Answer:
When balloon moves in the downward direction two forces acts on it.
i) Force exerted by air in the upward direction
ii) Weight
According to newton’s second law of motion:
Sum of forces = Ma
W – F = Ma
Mg – F = Ma …….. (i)
when some of the mass m is dropped and balloon is moving in upward direction with acceleration a/2 then,
F – W = (M-m)a/2
F – (M-m)g = (M-m)a/2
F – Mg + mg = Ma/2 – ma/2 ….. (ii)
Adding equation (i) and (ii)
mg = M(3a/2) – ma/2
m(g + a/2) = M(3a/2)
m = M(3a/2)/(g + a/2)
Answer:
A
Explanation:
Newton third law of motion state that action and reaction are equal and opposite
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>
Answer:
w = 25.05 rad / s
, α = 0.7807 rad / s²
, θ = 1972.75
Explanation:
This is a kinematic rotation exercise, let's start by looking for the acceleration when the engine is off
θ = w₀ t - ½ α t²
α = (w₀t - θ) 2/t²
let's reduce the magnitudes to the SI system
w₀ = 530 rev / min (2pi rad / 1 rev) (1 min / 60 s) = 55.5 rad / s
θ = 250 rev (2pi rad / 1 rev) = 1570.8 rad
let's calculate the angular acceleration
α = (55.5 39 - 1570.8) 2/39²
α = 0.7807 rad / s²
having the acceleration we can calculate the final speed
w = w₀ - ∝ t
w = 55.5 - 0.7807 39
w = 25.05 rad / s
the time to stop w = 0
0 = wo - alpha t
t = wo / alpha
t = 55.5 / 0.7807
t = 71.09 s
the angle traveled
w² = w₀⁹ - 2 α θ
w = 0
θ = w₀² / 2α
let's calculate
θ = 55.5 2 / (2 0.7807)
θ = 1972.75
