A2.4-m-diameter merry-go-round with a mass of 270 kg is spinning at 20 rpm. john runs around the merry-go-round at 5.0 m/s, in the same direction that it is turning, and jumps onto the outer edge. john’s mass is 34 kg . part a part complete what is the merry-go-round's angular speed, in rpm, after john jumps on?

23.98 rpm

Explanation:

d = diameter of merry-go-round = 2.4 m

r = radius of merry-go-round = (0.5) d = (0.5) (2.4) = 1.2 m

m = mass of merry-go-round = 270 kg

I = moment of inertia of merry-go-round

Moment of inertia of merry-go-round is given as  

I = (0.5) m r² = (0.5) (270) (1.2)² = 194.4 kgm²

M = mass of john = 34 kg

Moment of inertia of merry-go-round and john together after jump is given as

  I' = (0.5) m r² + M r² = 194.4 + (34) (1.2)² = 243.36 kgm²

w = final angular speed

w₀ = initial angular speed of merry-go-round = 20 rpm = 2.093 rad/s

v = speed of john before jump

using conservation of angular momentum

Mvr + I w₀ = I' w

(34) (5) (1.2) + (194.4) (2.093) = (243.36) w

w = 2.51 rad/s

w = 23.98 rpm

i think it’s a. this is because if they measured the mass of something with a mass of 150g, then their final answer would have to be approximately 150g. though all of their answers are almost 150, rosa and tran have the closest answers (or the exact answer) to the accurate one.

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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