Answer:
(a)2.7 m/s
(b) 5.52 m/s
Explanation:
The total of the system would be conserved as no external force is acting on it.
Initial momentum = final momentum
⇒(4.30 g × 943 m/s) + (730 g × 0) = (4.30 g × 484 m/s) + (730 g × v)
⇒ 730 ×v = (4054.9 - 2081.2) =1973.7
⇒v=2.7 m/s
Thus, the resulting speed of the block is 2.7 m/s.
(b) since, the momentum is conserved, the speed of the bullet-block center of mass would be constant.

Thus, the speed of the bullet-block center of mass is 5.52 m/s.
The closer to the equator, the hotter the climate will be.
Answer:
10 newtons, because the gravitaional force willbe stronger the closer it gets.
Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
Answer:
total momentum = 8.42 kgm/s
velocity of the first cart is 3.660 m/s
Explanation:
Given data
mass m1 = 2.3 kg
mass m2 = 1.5 kg
final velocity V2 = 4.9 m/s
final velocity V3 = - 1.9 m/s
to find out
total momentum and velocity of the first cart
solution
we know mass and final velocty
and initial velocity of second cart V1 = 0
so now we can calculate total momentum that is m1 v2 + m2 v2
total momentum = 2.3 ×4.9 + 1.5 ×(-1.9)
total momentum = 8.42 kgm/s
and
conservation of momentum is
m1 V + m2 v1 = m1 v2 + m2 v3
put all value and find V
2.3 V + 1.5 ( 0) = 2.3 ( 4.9 ) + 1.5 ( -1.9)
V = 8.42 / 2.3
V = 3.660 m/s
so velocity of the first cart is 3.660 m/s