(0.7) - \Phi(0)}{0.7} \approx\rm 1.62\times 10^{-4}\; V$ .

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