Answer:
v2 = 27.3m/s
Explanation:
Assuming forward as positive.
Mass = m1 = 64kg
Let v be the common velocity of the student and the skateboard.
mass of skateboard = m2 = 5.94kg
v = 1.4m/s
Since the skateboard and the student are initially moving together at the same velocity their momentum together is
(m1 + m2)v
Let the final velocity of the student be v1 and the final velocity of the skateboard be v2
v1 = – 1.0m/s (falls backwards that's why the velocity is negative since we are assuming forward as positive)
Then from conservation of momentum, momentum before is equal to momentum after.
(m1 + m2)v = m1v1 + m2v2
m2v2= (m1 + m2)v – m1v1
v2 = ( (m1 + m2)v – m1v1)/m2
v2 = ( (64 + 5.94)×1.4 – 64×(-1.0))/5.94
v2 = ( (64 + 5.94)×1.4 + 64×1.0)/5.94
v2 = 27.3m/s
Answer: [B]: The letter, "<em /> I " ; for current; in units of "Amps" .
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Answer:

Explanation:
Given:
- mass of the object on a horizontal surface,

- coefficient of static friction,

- coefficient of kinetic friction,

- horizontal force on the object,

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

where:
normal force of reaction acting on the body= weight of the body


As we know that the frictional force acting on the body is always in the opposite direction:
So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.
so, the frictional force will be:
