Answer:
K_b = 78 J
Explanation:
For this exercise we can use the conservation of energy relations
starting point. Lowest of the trajectory
Em₀ = K = ½ mv²
final point. When it is at tea = 50º
Em_f = K + U
Em_f = ½ m v_b² + m g h
where h is the height from the lowest point
h = L - L cos 50
Em_f = ½ m v_b² + mg L (1 - cos50)
energy be conserve
Em₀ = Em_f
½ mv² = ½ m v_b² + mg L (1 - cos50)
K_b = ½ m v_b² + mg L (1 - cos50)
let's calculate
K_b = ½ 2.0 6.0² + 2.0 9.8 6.0 (1 - cos50)
K_b = 36 +42.0
K_b = 78 J
Filling balloons , for air balloons to float , can cool cylinder files and superconductors
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Explanation:
physical quantity is any physical property that can be qualified that,is, be measured using numbers e.g mass, amount of substance,time and length
Complete question:
A fireman of mass 80 kg slides down a pole. When he reaches the bottom, 4.2 m below his starting point, his speed is 2.2 m/s. By how much has thermal energy increased during his slide?
Answer:
The thermal energy increased by 3,099.2 J
Explanation:
Given;
mass of the fireman, m = 80 kg
initial position of the fireman, hi = 4.2 m
final speed, v = 2.2 m/s
The change in the thermal energy is calculated as;
ΔE + (K.Ef - K.Ei) + (Uf - Ui) = 0
where;
ΔE is the change in the thermal energy
K.Ef is the final kinetic energy
K.Ei is the initial kinetic energy
Uf is the final potential energy
Ui is the initial potential energy
