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
No one expected violet & ultraviolet spectral lines to be shifted towards the red.
Answer:
yes independent of the sign or valve of Q
Explanation:
<span>Storm cells in a squall line typically move from the southwest to the northeast, and as the mature cells in the northeast begin to die off, new ones are formed at the opposite end to advance the line. The air in the southwest corner has strong vertical updrafts that allow new cells to grow and develop into thunderstorms.</span>
Answer:
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
Explanation:
The density of a liquid is inversely proportional to the volume (height) of object submerged in it.
High density liquid possess higher buoyant force preventing objects from submerging.
p ∝ 1/V ∝ 1/h
since V = Ah
pu/pw = hw/hu
pu = pwhw/hu
Where;
p = density
h = height submerged
pu and pw is the density of unknown liquid and water respectively
hu and hw is the height of object submerged in unknown liquid and water respectively
pw = 1000kg/m^3
hu = 4.6cm = 0.046m
hw = 5.8cm = 0.058m
Substituting the given values;
pu = 1000×0.058/0.046
pu = 1260.9kg/m^3
the density of the unknown liquid is 1260.9kg/m^3
