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:
1.5 m
Explanation:
H = actual height of the superhero = ?
H₀ = height of the superhero as observed = 1.73 m
v = speed of the superhero = 0.50 c
Using the equation

Inserting the values

H = 1.5 m
Answer:
a) y = 16.51 [m]
b) t = 1.83 [s]
Explanation:
To solve this problem we must use two kinematics equations, the first to determine the height to which the ball reaches, and the second equation to determine how long it lasts in the air.

where:
Vf = final velocity = 0
Vi = initial velocity = 18 [m/s]
g = gravity acceleration = 9.81[m/s^2]
t = time [s]
Note: the negative sign of the Equation indicates that the acceleration of gravity acts in the opposite direction to the movement of the ball. The final velocity is zero, since the ball reaches its maximum altitude when the velocity is zero.
Now replacing:
0 = (18)^2 - (2*9.81*y)
y = 16.51 [m]
b)

0 = 18 - (9.81*t)
t = 1.83 [s]
Explanation:
a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:


From Eqn(2), we see that

so using Eqn(3) on Eqn(1), we get

Solving for the acceleration, we see that


b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

Since the crate started from rest,
Thus our equation reduces to



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