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
The product of √30 and √610 is 10√183.
√30 = √(2×3×5)
and √610 = √(2×5×61
Since 61 can't be factorised further.
Therefore, the value of √30×√610 is
= √(2×3×5×2×5×61)
= 2×5×√(3×61)
=10√183
C the atmosphere provides warmth
Answer:
Explanation:
v = I / (n q A)
where A is the cross sectional area (you are given diameter, so A = (pi/4) * d^2), n is the number of charge carriers in a volume, and q is the charge of each carrier (-1.60x10^-19). I is the current, in Amps.
You will need to either convert your diameter from inches to centimeters or you will need to convert your volume in cc to cubic inches. I'd convert diameter.
