Same temperature and difference in air pressure
Answer:
v_f = 24.3 m / s
Explanation:
A) In this exercise there is no friction so energy is conserved.
Starting point. On the roof of the building
Em₀ = K + U = ½ m v₀² + m g y₀
Final point. On the floor
Em_f = K = ½ m v_f²
Emo = Em_g
½ m v₀² + m g y₀ = ½ m v_f²
v_f² = v₀² + 2 g y₀
let's calculate
v_f = √(10² + 2 9.8 25)
v_f = 24.3 m / s
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F = 52000 N
m = 1060 kg
a= F/m = 52000 N/1060 kg = 49.0566 m/s^2
Answer:
Correct answer: E total = 2,800 J
Explanation:
Given:
m = 4 kg the mass of the object
V = 20 m/s the speed (velocity) of the object
H = 50 m the height of the object above the surface
E total = ? J
The total energy of an object is equal to the sum of potential and kinetic energy
E total = Ep + Ek
Ep = m g H we take g = 10 m/s²
Ep = 4 · 10 · 50 = 2,000 J
Ek = m V² / 2
Ek = 4 · 20² / 2 = 2 · 400 = 800 J
E total = 2,000 + 800 = 2,800 J
E total = 2,800 J
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