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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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The 12.2-m crane weighs 18 kn and is lifting a 67-kn load. the hoisting cable (tension t1) passes over a pulley at the top of th
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What is the mass and density of 237 mL of water
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An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The el
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Complete Question

An aluminum "12 gauge" wire has a diameter d of 0.205 centimeters. The resistivity ρ of aluminum is 2.75×10−8 ohm-meters. The electric field in the wire changes with time as E(t)=0.0004t2−0.0001t+0.0004 newtons per coulomb, where time is measured in seconds.

I = 1.2 A at time 5 secs.

Find the charge Q passing through a cross-section of the conductor between time 0 seconds and time 5 seconds.

Answer:

The charge is  Q =2.094 C

Explanation:

From the question we are told that

    The diameter of the wire is  d =  0.205cm = 0.00205 \ m

     The radius of  the wire is  r =  \frac{0.00205}{2} = 0.001025  \ m

     The resistivity of aluminum is 2.75*10^{-8} \ ohm-meters.

       The electric field change is mathematically defied as

         E (t) =  0.0004t^2 - 0.0001 +0.0004

     

Generally the charge is  mathematically represented as

       Q = \int\limits^{t}_{0} {\frac{A}{\rho} E(t) } \, dt

Where A is the area which is mathematically represented as

       A =  \pi r^2 =  (3.142 * (0.001025^2)) = 3.30*10^{-6} \ m^2

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       \frac{A}{\rho} =  \frac{3.3 *10^{-6}}{2.75 *10^{-8}} =  120.03 \ m / \Omega

Therefore

      Q = 120 \int\limits^{t}_{0} { E(t) } \, dt

substituting values

      Q = 120 \int\limits^{t}_{0} { [ 0.0004t^2 - 0.0001t +0.0004] } \, dt

     Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | t} \atop {0}} \right.

From the question we are told that t =  5 sec

           Q = 120 [ \frac{0.0004t^3 }{3} - \frac{0.0001 t^2}{2} +0.0004t] }  \left | 5} \atop {0}} \right.

          Q = 120 [ \frac{0.0004(5)^3 }{3} - \frac{0.0001 (5)^2}{2} +0.0004(5)] }

         Q =2.094 C

     

5 0
2 years ago
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