Answer:
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A hillside of course my friend
In fresh water sound waves travel at 1497m/s at 25 degrees, I'll assume that's the characteristics of the water.
If it's 0.01s then you need to divide the speed by 100 to get the, 14.97, however it gets there and back in that time so you need to halve it.
<u>7.485m</u>
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ
............1
put here value
= 1.75×
× 
= 0.175 v
and
area between air and puck is given by
Area =
area =
area = 7.85 ×
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 ×
v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 ×
v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds
Work done = Force x Distance
Force = 10 lb = 44.5 N
Work Done = 44.5 N x 15 m
= 667.5 N-m