A 2.03 kg book is placed on a flat desk. Suppose the coefficient of static friction between the book and the desk is 0.602 and t
1 answer:
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Answer:
(a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Explanation:
Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.
Given that,
Mass of steel block = 2.30 kg
Mass of ball = 0.500 kg
Length of cord = 50.0 cm
We need to calculate the initial speed of the ball
Using conservation of energy
Put the value into the formula
The initial speed of the ball
The initial speed of the block
(a). We need to calculate the speed of the ball after collision
Using formula of collision
Put the value into the formula
Negative sign shows the opposite direction of initial direction.
(b). We need to calculate the speed of the block after collision
Using formula of collision
Put the value into the formula
Hence, (a). The speed of the ball after collision is 2.01 m/s.
(b). The speed of the block after collision 1.11 m/s.
Answer:
11.25 amps
Explanation:
For transformers, the magnetic flux
Therefore;
Ф = Фmax (cosωt) = 0.21·(cos(5·t))
From Faraday's law of induction, we have;
ε = -N × dΦ/dt
Which gives;
dΦ/dt = -1.05(sin (5t) )
ε = -N × dΦ/dt = -50× -1.05(sin (5t) )
ε = 52.5(sin (5t) )
I = ε/R = 52.5(sin (5t) )/3.3 = 15.9091(sin (5t) ) amps
The peak current is therefore = 15.9091 amps
The rms current = Peak current /√2 = 15.9091/(√2) = 11.25 amps.