Resultant force= (2*6^2)^(1/2)
=8.5m/s
answer is B.
Answer:
15
Explanation:
mass, M = 5Kg
horizontal force, F_h = 40N
acceleration, a =5 m/s^2
frictional force, F_f =?
net force = ma
net force = F_h - F_f = 40N - F_f
40 - F_f = 5 x 5
- F_f = 25 - 40
multiply both side by -1
F_f = 40 - 25 = 15
the frictional force is 15N
Well, im pretty sure that when we do touch eachother, the atoms themselves are touching. idk if this is what ur looking for but hope this helps.
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
