Answer:
The work done by gravity is 784 J.
Explanation:
Given:
Mass of the block is,
Height to which it is raised is,
Acceleration due to gravity is,
Now, work done by gravity is equal to the product of force of gravity and the distance moved in the direction of gravity. So,
Force of gravity is given as the product of mass and acceleration due to gravity.
. Now,
Therefore, the work done by gravity is 784 J.
Answer:
a. The volume V₁ and V₂
b. The case that involves the greatest momentum change = Case B
c. The case that involves the greatest impulse = Case B
d. b. The case that involves the greatest force = Case B
Explanation:
Here we have
Case A: V₁ = 150-mL, v₁ = 8 m/s
Case B: V₂ = 600-mL, v₁ = 8 m/s
a. The variable that is different for the two cases is the volume V₁ and V₂
b. The momentum change is by the following relation;
ΔM₁ = Mass, m × Δv₁
The mass of the balloon are;
Δv₁ = Change in velocity = Final velocity - Initial velocity
Mass = Density × Volume
Density of water = 0.997 g/mL
Case A, mass = 150 × 0.997 = 149.55 g
Case B, mass = 600 × 0.997 = 598.2 g
The momentum change is;
Case A: Mass, m × Δv₁ = 149.55 g/1000 × 8 m/s = 1.1964 g·m/s
Case B: Mass, m × Δv₁ = 598.2/1000 × 8 = 4.7856 g·m/s
Therefore Case B has the greatest momentum change
The case that has the gretest momentum change = Case B
c. The momentum change = impulse therefore Case B involves the greatest impulse
d. Here we have;
Impulse = Momentum change = × Δt = mΔV
∴ = m·ΔV/Δt
∴ For Case A = 149.55×8/Δt = 1196.4/Δt N
For Case B = 598.2×8/Δt = 4785.6/Δt
Where Δt is the same for Case A and Case B, for Case B >>
for Case B
Therefore, Case B involves the greatest force.