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Travka [436]
3 years ago
14

If a train is traveling eastward with a constant velocity of 15 m/s, what is the net external force action on it? There is not e

nough information, so it cannot be solved. There is no net force acting on the train. the force from the engines
Physics
2 answers:
sammy [17]3 years ago
6 0

Answer:

There is no net force acting on the train.

Explanation:

The net force equals mass times acceleration:

∑F = ma

The train moves at constant velocity, so its acceleration is 0.  Therefore, the net force is 0.

ExtremeBDS [4]3 years ago
3 0

If a train is travelling eastward with a constant velocity of 15 m/s, there is no net force acting on the train.

Answer: Option B

<u>Explanation: </u>

According to Newton 2nd law, the external force is proportionate to the rates of change of momentum which states that force will be indirectly proportional to the rates of variation in velocity. So, as the velocity is constant for the train, the rate of variation in velocity or the acceleration will be zero. Thus the total external forces acting on the train will be zero. But the train will be moving continuously due to the energy obtained from the engine.

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A cart on a horizontal, linear track has a fan attached to it. The cart is positioned at one end of the track, and the fan is tu
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Answer:

a

  F = 0.0566 \  N  

b

   t =  6.147 \  s

Explanation:

From the question we are told that

     The distance travel in  4.22 s is  s =  1.44 \ m

     The mass of the cart plus the fan is  m  =  350 \  g =  0.35 \  kg

Generally from kinematic equation we have that

        s =  ut + \frac{1}{2}  * a * t^2

Here  u is the initial  velocity with value  u =  0 \ m/s

So  

         1.44=  0 * t + \frac{1}{2}  *  a * 4.22^2      

=>      a =  0.1617 \  m/s^2

Generally the net force is  

         F = m * a

=>      F = 0.35  *  0.1617  

=>      F = 0.0566 \  N  

Gnerally the new mass of the cart plus the fan is  M  =  656 \  g  = 0.656 \  kg

    The distance considered is s_1 = 1.63 \  m

     Generally the new acceleration of the cart is mathematically represented as

        F =  M  *  a_1

=>      a_1 =  \frac{F}{M}

=>      a_1 =  \frac{0.0566}{0.656}

=>      a_1 = 0.08628 \  m/s^2

Gnerally from kinematic equation we have

          s =  ut + \frac{1}{2} *  a_1 *  t ^2

Here u  is the initial velocity and the value is zero because it started from rest  

=>       1.63 =  0 * t + \frac{1}{2} *  0.08628*  t ^2

=>        t =  6.147 \  s

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Answer:

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Explanation:

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