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
By definition, the kinetic energy is given by:
K = (1/2) * m * v ^ 2
where
m = mass
v = speed
We must then find the speed of both objects:
blue puck
v = root ((0) ^ 2 + (- 3) ^ 2) = 3
gold puck
v = root ((12) ^ 2 + (- 5) ^ 2) = 13
Then, the kinetic energy of the system will be:
K = (1/2) * m1 * v1 ^ 2 + (1/2) * m2 * v2 ^ 2
K = (1/2) * (4) * (3 ^ 2) + (1/2) * (6) * (13 ^ 2)
K = <span>
525</span> J
answer
The kinetic energy of the system is<span>
<span>525 </span></span>J
Answer:
a = 15.1 g
Explanation:
The relation between mass and acceleration is given by :

If a₁ = 0.80g, m₁ = 1510 kg, m₂ = 80 kg, we need to find a₂
So,

So, the car's acceleration would be 15.1 g.
Answer:
Terminal velocity of object = 12.58 m/s
Explanation:
We know that the terminal velocity is attained when drag force and gravitational force are of the same magnitude.
Gravitational force = mg = 80 * 9.8 = 784 N
Drag force = 
Equating both, we have

So v = 12.58 m/s or v = -15.58 m/s ( not possible)
So terminal velocity of object = 12.58 m/s
Answer:
14160 kg/m^3
Explanation:
First of all, we need to find the volume of the cylinder.
The volume of the cylinder is given by:

where:
is the radius
is the height
Substituting, we find

And the density is given by

where m = 1 kg is the mass. Substituting, we find
