Answer:
Explanation:
given,
In first case Volume remains constant.
Work done in the first case is zero.
In Second case Volume change
V₁ = 0.2 m³
V₂ = 0.11 m³
Pressure, P = 5.5 x 10⁵ Pa
Work done = Pressure x change in volume
W = P ΔV
Hence, Work done when volume changes is equal to
It’s a force which acts on an object without coming physically in contract with it.
This question is a critical question. as we all know, when energy is added to any state of water, the particles move faster. and when energy is taken away from any state of water, the particles reduce speed. same with the particles of air. when energy is added; they move faster. when energy is removed; they move slower. so the answer is they move faster
the answer of this question is helium
Answer:
1.04 s
Explanation:
The computation is shown below:
As we know that
t = t' × 1 ÷ (√(1 - (v/c)^2)
here
v = 0.5c
t = 1.20 -s
So,
1.20 = t' × 1 ÷ (√(1 - (0.5/c)^2)
1.20 = t' × 1 ÷ (√(1 - (0.5)^2)
1.20 = t' ÷ √0.75
1.20 = t' ÷ 0.866
t' = 0.866 × 1.20
= 1.04 s
The above formula should be applied
