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3 years ago

7

If the coefficient of kinetic friction between tires and dry pavement is 0.84, what is the shortest distance in which you can st

op an automobile by locking the brakes when traveling at 29.0 m/s

1 answer:

liberstina [14]3 years ago

4 0

Answer:

The shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

Explanation:

Given;

coefficient of kinetic friction, μ = 0.84

speed of the automobile, u = 29.0 m/s

To determine the the shortest distance in which you can stop an automobile by locking the brakes, we apply the following equation;

v² = u² + 2ax

where;

v is the final velocity

u is the initial velocity

a is the acceleration

x is the shortest distance

First we determine a;

From Newton's second law of motion

∑F = ma

F is the kinetic friction that opposes the motion of the car

-Fk = ma

but, -Fk = -μN

-μN = ma

-μmg = ma

-μg = a

- 0.8 x 9.8 = a

-7.84 m/s² = a

Now, substitute in the value of a in the equation above

v² = u² + 2ax

when the automobile stops, the final velocity, v = 0

0 = 29² + 2(-7.84)x

0 = 841 - 15.68x

15.68x = 841

x = 841 / 15.68

x = 53.64 m

Thus, the shortest distance in which you can stop the automobile by locking the brakes is 53.64 m

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Answer:

Therefore the amplitude of the resultant wave is $=0.95 y_m$

Explanation:

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For wave 1:

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Where A= amplitude= $y_m$

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$k=\frac{2\pi}{\lambda}$ , $\lambda$ = wave length.

t= time

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The resultant wave will be

y = y₁ + y₂

= $y_m$ sin (kx-ωt) + $y_m$ sin (kx-ωt+Φ)

$=y_m$ {sin (kx-ωt) + sin (kx-ωt+Φ)}

$=y_m\ sin(\frac{kx-\omega t +\phi + kx-\omega t }2)\ cos(\frac{kx-\omega t +\phi -kx+\omega t}2)$

$=y_m\ sin({kx-\omega t +\frac\phi 2)\ cos(\frac{\phi }2)$

$=y_m\ cos(\frac{\phi }2) sin({kx-\omega t +\frac\phi 2)$

Therefore the amplitude of the resultant wave is

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Given :

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If the two cars crashed and stuck together, what would their combined momentum be afterwards and what direction would they be moving.

Solution :

Let, combined velocity of the car is v.

Also, let us assume that mass of both the cars are equal and it is m.

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4 0

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Answer:

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solution

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Answer:

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<u>Given the following data;</u>

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Time = 3.2 seconds

To show that the speed of the stone when it hits the ground is 32 m/s, we would use the first equation of motion;

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