Linear momentum (mass x speed) has to be conserved.
-- Momentum before the jump:
(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s
(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero
Total momentum before the jump: (100 kg-m/s) + (zero) = (100 kg-m/s)
-- Momentum after the jump:
(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)
-- Momentum after the jump = momentum before the jump
(40 kg) x (speed) = 100 kg-m/s
Divide each side by 40 kg:
Speed = (100 kg-m/s) / (40 kg)
<em>Speed = 2.5 m/s</em> (d)
Answer:
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Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:

where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
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