Answer:
option C
Explanation:
given,
energy dissipated by the system to the surrounding = 12 J
Work done on the system = 28 J
change in internal energy of the system
Δ U = Q - W
system losses energy = - 12 J
work done = -28 J
Δ U = Q - W
Δ U = -12 -(-28)
Δ U = 16 J
hence, the correct answer is option C
Answer:
position 9.58 m
Explanation:
In impulse exercises and amount of movement, we always assume that the contact time is small,
I = Δp
With this expression we can calculate the final speed
I = m Vf - m Vo
Vf = (I + mVo) / m
Vf = (1.8 + 0.35 1.8) /0.35
Vf = 6.94 m / s
To calculate the acceleration of the ball we use Newton's second law, after finishing the impulse
∑ F = m a
fr = m a
a = fr / m
a = -0.26 / 0.35
a = -0.74 m/s²
A negative sign indicates that this acceleration is slowing the ball
Now we have speed and time acceleration, so we can use the kinematic equations to find the position at 1.5 s
X = Vo t + ½ to t²
In this case Vo is the speed with which the ball comes out after the impulse 6.94
X = 6.94 1.5 + ½ (-0.74) 1.522
X = 9.58 m
Answer:
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