Answer:
Given, Apparent weight(W₂)=4.2N
Weight of liquid displaced (u)=2.5N
Let weight of body in air = W₁
Solution,
U=W₁-W₂
W₁=4.2=2.5=6.7N
∴Weight of body in air is 6.7N
Answer:
U₂ = 20 J
KE₂ = 40 J
v= 12.64 m/s
Explanation:
Given that
H= 12 m
m = 0.5 kg
h= 4 m
The potential energy at position 1
U₁ = m g H
U₁ = 0.5 x 10 x 12 ( take g= 10 m/s²)
U₁ = 60 J
The potential energy at position 2
U₂ = m g h
U ₂= 0.5 x 10 x 4 ( take g= 10 m/s²)
U₂ = 20 J
The kinetic energy at position 1
KE= 0
The kinetic energy at position 2
KE= 1/2 m V²
From energy conservation
U₁+KE₁=U₂+KE₂
By putting the values
60 - 20 = KE₂
KE₂ = 40 J
lets take final velocity is v m/s
KE₂= 1/2 m v²
By putting the values
40 = 1/2 x 0.5 x v²
160 = v²
v= 12.64 m/s
Answer:0.43
Explanation:
Given
mass of car 
Speed of car 
Distance traveled before coming to halt 
Let
the coefficient of friction
Maximum deceleration road can provide during motion is

using 



Answer:
340.67 kgm²/s
Explanation:
R = Radius of merry-go-round = 1.9 m
I = Moment of inertia = 209 kgm²
= Initial angular velocity = 1.63 rad/s
m = Mass of person = 73 kg
v = Velocity = 4.8 m/s
Initial angular momentum is given by

The initial angular momentum of the merry-go-round is 340.67 kgm²/s