Answer:
Same reading.
Explanation:
Assume that after the string breaks the ball falls through the liquid with constant speed. If the mass of the bucket and the liquid is 1.20 kg, and the mass of the ball is 0.150 kg,
A.) Before the string break, the total weight = weight of the can + weight of the water.
According to Archimedes' Principle which state that: “A body immersed in a liquid loses weight by an amount equal to the weight of the liquid displaced.” Archimedes principle also states that: “When a body is immersed in a liquid, an upward thrust, equal to the weight of the liquid displaced, acts on it
B.) After the string break.
The scale will have the same reading as before the string break.
Answer:
v = 0.84m/s, v(max)= 0.997m/s
Explanation:
Initial work done by the spring, where c is the compression = 0.28m:

Work lost to friction:

Energy:

(a) Solve for v:

(b) Solve
for x:

if:



Answer: 
Explanation:
Given
Length of plank is 1.6 m
Force
is applied on the left side of plank
Force
is applied 43 cm from the left end O.
Mass of the plank is 
for equilibrium
Net torque must be zero. Taking torque about left side of the plank

Net vertical force must be zero on the plank
