Because the elevator moves at a constant speed, it's in equilibrium and the net force acting on it is zero. Then the tension in the cable exactly equals the magnitude of the elevator's weight, which is
(3000 kg) (9.80 m/s²) = 29,400 N
First method
initial distance = 16m
final distance= 43 m
total distance covered= final -initial
=43m -16m
=27m
Second method
Si= 16m
Sf =43 m
t= 12 s
first we will find V
V = (Sf-Si)/ t
V =( 43- 16)/ 12
V = 27/12 ⇒ V= 9/4
V= distance / time
distance= V×time
distance = (9/4) ×12
distance =27
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