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3 years ago

12

Boy pulls a 5.0-kg sled with a rope that makes a 60.0° angle with respect to the horizontal surface of a frozen pond. The boy pu

lls on the rope with a force of 10.0 N; and the sled moves with constant velocity. What is the coefficient of friction between the and the ice?
(a) 0.09
(b) 0.12
(c) 0.18
(d) 0.06
(e) 0.24

1 answer:

kykrilka [37]3 years ago

4 0

Answer:

0.1

Explanation:

mass, m = 5 kg

θ = 60°

Force, F = 10 N

velocity is constant , it means the net force is zero.

So, the component of force along the surface is equal to the friction force

FCosθ = friction force

10 x cos 60 = μ x m x g

where, μ is the coefficient of friction

5 = μ x 5 x 9.8

μ = 0.1

Thus, the coefficient of friction is 0.1

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Complete Question

The complete question is shown on the first uploaded image

Answer:

a

$I(\tau)=0.051 A$

b

$I(3 \tau)=0.076 A$

c

$I_c= 0.08 A$

Explanation:

From the question we are told that

$I(t) = \frac{e}{R}(1-e^{\frac{t}{\tau} }) ; \ Where \ \tau = L/R$

From the question we are told to find $I(\tau)$ when t=0 equals the time constant ( $\tau$ )

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Substituting 12 V for $\epsilon$ and 150Ω for R

$I(\tau) = \frac{12}{150} (1- e^{-1})$

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From the question we are told to find $I(3 \tau)$ when t=0 equals the 3 times the time constant ( $\tau$ )

That is to obtain $I(3\tau)$ .This is mathematically represented as

$I(\tau = t) = \frac{\epsilon}{R} (1- e^{-\frac{3\tau}{\tau} })$

$I(\tau) = \frac{12}{150} (1- e^{-3})$

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Answer:

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2) 2.7 m/s²

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