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rewona [7]
3 years ago
9

The 2.5-Mg van is traveling with a speed of 100 km>h when the brakes are applied and all four wheels lock. If the speed decre

ases to 40 km>h in 5 s, determine the coefficient of kinetic friction between the tires and the road.
Physics
1 answer:
Andrew [12]3 years ago
4 0

Answer:

0.34

Explanation:

2.5 Mg = 2500 kg

The change in speed from 100 km/h to 40 km/h is

\Delta v = 100 - 40 = 60 km/h = 60 * 1000 / (60 * 60) = 16.67 m/s

The deceleration caused by friction force is the change in speed per unit of time

a = \Delta v / \Delta t = 16.67 / 5 = 3.33 m/s^2

Using Newton 2nd law we can calculate the friction force that caused this deceleration:

F = ma = 2500 * 3.33 = 8333 N

Let g = 9.8m/s2. Friction force is the product of normal (gravity) force and friction coefficient

F = mg\mu

8333 = 2500*9.8\mu

\mu = \frac{8333}{2500 * 9.8} = 0.34

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Answer:

62.64 RPM.

Explanation:

Given that

m= 4.6 g

r= 19 cm

μs = 0.820

μk = 0.440.

The angular speed of the turntable = ω rad/s

Condition just before the slipping starts

The maximum value of the static friction force =Centripetal force

\mu_s\ m g=m\ \omega^2\ r\\ \omega^2=\dfrac{\mu_s\ m g}{m r}\\ \omega=\sqrt{\dfrac{\mu_s\  g}{ r}}\\ \omega=\sqrt{\dfrac{0.82\times 10}{ 0.19}}\ rad/s\\\omega= 6.56\ rad/s

\omega=\dfrac{2\pi N}{60}\\N=\dfrac{60\times \omega}{2\pi }\\N=\dfrac{60\times 6.56}{2\pi }\ RPM\\N=62.64\ RPM

Therefore the speed in RPM will be 62.64 RPM.

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Kelvin, Celsius, and Fahrenheit are three types of<br> scales
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Suppose a man pushes a crate with a force of 20 N north. What is the magnitude and direction of the reaction force?
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20 N north because He is pushing it 20 newtons north.

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A yo-yo has a string that is 0.95 m in length. What is the period of oscillation if the yo-yo is allowed to swing back and forth
yulyashka [42]

Answer:

Explanation:

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Sarah, whose mass is 40 kg, is on her way to school after a winter storm when she accidentally slips on a patch of ice whose coe
RideAnS [48]

Sarah's acceleration is -0.49 m/s^2

Explanation:

The force of kinetic friction acting on Sarah has a magnitude which is given by:

F_f = \mu mg

where

\mu is the coefficient of kinetic friction

m is Sarah's mass

g is the acceleration of gravity

Moreover, according to Newton's second law of motion, we know that the net force on Sarah is equal to its mass times its acceleration:

F=ma

where a is the acceleration

Since the force of friction is the only force acting on Sarah, we can say that the net force is equal to the force of friction, therefore:

F=-\mu mg = ma

where the negative sign is due to the fact that the force of friction has a direction opposite to the motion of Sarah. Solving for a, we find

a=-\mu g

And substituting the following values:

\mu = 0.05 (coefficient of friction)

g=9.81 m/s^2 (acceleration of gravity)

we find:

a=-(0.05)(9.81)=-0.49 m/s^2

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