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3 years ago

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A 16.0 kg sled is being pulled along the horizontal snow-covered ground by a horizontal force of 25.0 N. Starting from rest, the

sled attains a speed of 1.00 m/s in 8.00 m. Find the coefficient of kinetic friction between the runners of the sled and the snow. You Answered

1 answer:

Fudgin [204]3 years ago

4 0

Answer:

$\mu_k = 0.15$

Explanation:

according to the kinematic equation

$v^{2} - u^{2} = 2aS$

Where

u is initial velocity = 0 m/s

a = acceleration

S is distance = 8.00 m

final velocity = 1.0 m/s

$a = \frac {v^{2}}{2S}$

$a = \frac {1{2}}{2*8.6}$

a = 0.058 m/s^2

from newton second law

Net force = ma

$f_{net} = ma$

F - f = ma

2 $5 - \mu_kN = ma$

$25 - \mu_kmg = ma$

$\frac {25 - ma}{mg} =\mu_k$

$\frac {25 - 16*0.058}{16*9.81} = 0.15$

$\mu_k = 0.15$

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3 years ago

The rocket is fired vertically and tracked by the radar station shown. When θ reaches 66°, other corresponding measurements give

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Answer:

velocity = 1527.52 ft/s

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We are given;

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