Answer:
Explanation:
From the Question we are told that
Mass of A and B is 60kg
Speed of A=2m/s
Speed of B=1m/s
Mass of bag =5kg
Generally the momentum of the astronaut A and bag is mathematically given as
Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag
Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1
Therefore
Answer:
34.17°C
Explanation:
Given:
mass of metal block = 125 g
initial temperature = 93.2°C
We know
..................(1)
Q= Quantity of heat
m = mass of the substance
c = specific heat capacity
c = 4.19 for H₂O in
= change in temperature
Now
The heat lost by metal = The heat gained by the metal
Heat lost by metal =
Heat gained by the water =
thus, we have
=
⇒
Therefore, the final temperature will be = 34.17°C
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