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DedPeter [7]
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
Physics
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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Answer:

2.43J

Explanation:

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Unknown:

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Solution:

The kinetic energy of a body is the energy in motion of the body;

 it can be derived using the expression below:

 

   K.E  = \frac{1}{2}  m v²

m is the mass

v is the velocity

 Solve for K.E;

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Steam enters a well-insulated nozzle at 200 lbf/in.2 , 500F, with a velocity of 200 ft/s and exits at 60 lbf/in.2 with a velocit
Ede4ka [16]

Answer:

386.2^{\circ}F

Explanation:

We are given that

P_1=200lbf/in^2

P_2=60lbf/in^2

v_1=200ft/s

v_2=1700ft/s

T_1=500^{\circ}F

Q=0

C_p=1BTU/lb^{\circ}F

We have to find the exit temperature.

By steady energy flow equation

h_1+v^2_1+Q=h_2+v^2_2

C_pT_1+\frac{P^2_1}{25037}+Q=C_pT_2+\frac{P^2_2}{25037}

1BTU/lb=25037ft^2/s^2

Substitute the values

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