Answer:
maximum speed of the bananas is 18.8183 m/s
Explanation:
Given data
amplitude A = 23.195 cm
spring constant K = 15.2676 N/m
mass of the bananas m = 56.9816 kg
to find out
maximum speed of the bananas
solution
we know that radial oscillation frequency formula that is = √(K/A)
radial oscillation frequency = √(15.2676/23.195)
radial oscillation frequency is 0.8113125 rad/s
so maximum speed of the bananas = radial oscillation frequency × amplitude
maximum speed of the bananas = 0.8113125 × 23.195
maximum speed of the bananas is 18.8183 m/s
True, the measurement shown is a derived unit.
Answer:
C = 771.35 J/kg°C
Explanation:
Here, e consider the conservation of energy equation. The conservation of energy principle states that:
Heat Given by Metal Piece = Heat Absorbed by Water + Heat Absorbed by Container
Since,
Heat Given or Absorbed by a material = m C ΔT
Therefore,
m₁CΔT₁ = m₂CΔT₂ + m₃C₃ΔT₃
where,
m₁ = Mass of Metal Piece = 2.3 kg
C = Specific Heat of Metal = ?
ΔT₁ = Change in temperature of metal piece = 165°C - 18°C = 147°C
m₂ = Mass of Metal Container = 3.8 kg
ΔT₂ = Change in temperature of metal piece = 18°C - 15°C = 3°C
m₃ = Mass of Water = 20 kg
C₃ = Specific Heat of Water = 4200 J/kg°C
ΔT₃ = Change in temperature of water = 18°C - 15°C = 3°C
Therefore,
(2.3 kg)(C)(147°C) = (3.8 kg)(C)(3°C) + (20 kg)(4186 J/kg°C)(3°C)
C[(2.3 kg)(147°C) - (3.8 kg)(3°C)] = 252000 J
C = 252000 J/326.7 kg°C
<u>C = 771.35 J/kg°C</u>