The work done by the elephant to lift one log is the force multiplied by the height at which the log has been lifted:
And so, the total work done to lift the 7 logs is 7 times the work done to lift each log:
Answer:
a. 37.7 kgm/s b. 0.94 m/s c. -528.85 J
Explanation:
a. The initial momentum of block 1 of m₁ = 1.30 kg with speed v₁ = 29.0 m/s is p₁ = m₁v₁ = 1.30 kg × 29.0 m/s = 37.7 kgm/s
The initial momentum of block 2 of m₁ = 39.0 kg with speed v₂ = 0 m/s since it is initially at rest is p₁ = m₁v₁ = 39.0 kg × 0 m/s = 0 kgm/s
So, the magnitude of the total initial momentum of the two-block system = (37.7 + 0) kgm/s = 37.7 kgm/s
b. Since the blocks stick together after the collision, their final momentum is p₂ = (m₁ + m₂)v where v is the final speed of the two-block system.
p₂ = (1.3 + 39.0)v = 40.3v
From the principle of conservation of momentum,
p₁ = p₂
37.7 kgm/s = 40.3v
v = 37.7/40.3 = 0.94 m/s
So the final velocity of the two-block system is 0.94 m/s
c. The change in kinetic energy of the two-block system is ΔK = K₂ - K₁ where K₂ = final kinetic energy of the two-block system = 1/2(m₁ + m₂)v² and K₁ = final kinetic energy of the two-block system = 1/2m₁v₁²
So, ΔK = K₂ - K₁ = 1/2(m₁ + m₂)v² - 1/2m₁v₁² = 1/2(1.3 + 39.0) × 0.94² - 1/2 × 1.3 × 29.0² = 17.805 J - 546.65 J = -528.845 J ≅ -528.85 J
Lets say sphere 1 has a charge of 12 + and sphere 2 has a charge of 0 +. After they are touched Sphere 1 becomes 6 + and sphere 2 6 +. So 6 - 12 = a change of 6 -, while 6 - 0 = a change of 6 + Therfore,
Answer: The sign of the charge change / transfered are opposites.
None of the above. 1 mL= 1 cubic centimeter
mm is the smallest.
