A 7.0 kg bowling ball moves at 2.45 m/s. How fast must a 2.60 g ping pong ball move so that the two balls have the same kinetic energy

127.12 m/s

Explanation:

From the question,

Kinetic energy of the bowling ball = Kinetic energy of the ping pong ball

1/2mv² = 1/2m'v'²

Where m = mass of the bowling ball, v = velocity of the bowling ball, m' = mass of the ping pong ball v' = velocity of the ping pong ball

make v' the subject of the equation

v' = √(mv²/m') Equation 2

Given: m = 7.0 kg, v = 2.45 m/s, m' = 2.60 g = 0.0026 kg

Subtitute into equation 2

v' = √[(7×2.45²)/0.0026]

v' = √16160.57

v' = 127.12 m/s

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete

initial speed when it is dropped is 150 ft/s

now at the same moment the acceleration due to gravity is perpendicular to its velocity

so as we know that

now we have centripetal acceleration a = 32 ft/s^2 as its perpendicular to velocity

while tangential acceleration at the same moment will be zero

so

now radius of curvature is given as

so the radius of curvature just after it is dropped is 703.125 ft