Answer:
0.102 m
Explanation:
k = spring constant of the spring = 125 N/m
m = mass of the block attached to the spring = 650 g = 0.650 kg
x = maximum extension of the spring
h = height dropped by the block = x
Using conservation of energy
Spring potential energy gained = Gravitational potential energy lost
(0.5) k x² = mgh
(0.5) k x² = mgx
(0.5) (125) x = (0.650) (9.8)
x = 0.102 m
Answer: Remain unchanged
Explanation:
The boat with water barrel overboard floats in swimming pool when weight of the water displaced by the boat is equal to the buoyant force acting on the boat.
When the water in the barrel is poured overboard, the level of the swimming pool level would remain unchanged as the weight of the boat with the water and barrel would remain unchanged ( as the density and volume of the whole system remains same) and hence, the weight of the water (of the swimming pool) displaced by the boat would remain same.
A boat loaded with a barrel of water floats in a swimming pool. When the water in the barrel is poured overboard, the swimming pool level will <u>remain unchanged. </u>
Well, it happens a few weeks ahead, then for a total of 3 hours and 40 minutes.
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Answer:
44.6 N
Explanation:
Draw a free body diagram of the block. There are four forces on the block:
Weight force mg pulling down,
Normal force N pushing up,
Friction force Nμ pushing left,
and applied force F pulling right 30° above horizontal.
Sum of forces in the y direction:
∑F = ma
N + F sin 30° − mg = 0
N = mg − F sin 30°
Sum of forces in the x direction:
∑F = ma
F cos 30° − Nμ = 0
F cos 30° = Nμ
N = F cos 30° / μ
Substitute:
mg − F sin 30° = F cos 30° / μ
mg = F sin 30° + (F cos 30° / μ)
Plug in values:
mg = 20 N sin 30° + (20 N cos 30° / 0.5)
mg = 44.6 N
