Answer:
The correct answers are
(a) It decreases to 1/3 L
(ii) is (c) It is constant
Explanation:
to solve this, we list out the number of knowns and unknowns so as to determine the correct equation to solve the problem
The given variables are as follows
Initial volume V1 = 1L
V2 = Unknown
Initial Temperature T1 = 300K
let us assume that the balloon is perfectly elastic
At 300K the balloon is filled and it stretches to maintain 1 atmosphere
at 100K the content of the balloon cools reducing the excitement of the gas content which also reduces the pressure, however, the balloon being perfectly elastic, contracts to maintain the 1 atmospheric pressure, hence the answer to (ii) is (c) It is constant,
For (i) since we know that the pressure of the balloon is constant
by Charles Law V1/T1 =V2/T2
or V2 = (V1/T1)×T2 =× = × L = L/3 hence the correct answer to (i) is 1/3L
Answer:
0.3 m
Explanation:
Initially, the package has both gravitational potential energy and kinetic energy. The spring has elastic energy. After the package is brought to rest, all the energy is stored in the spring.
Initial energy = final energy
mgh + ½ mv² + ½ kx₁² = ½ kx₂²
Given:
m = 50 kg
g = 9.8 m/s²
h = 8 sin 20º m
v = 2 m/s
k = 30000 N/m
x₁ = 0.05 m
(50)(9.8)(8 sin 20) + ½ (50)(2)² + ½ (30000)(0.05)² = ½ (30000)x₂²
x₂ ≈ 0.314 m
So the spring is compressed 0.314 m from it's natural length. However, we're asked to find the additional deformation from the original 50mm.
x₂ − x₁
0.314 m − 0.05 m
0.264 m
Rounding to 1 sig-fig, the spring is compressed an additional 0.3 meters.
Answer: to avoid problems with water supply
Explanation: power plant needs water to run