3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Answer:
the moon lacks a sizeable iron core
Explanation:
Answer:
Any incident ray traveling parallel to the principal axis of a converging lens will refract through the lens and travel through the focal point on the opposite side of the lens. ... These rays of light will refract when they enter the lens and refract when they leave the lens.
Hope this helps...
Answer: 3 m.
Explanation:
Neglecting the mass of the seesaw, in order the seesaw to be balanced, the sum of the torques created by gravity acting on both children must be 0.
As we are asked to locate Jack at some distance from the fulcrum, we can take torques regarding the fulcrum, which is located at just in the middle of the length of the seesaw.
If we choose the counterclockwise direction as positive, we can write the torque equation as follows (assuming that Jill sits at the left end of the seesaw):
mJill* 5m -mJack* d = 0
60 kg*5 m -100 kg* d =0
Solving for d:
d = 3 m.
