Answer:
D. 18.60
Explanation:
By the law of conservation, the momentum is neither loss nor gained but instead transfered. When they crash into each other, and stick, they combine to create a total mass of 215 kg. Since the momentum is transfered, the two objects, combined, have a total momentum of 4000 kg-m/s. We know that momentum equals mass times velocity. You then divide 4000 by 215 and get approximately 18.6 m/s
Answer:
- 670 kg.m/s
Explanation:
Newton's third law states that to every action, there is equal and opposite reaction force. Since the force will be same but different in direction and acted in the same time then the impulses ( force multiply by time) of the two car be same in magnitude but different in direction - 670 kg.m/s
Answer:
θ = 4.78º
with respect to the vertical or 4.78 to the east - north
Explanation:
This is a velocity compound exercise since it is a vector quantity.
The plane takes a direction, the air blows to the west and the result must be to the north, let's use the Pythagorean theorem to find the speed
v_fly² = v_nort² + v_air²
v_nort² = v_fly² + - v_air²
Let's use trigonometry to find the direction of the plane
sin θ = v_air / v_fly
θ = sin⁻¹ (v_air / v_fly)
let's calculate
θ = sin⁻¹ (10/120)
θ = 4.78º
with respect to the vertical or 4.78 to the north-east
This problem is a piece o' cake, IF you know the formulas for both kinetic energy and momentum. So here they are:
Kinetic energy = (1/2) · (mass) · (speed²)
Momentum = (mass) · (speed)
So, now ... We know that
==> mass = 15 kg, and
==> kinetic energy = 30 Joules
Take those pieces of info and pluggum into the formula for kinetic energy:
Kinetic energy = (1/2) · (mass) · (speed²)
30 Joules = (1/2) · (15 kg) · (speed²)
60 Joules = (15 kg) · (speed²)
4 m²/s² = speed²
Speed = 2 m/s
THAT's all you need ! Now you can find momentum:
Momentum = (mass) · (speed)
Momentum = (15 kg) · (2 m/s)
<em>Momentum = 30 kg·m/s</em>
<em>(Notice that in this problem, although their units are different, the magnitude of the KE is equal to the magnitude of the momentum. When I saw this, I wondered whether that's always true. So I did a little more work, and I found out that it isn't ... it's a coincidence that's true for this problem and some others, but it's usually not true.)</em>
