When a golfer tees off, the head of her golf club which has a mass of 152 g is traveling 44.8 m/s just before it strikes a 46.0 g golf ball at rest on a tee. Immediately after the collision, the club head continues to travel in the same direction but at a reduced speed of 27.7 m/s. Neglect the mass of the club handle and determine the speed of the golf ball just after impact.

56.5 m/s²

Explanation:

From the law of conservation of momentum,

mu+m'u' = mv+m'v' Equation 1

Where m = mass of the golf club, u = initial velocity of the golf club, m' = mass of the golf ball, u' = initial velocity of the golf ball, v = final velocity of the golf club, v' = final velocity of the golf ball.

From the question,

The golf ball is at rest, Hence u' = 0 m/s

mu = mv+m'v'

Make v' the subject of the equation

v' = (mu-mv)/m' Equation 2

Given: m = 152 g = 0.152 kg, u = 44.8 m/s, v = 27.7 m/s, m' = 46 g = 0.046 kg.

Substitute into equation 2

v' = (0.152×44.8+0.152×27.7)/0.046

v' = (6.8096-4.2104)/0.046

v' = 2.5992/0.046

v' = 56.5 m/s²

i think they are all correct

explanation:

they seem good

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the shielding can reduce the coupling of radio waves, electromagnetic fields and electrostatic fields. a conductive enclosure used to block electrostatic fields is also known as a faraday cage. the amount of reduction depends very much upon the material used, its thickness, the size of the shielded volume and the frequency of the fields of interest and the size, shape and orientation of apertures in a shield to an incident electromagnetic

from wikipedia