Answer:
0.167m/s
Explanation:
According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.
Given momentum = Maas × velocity.
Momentum of glider A = 1kg×1m/s
Momentum of glider = 1kgm/s
Momentum of glider B = 5kg × 0m/s
The initial velocity of glider B is zero since it is at rest.
Momentum of glider B = 0kgm/s
Momentum of the bodies after collision = (mA+mB)v where;
mA and mB are the masses of the gliders
v is their common velocity after collision.
Momentum = (1+5)v
Momentum after collision = 6v
According to the law of conservation of momentum;
1kgm/s + 0kgm/s = 6v
1 =6v
V =1/6m/s
Their speed after collision will be 0.167m/s
Answer:
M1 V1 = M1 V2 + M2 V3 conservation of momentum
V2 = (M1 V1 - M2 V3) / M1 where V2 = speed of M1 after impact
V2 = (3 * 9 - 1.5 * 5) / 9 = (27 - 7.5) / 9 = 2.17 m/s
Note: All speeds are in the same direction and have the same sign
False, we lack cell walls whereas they have cell walls.
Answer:
wave A
Explanation:
shortest wavelength carry the most energy
more energy in a wave, the higher its frequency. The lower the frequency is, the less energy in the wave.
