Answer:
9.6 Ns
Explanation:
Note: From newton's second law of motion,
Impulse = change in momentum
I = m(v-u).................. Equation 1
Where I = impulse, m = mass of the ball, v = final velocity, u = initial velocity.
Given: m = 2.4 kg, v = 2.5 m/s, u = -1.5 m/s (rebounds)
Substitute into equation 1
I = 2.4[2.5-(-1.5)]
I = 2.4(2.5+1.5)
I = 2.4(4)
I = 9.6 Ns
Answer: part a: 19.62m
part b: 19.62 m/s
part a: 2.83 secs
Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence
u=0
a=9.81 m/s2
t=2 secs

s=h

Part b

Part c
now we have h=2*19.62=39.24

Given that,
Frequency emitted by the bat, f = 47.6 kHz
The speed off sound in air, v = 413 m/s
We need to find the wavelength detected by the bat. The speed of a wave is given by formula as follows :

or

So, the bat can detect small objects such as an insect whose size is approximately equal to the wavelength of the sound the bat makes i.e. 8.67 mm.
The frequency of the wave is 4 Hz