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Answer:
the frequency of the oscillation is 1.5 Hz
Explanation:
Given;
mass of the spring, m = 1500 kg
extention of the spring, x = 5 mm = 5 x 10⁻³ m
mass of the driver = 68 kg
The weight of the driver is calculated as;
F = mg
F = 68 x 9.8 = 666.4 N
The spring constant, k, is calculated as;
k = F/m
k = (666.4 N) / (5 x 10⁻³ m)
k = 133,280 N/m
The angular speed of the spring is calculated;

The frequency of the oscillation is calculated as;
ω = 2πf
f = ω / 2π
f = (9.426) / (2π)
f = 1.5 Hz
Therefore, the frequency of the oscillation is 1.5 Hz
Explanation:
equating the parameters into the formula, it's gonna be
= ½ × 60 × 20²
= ½ × 60 × 400
= ½ × 24000
K.E = 12000J
Answer:
F = 4000 N
Explanation:
given,
mass of rocket (M)= 5000 Kg
10 Kg gas burns at speed (m)= 4000 m/s
time = 10 s
average force = ?
at the end the rocket is at rest
by conservation of momentum
M v + m v' = 0
5000 x v - 10 x 4000 = 0
5000 v = 40000
v = 8 m/s
speed of rocket = 8 m/s
now,
we know
change in momentum = F x Δ t


F = 4000 N
Hence, the average force applied to the rocket is equal to F = 4000 N