Answer:
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Explanation:
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
Answer:
3.88m/s
Explanation:
Using the law of conservation of momentum
m1u1+m2u2 = (m1+m2)v
m1 and m2 are the masses
u1 and 2 are the initial velocities
v is the final velocity
Given
m1 = 64kg
u1 = 4.2m/s
m2 = 25kg
u2 = 3.2m/s
Required
Final velocity v
Substitute the given values into the formula
64(4.2)+25(3.2) = (65+25)v
268.8+80 = 90v
348.8 = 90v
v = 348.8/90
v = 3.88m/s
Hence the velocity of the kayak after the swimmer jumps off is 3.88m/s