Answer
given,
mass of ball, m = 57.5 g = 0.0575 kg
velocity of ball northward,v = 26.7 m/s
mass of racket, M = 331 g = 0.331 Kg
velocity of the ball after collision,v' = 29.5 m/s
a) momentum of ball before collision
P₁ = m v
P₁ = 0.0575 x 26.7
P₁ = 1.535 kg.m/s
b) momentum of ball after collision
P₂ = m v'
P₂ = 0.0575 x (-29.5)
P₂ = -1.696 kg.m/s
c) change in momentum
Δ P = P₂ - P₁
Δ P = -1.696 -1.535
Δ P = -3.231 kg.m/s
d) using conservation of momentum
initial speed of racket = 0 m/s
M u + m v = Mu' + m v
M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)
0.331 u' = 3.232
u' = 9.76 m/s
change in velocity of the racket is equal to 9.76 m/s
Answer:

Explanation:
Given that
x= 150 ft

y= 14 ft
From the diagram

When ,x= 150 ft and y= 14 ft


z=150.74 ft

By differentiating with respect to time t


Here x is constant that is why


Now by putting the values in the above equation we get



Therefore the distance between balloon and observer increasing with 0.65 ft/s.
Answer:
The work done on the box is 80 J.
Explanation:
Given that,
Weight of box = 40 N
Distance = 2 meter
We need to calculate the work done
Using formula of work done


Where, x = distance
mg = weight
Put the value into the formula



Hence, The work done on the box is 80 J.
When a crest-trough meet the interference produced will be destructive in nature hence they both will cancel out and the amplitude produced will be equal to zero hence the loudness will reduce to zero.