Answer:
Explanation:
Hello,
Let's get the data for this question before proceeding to solve the problems.
Mass of flywheel = 40kg
Speed of flywheel = 590rpm
Diameter = 75cm , radius = diameter/ 2 = 75 / 2 = 37.5cm.
Time = 30s = 0.5 min
During the power off, the flywheel made 230 complete revolutions.
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + ω₂) / 2] × 0.5
But ∇θ = 230 revolutions
∇θ/t = (530 + ω₂) / 2
230 / 0.5 = (530 + ω₂) / 2
Solve for ω₂
460 = 295 + 0.5ω₂
ω₂ = 330rpm
a)
ω₂ = ω₁ + αt
but α = ?
α = (ω₂ - ω₁) / t
α = (330 - 590) / 0.5
α = -260 / 0.5
α = -520rev/min
b)
ω₂ = ω₁ + αt
0 = 590 +(-520)t
520t = 590
solve for t
t = 590 / 520
t = 1.13min
60 seconds = 1min
X seconds = 1.13min
x = (60 × 1.13) / 1
x = 68seconds
∇θ = [(ω₂ + ω₁) / 2] × t
∇θ = [(590 + 0) / 2] × 1.13
∇θ = 333.35 rev/min
Answer:
21s
Explanation:
Given parameters;
Radius = 10m
Speed or velocity = 3m/s
Unknown:
Period = ?
Solution:
To solve this problem, use the expression:
v = 
r is the radius
T is the unknown
Input the parameters and solve for T;
3 = 
62.84 = 3T
T = 21s
Answer:
That is true. They share atoms with each other. I hope this helps. Comment if you have any question
Explanation:
Answer:
Explanation:
a ) Entropy change dS = dQ/T
= mcdT /T
Integrating both sides
S₂ - S₁ = - mclnT₂ /T₁
= - .5 X 4200 ln (85+273) /( 20 + 273 )
.5 X 4200 ln 358/ 293
= - 417.6 J/K
Entropy change will be negative as heat is lost by the system .
b ) Sine there is no change in the temperature of air , This heat will enter air at temperature ( 20+ 273) K = 293 K
Heat entering air
= .5 x 4200 x 65
= 136500 J
Change in entropy
136500 / 293 ( room temperature is constant at 293k
= + 465.87 J/K
Entropy change will be positive as heat is gained by the system .
Total change in the entropy of the system (tea + air )
= +465.87 - 417.6
= 48.27 J/K
Entropy change will be negative as heat is lost by the system .
