Here stress is parallel to the surface of the body. So it's a Shear stress.
Add all the sec. and all the meter's and then add the meter's and sec. together
The same braking force does work on these objects to slow them down. The work done is equal to their change in kinetic energy:
FΔx = 0.5mv²
F = force, Δx = distance traveled, m = mass, v = speed
Isolate Δx:
Δx = 0.5mv²/F
Calculate Δx for each object.
Object 1: m = 4.0kg, v = 2.0m/s
Δx = 0.5(4.0)(2.0)²/F = 8/F
Object 2: m = 1.0kg, v = 4.0m/s
Δx = 0.5(1.0)(4.0)²/F = 8/F
The two objects travel the same distance before stopping.
Answer:
0.48 m
Explanation:
I'm assuming that this takes place in an ideal situation, where we neglect a host of factors such as friction, weight of the spring and others
If the mass is hanging from equilibrium at 0.42 m above the floor, from the question, and it is then pulled 0.06 m below that particular position. This pulling is a means of adding more energy into the spring, when it is released, the weight compresses the spring and equals its distance (i.e, 0.06 m) above the height.
0.42 m + 0.06 m = 0.48 m
At the highest point thus, the height is 0.48 m above the ground.
Answer:increased pressure to increase solubility
Explanation:B on ed
