Which box has a larger upward force (normal force) acting on it?
1 answer:
You might be interested in
Answer:
(A) V = 9.89m/s
(B) U = -2.50m/s
(C) ΔK.E = –377047J
(D) ΔK.E = –257750J
Explanation:
The full solution can be found in the attachment below. The east has been chosen as the direction for positivity.
This problem involves the principle of momentum conservation. This principle states that the total momentum before collision is equal to the total momentum after collision. This problem is an inelastic kind of collision for which the momentum is conserved but the kinetic energy is not. The kinetic energy after collision is always lesser than that before collision. The balance is converted into heat by friction, and also sound energy.
See attachment below for full solution.
Answer:
Explanation:
From the given information:
The coordinate axis is situated in the east and north direction.
So, the north will be the y-axis and the east will be the x-axis
Similarly, the velocity of the plane in regard to the air in the coordinate system will be
where:
= velocity of the plane in regard to the air
v = velocity
θ = angle of inclination of the plane with respect to the horizontal
replacing v = 180 km/ and θ = 20° in above equation, then:
The velocity of the airplane in the coordinate system as:
where;
= velocity of the airplane
= velocity
∅ = angle of inclination with regard to the base axis;
Then; replacing = 150 km/h and ∅ = 30°
Therefore, the velocity of the plane in the system is :
--- (1)
( 150 cos 30° - 180 cos 20°)i + ( 150 sin 30° - 180sin 20°)j
(-39.24 km/h)i + (13.44 km/h) j
The magnitude is:
= 41.48 km/h
The airplane is moving at an angle of the inverse tangent to the abscissa and ordinate.
The angle of motion is:
tan θ = 39.24/13.44
tan θ = 2.9
θ =
θ = 70.97°
The angle of motion is 70.97° from west of north with a velocity of 41.48 km/h.