A baseball with a mass of 0.3kg is thrown into the air with a speed of 50 m/s and is at a height of 10m above. What is the speed
1 answer:
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Answer:
W = ½ m v²
Explanation:
In this exercise we must solve it in parts, in a first part we use the conservation of the moment to find the speed after the separation
We define the system formed by the two parts of the rocket, therefore the forces during internal separation and the moment are conserved
initial instant. before separation
p₀ = m v
final attempt. after separation
= m /2 0 + m /2 v_{f}
p₀ = p_{f}
m v = m /2
v_{f}= 2 v
this is the speed of the second part of the ship
now we can use the relation of work and energy, which establishes that the work is initial to the variation of the kinetic energy of the body
initial energy
K₀ = ½ m v²
final energy
= ½ m/2 0 + ½ m/2 v_{f}²
K_{f} = ¼ m (2v)²
K_{f} = m v²
the expression for work is
W = ΔK = K_{f} - K₀
W = m v² - ½ m v²
W = ½ m v²
Answer:
T = 20.84°C
Explanation:
From the law of conservation of energy:
Heat Lost by Copper Block = Heat Gained by Aluminum Calorimeter + Heat Gained by Water
where,
= mass of copper = 227 g
= mass of water = 844 g
= mass of aluminum = 155 g
= specific heat capacity of calorimeter = 385 J/kg.°C
= specific heat capacity of water = 4200 J/kg.°C
= specific heat capacity of aluminum = 890 J/kg.°C
= change in temperature of copper = 283°C - T
= change in temperature of water = T - 14.6°C
= change in temperature of aluminum = T - 14.6°C
T = equilibrium temperature = ?
Therefore,
<u>T = 20.84°C</u>