I think it might be heat energy. light transforms into heat energy
Answer:
(a) A = 0.650 m
(b) f = 1.3368 Hz
(c) E = 17.1416 J
(d) K = 11.8835 J
U = 5.2581 J
Explanation:
Given
m = 1.15 kg
x = 0.650 cos (8.40t)
(a) the amplitude,
A = 0.650 m
(b) the frequency,
if we know that
ω = 2πf = 8.40 ⇒ f = 8.40 / (2π)
⇒ f = 1.3368 Hz
(c) the total energy,
we use the formula
E = m*ω²*A² / 2
⇒ E = (1.15)(8.40)²(0.650)² / 2
⇒ E = 17.1416 J
(d) the kinetic energy and potential energy when x = 0.360 m.
We use the formulas
K = (1/2)*m*ω²*(A² - x²) (the kinetic energy)
and
U = (1/2)*m*ω²*x² (the potential energy)
then
K = (1/2)*(1.15)*(8.40)²*((0.650)² - (0.360)²)
⇒ K = 11.8835 J
U = (1/2)*(1.15)*(8.40)²*(0.360)²
⇒ U = 5.2581 J
Answer:
( a ) The specific volume by ideal gas equation = 0.02632 
% Error = 20.75 %
(b) The value of specific volume From the generalized compressibility chart = 0.0142
% Error = - 34.85 %
Explanation:
Pressure = 1 M pa
Temperature = 50 °c = 323 K
Gas constant ( R ) for refrigerant = 81.49 
(a). From ideal gas equation P V = m R T ---------- (1)
⇒ 
= 
⇒ Here = Specific volume = v
⇒ v =
Put all the values in the above formula we get
⇒ v = 
×81.49
⇒ v = 0.02632
This is the specific volume by ideal gas equation.
Actual value = 0.021796
Error = 0.02632 - 0.021796 = 0.004524
% Error = 
× 100
% Error = 20.75 %
(b). From the generalized compressibility chart the value of specific volume
= v = 0.0142
The actual value = 0.021796
Error = 0.0142 - 0.021796 =
% Error = 
× 100
% Error = - 34.85 %
Answer:
0.83 ω
Explanation:
mass of flywheel, m = M
initial angular velocity of the flywheel, ω = ωo
mass of another flywheel, m' = M/5
radius of both the flywheels = R
let the final angular velocity of the system is ω'
Moment of inertia of the first flywheel , I = 0.5 MR²
Moment of inertia of the second flywheel, I' = 0.5 x M/5 x R² = 0.1 MR²
use the conservation of angular momentum as no external torque is applied on the system.
I x ω = ( I + I') x ω'
0.5 x MR² x ωo = (0.5 MR² + 0.1 MR²) x ω'
0.5 x MR² x ωo = 0.6 MR² x ω'
ω' = 0.83 ω
Thus, the final angular velocity of the system of flywheels is 0.83 ω.
Answer:
the correct answer is E
A graph of the cart's maximum speed squared as a function of x^3
Explanation:
For this exercise let's use Newton's second law
F = m a
force has the form
F = k x²
and acceleration is related to velocity
a = dv / dt
Let's use the chain rule or L'Hospital
a = dv /dx dx/dt
a = dv /dx v
let's substitute
k x² = m v dv / dx
k /m x² dx = v dv
we integrate
k /m x³ /3 = v² / 2
v² = (2k /3m) x³
This is the expression for the variation of the speed as a function of the position, to make a linear graph realism the changes of variable
y = v²
x´ = x³
y = (2k/3m) x´
if we graph y vs x 'we have a linear graph whose slope is
m = 2k / 3m
By reviewing the different answers, the correct answer is E
