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3 years ago

11

While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock an

d the car begins to slide against the road. The car collides with the rabbit at a final speed of vf, after T seconds of braking. What is the coefficient of kinetic friction of the tires against the road?

1 answer:

MA_775_DIABLO [31]3 years ago

4 0

Answer:

μk = (Vf - Vc)/(T×g)

Explanation:

Given

Vi = initial velocity of the car

Vf = final velocity of the car

T = Time of application of brakes

g = acceleration due to gravity (known constant)

Let the mass of the car be Mc

Assuming only kinetic frictional force acts on the car as the driver applies the brakes,

The n from Newtown's second law of motion.

Fk = Mc×a

Fk = μk×Mc×g

a = (Vf - Vc)/T

Equating both preceding equation.

μk×Mc×g = Mc × (Vf - Vc)/T

Mc cancels out.

μk = (Vf - Vc)/(T×g)

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$x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ\,t+\dfrac12\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t^2$

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and its velocity vector has components

$v_x=\left(103\,\frac{\rm m}{\rm s}\right)\cos53.0^\circ+\left(32.0\,\frac{\rm m}{\mathrm s^2}\right)\cos53.0^\circ t$

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After $t=3.00\,\rm s$ , its position is

$x=273\,\rm m$

$y=362\,\rm m$

and the rocket's velocity vector has horizontal and vertical components

$v_x=120\,\frac{\rm m}{\rm s}$

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After the engine failure $(t>3.00\,\rm s)$ , the rocket is in freefall and its position is given by

$x=273\,\mathrm m+\left(120\,\frac{\rm m}{\rm s}\right)t$

$y=362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2$

and its velocity vector's components are

$v_x=120\,\frac{\rm m}{\rm s}$

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$159\,\frac{\rm m}{\rm s}-gt=0\implies t=16.2\,\rm s$

At this point, the rocket has an altitude of

$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)(16.2\,\rm s)-\dfrac g2(16.2\,\rm s)^2=1650\,\rm m$

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$362\,\mathrm m+\left(159\,\frac{\rm m}{\rm s}\right)t-\dfrac g2t^2=0\implies t=34.6\,\rm s$

So the rocket stays in the air for a total of $37.6\,\rm s$ .

c. After the engine failure, the rocket traveled for about 34.6 seconds, so we evalute $x$ for this time $t$ :

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Answer:

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Explanation:

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In the old transformer the ratio of voltages was:

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In the new transformer the ratio of voltages is:

$\frac{n_1}{n_2}=\frac{V_1}{V_2}=\frac{12}{500} =0.024\\\\n_2=n_1/0.24=41.7n_1$

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$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{500}= 0.024\\\\I_2=0.024I_1$

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$\frac{V_1}{V_2}=\frac{I_2}{I_1}=\frac{12}{425}= 0.028\\\\I_2=0.028I_1$

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$P_L=I^2\cdot R$

Then, the ratio of losses is (R is constant for both power losses):

$\frac{P_n}{P_o} =\frac{I_n^2R}{I_o^2R} =(\frac{I_n}{I_o} )^2=0.86^2=0.74$

The losses in the new line are 74% the losses in the old line.

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