Answer:
3.974 Joule
Explanation:
Diameter of ring = 7.7 cm
a = Distance from the center = d/2 = 3.85 cm = 0.0385 m
Q = Charge = 5 mC
q = Charge to move = 3.4 mC
k = Coulomb constant = 9×10⁹ Nm²/C²
Work done will be equal to Potential energy when mass is at center
∴ Work to move a tiny 3.4 mC charge from very far away to the center of the ring is 3.974 Joule
i would sat C then the impact is weaker
Given
m1(mass of the first object): 55 Kg
m2 (mass of the second object): 55 Kg
v1 (velocity of the first object): 4.5 m/s
v2 (velocity of the second object): ?
m3(mass of the object dropped): 2.5 Kg
The law of conservation of momentum states that when two bodies collide with each other, the momentum of the two bodies before the collision is equal to the momentum after the collision. This can be mathemetaically represented as below:
Pa= Pb
Where Pa is the momentum before collision and Pb is the momentum after collision.
Now applying this law for the above problem we get
Momentum before collision= momentum after collision.
Momentum before collision = (m1+m2) x v1 =(55+5)x 4.5 = 270 Kgm/s
Momentum after collision = (m1+m2+m3) x v2 =(55+5+2.5) x v2
Now we know that Momentum before collision= momentum after collision.
Hence we get
270 = 62.5 v2
v2 = 4.32 m/s
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Answer:
Its A. Walking on the beach
