Let the mass of 2500 kg car be 
and it's velocity be 
and the mass of 1500 kg car be 
and it's velocity be 
.
After the bumping the mass be M and it's velocity be V.
By law of conservation of momentum we have
2500 * 5 + 1500 * 1=4000 * V
V = 14000/4000 = 7/2 = 3.5 m/s
So the velocity of the two-car train = 3.5 m/s
Suppose car A is moving with a velocity Va, and car b with a velocity Vb,
According the principle of conservation of momentum:
Va x Ma + Vb x Mb = (Ma + Mb) V
V = (Va x Ma + Vb x Mb)/(Ma +Mb)
V = speed of cars after coupling
V = (Va x 20 mg + Vb x 15 mg)/(20 mg + 15 mg)
Put in the values of Va and Vb, and get the V
Answer:
Explanation:
5.non contact force
6.balanced force
7.unbalanced force
8.net force
9.zero
10.is the difference between two forces
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Answer:
The magnitude of the force is 12 N Upwards
Explanation:
The force on a positive charge will be in the same direction as the field, but the force on a negative charge will be in the opposite direction to the field. Thus the direction of the force is upward.
Given;
magnitude of charge, q = 0.06 C
magnitude of electric field, E = 200 N/C
The magnitude of the force is given by;
F = qE
F = 0.06 x 200 N/C
F = 12 N Upwards
Therefore, the magnitude of the force is 12 N Upwards
