Answer:
1.01 × 10⁵ Pa
Explanation:
At the surface, atmospheric pressure is 1.013 × 10⁵ Pa.
We need to find the total pressure on the air in the lungs of a person to a depth of 1 meter.
Pressure at a depth is given by :
Where
is the density of air,
So,
Total pressure, P = Atmospheric pressure + 12 Pa
= 1.013 × 10⁵ Pa + 12 Pa
= 1.01 × 10⁵ Pa
Hence, the total pressure is 1.01 × 10⁵ Pa.
Answer:
In an elastic collision:
- There is no external net force acting. Thus, Momentum before and after collision is equal. Momentum remains conserved.
- Total energy always remains conserved as energy cannot be created nor destroyed. It can change from one form to another.
- There is no lost due to friction in elastic collision. So the kinetic energy is also conserved.
- Velocities may change after collision. If the masses are equal, the velocities interchange.
When one object is stationary:
Final velocity of object 1:
v₁ = (m₁ - m₂)u₁/(m₁ +m₂)
Final velocity of object 2:
v₂ = (2 m₁ u₁)/(m₁+m₂) =
- Objects do not stick together in elastic collision. They stick together in inelastic collision.
- One object may be stationary before the elastic collision.
Thus, conditions for an elastic collision:
- Energy is conserved.
- Velocities may change.
- Momentum is conserved.
- Kinetic energy is conserved.
- One object may be stationary before the elastic collision.
Answer:
Explanation:
We shall apply law of conservation of momentum in space to know the velocity of combination after the impact
m₁v₁ = m₂v₂
.1 x 4 = ( 1 + .1 ) v₂
v₂ = .3636 m /s
1 )
Kinetic energy of the combination
= 1/2 x 1.1 x ( .3636)²
= 7.3 x 10⁻² J
2 )
Initial kinetic energy of the system
= 1/2 x 0.1 x 4²
= 0.8 J
Final kinetic energy of the system = 7.3 x 10⁻²
Loss of energy = .8 - .073
= .727 J
This energy was converted into internal energy of the system .
3 )
increase in entropy = dQ / T
Here dQ = .727 J
T = 300 ( Constant )
dQ / T = 2.42 X 10⁻³ J/K
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