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aleksklad [387]
4 years ago
9

In order to determine the coefficients of friction between rubber and various surfaces, a student uses a rubber eraser and an in

cline. In one experiment, the eraser begins to slip down the incline when the angle of inclination is 36.7� and then moves down the incline with constant speed when the angle is reduced to 30.1�. From these data, determine the coefficients of static and kinetic friction for this experiment.
Physics
1 answer:
Strike441 [17]4 years ago
8 0

Answer:

μs = 0.75

μk = 0.58

Explanation:

From a force diagram:

m*g*sin \theta - Ff=m*a     (1)

N-m*g*cos \theta = 0         (2)

When it starts slipping, friction force is the maximum and acceleration is 0. Replacing these conditions on (1):

m*g*sin \theta - \mu*m*g*cos \theta=0   Solving for μs:

\mu=tan \theta

μs = tan 36.7° = 0.75

When it moves at constant speed, friction force is kinetic friction and acceleration is 0. With these conditions the coefficient is:

μk = tan 30.1° = 0.58

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Answer:

1.78 rad/s

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Explanation:

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r_2 = New distance between center of mass of person and pole = 0.46 m

I = Moment of inertia

Angular speed is given by

\omega_1=\dfrac{v}{r_1}\\\Rightarrow \omega_1=\dfrac{0.8}{0.45}\\\Rightarrow \omega_1=1.78\ rad/s

The angular speed is 1.78 rad/s

In this system the angular momentum is conserved

L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow mr_1^2\omega_1=mr_2^2\omega_2\\\Rightarrow \omega_2=\dfrac{r_1^2\omega_1}{r_2^2}\\\Rightarrow \omega_2=\dfrac{0.45^2\times 1.78}{0.46^2}\\\Rightarrow \omega_2=1.70344\ rad/s

The new angular speed is 1.70344 rad/s

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3 years ago
A 2.0-kg object is attached to a spring (k = 55.6 N/m) that hangs vertically from the ceiling. The object is displaced 0.045 m v
rusak2 [61]

Answer:

Maximum acceleration will be 1.251m/sec^2

Explanation:

We have given mass of the object m = 2 kg

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We know that maximum acceleration in SHM is given by

Maximum acceleration =A\omega ^2

We know that \omega ^2=\frac{k}{m}=\frac{55.6}{2}=27.8

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<h2>A  or B</h2>

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The velocity of the water in the pipe at right is given by V1 = 0.5t m/s and V2 = 1.0t m/s, where t is in seconds. Determine the
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Answer:

A) At point 1, local acceleration = 0.5 m/s²

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Explanation:

Local acceleration at those points is the instantaneous acceleration at those points and it is given as

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Average acceleration = 0.5t/t = 0.5 m/s²

This value is positive indicating an increase in velocity and acceleration kf the fluid as the cross sectional Area of flow reduces.

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4 years ago
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