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
Answer:
<em>The internal resistance of an ideal ammeter will be zero since it should allow current to pass through it. Voltmeter measures the potential difference, it is connected in parallel. .</em>
Explanation:
<h3>
<em>I </em><em>hope</em><em> this</em><em> helps</em><em>!</em></h3>
Answer:
90 J
Explanation:
W=fd
W=(75)(1.2)
W= 90 J
F = mass x acceleration
We have mass = 200kg
and acceleration = 3 m/s^2 so...
F = (200)(3)
F = 600 N
