The acceleration at which all falling objects drop (neglecting air resistance) near the
1 answer:
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Answer:
D = 43 m
Explanation:
given,
initial velocity = 18 m/s
angle θ = 60°
total horizontal distance covered by the shell is
applying conservation of momentum in horizontal direction
m v₀ cos θ = m₁v₁ + m₂ v₂
m v₀ cos θ = 0.5 m v₂
v₂ = 2 v₀ cos θ.
distance covered by the shell from point of explosion
R' = v t
=
=
=
= R
total distance traveled by the shell is
D =
= 1.5 R
=
D =
= 42.9 ≅ 43 m
D = 43 m
Answer:
t = 17658 s = 294.3 min = 4.9 h
Explanation:
The general formula for power is:
where,
P = Power of the Motor = 50 KW = 50000 W
W = Work Done by Motor = Change in P.E of Water = mgh
g = acceleration due to gravity = 9.81 m/s²
m = mass of water = ρV
h = depth of water = 180 m
ρ = density of water = 1000 kg/m³
V = Volume of Water = 500 m³
t = time taken = ?
Therefore,
<u>t = 17658 s = 294.3 min = 4.9 h</u>