To answer this question, it helps enormously if you know
the formula for momentum:
Momentum = (mass) x (speed) .
Looking at the formula, you can see that momentum is directly
proportional to speed. So if speed doubles, so does momentum.
If the car's momentum is 20,000 kg-m/s now, then after its speed
doubles, its momentum has also doubled, to 40,000 kg-m/s.
Answer: question D is trough
Explanation: I learned this already
The kinetic energy (KE) of a 0.155 kg arrow that is shot from ground level, upward at 31.4 m/s, when it is 30.0 m above the ground is 30.85 J
Assuming air friction is negligible,
a = - 9.8 m / s²
u = 31.4 m / s
s = 30 m
v² = u² + 2 a s
v² = 31.4² + ( 2 * - 9.8 * 30 )
v² = 985.96 - 588
v² = 397.96 m / s
KE = 1 / 2 m v²
KE = 1 / 2 * 0.155 * 397.96
KE = 0.0775 * 397.96
KE = 30.85 J
Therefore, the kinetic energy ( KE ) when it is 30.0 m above the ground is 30.85 J
To know more about kinetic energy
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Explanation:
Given that,
Initial speed of the billiard ball 1, u = 30i cm/s
Initial speed of another billiard ball 2, u' = 40j cm/s
After the collision,
Final speed of first ball, v = 50 cm/s
Final speed of second ball, v' = 0 (as it stops)
Let us consider that both balls have same mass i.e. m
Initial kinetic energy of the system is :
Final kinetic energy of the system is :
The change in kinetic energy of the system is equal to the difference of final and initial kinetic energy as :
So, the change in kinetic energy of the system as a result of the collision is equal to 0.
