A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wha
1 answer:
This question involves the conservation of energy. There are two energy in this case, potential energy and kinetic energy. Let's divid the energy into three status.
1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
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The final velocity is 5.87 m/s
<u>Explanation:</u>
Given-
mass, = 72 kg
speed, = 5.8 m/s
,
= 45 kg
,
= 12 m/s
Θ = 60°
Final velocity, v = ?
Applying the conservation of momentum:
X
+
X
= (
+
) v
72 X 5.8 + 45 X 12 X cos 60° = (72 + 45) v
v = 417.6 + 540 X
v = 417.6 +
v = 5.87 m/s
The final velocity is 5.87 m/s