To solve this problem, we should recall the law of
conservation of energy. That is, the heat lost by the aluminium must be equal
to the heat gained by the cold water. This is expressed in change in enthalpies
therefore:
- ΔH aluminium = ΔH water
where ΔH = m Cp (T2 – T1)
The negative sign simply means heat is lost. Therefore we
calculate for the mass of water (m):
- 0.5 (900) (20 – 200) = m (4186) (20 – 0)
m = 0.9675 kg
Using same mass of water and initial temperature, the final
temperature T of a 1.0 kg aluminium block is:
- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)
- 900 T + 180,000 = 4050 T
4950 T = 180,000
T = 36.36°C
The final temperature of the water and block is 36.36°C
Answer:
I₁ > I₃ > I₂
Explanation:
Taking the pic shown, we have
m₁ = 10m₀
m₂ = 2m₀
m₃ = m₀
r₁ = r₀
r₂ = 2r₀
r₃ = 3r₀
We apply the formula
I = mr²
then
I₁ = m₁r₁² = (10m₀)(r₀)² = 10m₀r₀²
I₂ = m₂r₂² = (2m₀)(2r₀)² = 8m₀r₀²
I₃ = m₃r₃² = (m₀)(3r₀)² = 9m₀r₀²
finally we have
I₁ > I₃ > I₂
Complete Question
The angular speed of an automobile engine is increased at a constant rate from 1120 rev/min to 2560 rev/min in 13.8 s.
(a) What is its angular acceleration in revolutions per minute-squared
(b) How many revolutions does the engine make during this 20 s interval?
rev
Answer:
a

b

Explanation:
From the question we are told that
The initial angular speed is 
The angular speed after
is 
The time for revolution considered is
Generally the angular acceleration is mathematically represented as

=>
=> 
Generally the number of revolution made is
is mathematically represented as

=> 
=> 
Answer:the resistance decrease
Explanation:
2.27 mps repeating.
This is the last question ill ever answer here. Thanks for being the last.