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ICE Princess25 [194]
2 years ago
6

A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction

Physics
2 answers:
Leni [432]2 years ago
7 0

Answer:

The coefficient of friction of the sled is 0.115

Explanation:

It is given that,

Mass of the sled, m = 46.6 kg

Frictional force acting on it, F = 52.9 N

We need to find the coefficient of friction. Frictional force is given by :

F=\mu mg

\mu=\dfrac{F}{mg}

\mu=\dfrac{52.9\ N}{46.6\ kg\times 9.8\ m/s^2}

\mu = 0.115

So, the coefficient of friction of the sled is 0.115. Hence, this is the required solution.                     

Setler [38]2 years ago
4 0

Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

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An OTR is removing electrodes from a client who has just received iontophoresis. Within several minutes of removing the electrod
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Two forces F1 and F2 act on a 5.00 kg object. Taking F1=20.0N and F2=15.00N, find the acceleration of the object for the configu
Anit [1.1K]
A) mass m with F1 acting in the positive x direction and F2 acting perpendicular in the positive y direction<span>

m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... y direction

Net force ^2 = F1^2 + F2^2 = (20N)^2 + (15n)^2 =  625N^2 =>

Net force = √625 = 25N

F = m*a => a = F/m = 25.0 N /5.00 kg = 5 m/s^2

Answer: 5.00 m/s^2

b) mass m with F1 acting in the positive x direction and F2 acting on the object at 60 degrees above the horizontal. </span>

<span>m = 5.00 kg
F1=20.0N  ... x direction
F2=15.00N</span><span>  ... 60 degress above x direction

Components of F2

F2,x = F2*cos(60) = 15N / 2 = 7.5N

F2, y = F2*sin(60) = 15N* 0.866 = 12.99 N ≈ 13 N


Total force in x = F1 + F2,x = 20.0 N + 7.5 N = 27.5 N

Total force in y = F2,y = 13.0 N

Net force^2 = (27.5N)^2 + (13.0N)^2 = 925.25 N^2 = Net force = √(925.25N^2) =

= 30.42N

a = F /m = 30.42 N / 5.00 kg = 6.08 m/s^2

Answer: 6.08 m/s^2


</span>
8 0
2 years ago
Read 2 more answers
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