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
Answer:
It would point up.
Explanation:
Since I am at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward, the north pole of the compass would also point towards the earth's geographic north magnetic pole, since all other compasses point toward there.
Since the compass is free to swivel in any direction, the compass would point up, since it is at the earth's geographic north magnetic pole, the place on the earth's surface that compasses point toward.
So, the compass would point up.
Disposing of hazardous waste in a labeled container
Answer:
A pedigree chart is a diagram that shows the occurrence and appearance of phenotypes of a particular gene or organism and its ancestors from one generation to the next, most commonly humans, show dogs, and race horses.
so ye
Explanation:
As we know that sphere roll without slipping so there is no loss of energy in this case
so here we can say that total energy is conserved
Initial Kinetic energy + initial potential energy = final kinetic energy + final potential energy

as we know that ball start from rest

height of the ball initially is given as


also we know that

also for pure rolling

also we know that


now plug in all data in above equation




So speed at the bottom of the inclined plane will be 29 rad/s