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astra-53 [7]
2 years ago
5

A hockey puck is sliding across a frozen pond with an initial speed of 9.3 m/s. It comes to rest after sliding a distance of 42.

0 m. What is the coefficient of kinetic friction between the puck and the ice?
Physics
1 answer:
kondaur [170]2 years ago
3 0

Answer:

The coefficient of kinetic friction between the puck and the ice is 0.11

Explanation:

Given;

initial speed, u = 9.3 m/s

sliding distance, S = 42 m

From equation of motion we determine the acceleration;

v² = u² + 2as

0 = (9.3)² + (2x42)a

- 84a = 86.49

a = -86.49/84

|a| = 1.0296

F_k = \mu_k N = ma

where;

Fk is the frictional force

μk is the coefficient of kinetic friction

N is the normal reaction = mg

μkmg = ma

μkg = a

μk = a/g

where;

g is the gravitational constant = 9.8 m/s²

μk = a/g

μk = 1.0296/9.8

μk = 0.11

Therefore, the coefficient of kinetic friction between the puck and the ice is 0.11

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Answer:

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(A). We need to calculate the velocity

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(B). We need to calculate the radius

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Put the value into the formula

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6 0
3 years ago
The altitude of a hang glider is increasing at a rate of 6.75 m/s. At the same time, the shadow of the glider moves along the gr
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Answer:

16.45 m/s

Explanation:

Let y be the vertical distance and x be the horizontal distance

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The shadow of the glider moves along the ground at speed=v_x=\frac{dx}{dt}=15m/s

We have to find the magnitude of glider's velocity.

We know that

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Substitute the values

v=\sqrt{(15)^2+(6.75)^2}

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v=16.45m/s

Hence, the magnitude of glider's velocity=16.45 m/s

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3 years ago
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