Answer:
695800 N/m^2 or Pa
Explanation:
Height of the water from the ground H = 71 m
Acceleration due to gravity g =9.8 m/s^2
density of water ρ= 1000 kg/m^3
The minimum output gauge pressure to make water reach height H
P= ρgH
= 1000×9.8×71= 695800 N/m^2 or Pa
The point in which it originates.
Answer:
8.91 J
Explanation:
mass, m = 8.20 kg
radius, r = 0.22 m
Moment of inertia of the shell, I = 2/3 mr^2
= 2/3 x 8.2 x 0.22 x 0.22 = 0.265 kgm^2
n = 6 revolutions
Angular displacement, θ = 6 x 2 x π = 37.68 rad
angular acceleration, α = 0.890 rad/s^2
initial angular velocity, ωo = 0 rad/s
Let the final angular velocity is ω.
Use third equation of motion
ω² = ωo² + 2αθ
ω² = 0 + 2 x 0.890 x 37.68
ω = 8.2 rad/s
Kinetic energy,

K = 0.5 x 0.265 x 8.2 x 8.2
K = 8.91 J
The force needed to accelerate a vehicle with a mass of 1000kg at a rate of 5m/s2 would be 5000