Answer:
19.6 N of torque. The 2kg load is being affected by acceleration due to gravity which is 9.8 m/s^s
Explanation:
2×9.8=19.6
Velocity of submarine A is vs = 11.0m/s
frequency emitted by submarine A. F = 55.273 × 10∧3HZ
Velocity of submarine B = vO = 3.00m/s
The given equation is
f' = ((V + vO) ((v - vS)) × f
The observer on submarine detects the frequency f'.
The sign of vO should be positive as the observer of submarine B is moving away from the source of submarine A.
The speed of the sound used in seawater is 1533m/s
The frequency which is detected by submarine B is
fo = fs (V -vO/ v +vs)
= 53.273 × 10∧3hz) ((1533 m/s - 4.5 m/s)/ (1533 m/s +11 m/s)
fo = 5408 HZ
Acceleration, a = (v - u)/t
where v is the final velocity, u is the initial velocity, and t is the time.
This formula on a velocity time graph represents the slope of the graph.
Resolution in this sentence refers to a solution to the problem
