Answer:
f" = 40779.61 Hz
Explanation:
From the question, we see that the bat is the source of the sound wave and is initially at rest and the object is in motion as the observer, thus;
from the Doppler effect equation, we can calculate the initial observed frequency as:
f' = f(1 - (v_o/v))
We are given;
f = 46.2 kHz = 46200 Hz
v_o = 21.8 m/s
v is speed of sound = 343 m/s
Thus;
f' = 46200(1 - (21/343))
f' = 43371.4285 Hz
In the second stage, we see that the bat is now a stationary observer while the object is now the moving source;
Thus, from doppler effect again but this time with the source going away from the obsever, the new observed frequency is;
f" = f'/(1 + (v_o/v))
f" = 43371.4285/(1 + (21.8/343))
f" = 40779.61 Hz
Definitely not A or B. It really depends on the size of the boulder... It should be kilograms, unless it's a extremely huge boulder.
Answer:
In one rotation, the large wheel turns 4m.
Explanation:
The given values are:
Input distance,
= 0.64 m
Mechanical advantage,
= 0.16
As we know,
⇒ 
On putting the values, we get
⇒ 
⇒ 
Answer: Magnitude of the force exerted on the egg by the ground is 9.2N
Explanation:
Given the following :
Mass of egg (m) = 150g = 0.15kg
Height(h) from which egg is dropped = 3m
velocity of egg before hitting the ground (u) = 4.4m/s
Final velocity of egg (V) = 0
Time taken (t) = 0.072s
Magnitude of the force exerted on the egg by the ground can be found by applying Newton's 2nd law:
Momentum = mass × velocity
From Newton's second law:
Force = mass × change in Velocity with time ;
That is
F = m * ΔV / t
Inputting our values
F = 0.15 * (4.4 - 0) / 0.072
F = 0.15 × (4.4 / 0.072)
F = 0.15 × 61.11
F = 9.16N
F = 9.2N
