Answer:
<em>The internal energy change is 330.01 J</em>
Explanation:
Given
the initial volume = 5.75 L
the final volume = 1.23 L
is the external pressure = 1.00 atm
q the heat energy removed = -128 J (since is removed from the system)
expansion against a constant external pressure is an example of an irreversible pathway, here pressure in is greater than pressure out and can be obtained thus;
W = -ΔV
W = -1.00 x(1.23 - 5.75)
W = -1.00 x -4.52
W = 4.52 L atm
converting to joules we have
W = 4.52 L atm x 101.33 J/ L atm = 458.01 J
The internal energy change during compression can be calculated thus;
ΔU = q + W
ΔU = -128 J + 458.01 J
ΔU = 330.01 J
Therefore the internal energy change is 330.01 J
Answer:
Explanation:
Given a parallel plate capacitor of
Area=A
Distance apart =d
Potential difference, =V
If the distance is reduce to d/2
What is p.d
We know that
Q=CV
Then,
V=Q/C
Then this shows that the voltage is inversely proportional to the capacitance
Therefore,
V∝1/C
So, VC=K
Now, the capacitance of a parallel plate capacitor is given as
C= εA/d
When the distance apart is d
Then,
C1=εA/d
When the distance is half d/2
C2= εA/(d/2)
C2= 2εA/d
Then, applying
VC=K
V1 is voltage of the full capacitor V1=V
V2 is the required voltage let say V'
Then,
V1C1=V2C2
V × εA/d=V' × 2εA/d
VεA/d = 2V'εA/d
Then the εA/d cancels on both sides and remains
V=2V'
Then, V'=V/2
The potential difference is half when the distance between the parallel plate capacitor was reduce to d/2
Answer:
a) 6.9*10^14 Hz
b) 9*10^-12 T
Explanation:
From the question, we know that
435 nm is given as the wavelength of the wave, at the same time, we also know that the amplitude of the electric field, E(max) has been given to be 2.7*10^-3 V/m
a)
To find the frequency of the wave, we would be applying this formula
c = fλ, where c = speed of light
f = c/λ
f = 3*10^8 / 435*10^-9
f = 6.90*10^14 Hz
b) again, to find the amplitude of the magnetic field, we would use this relation
E(max) = B(max) * c, magnetic field amplitude, B(max) =
B(max) = E(max)/c
B(max) = 2.7*10^-3 / 3*10^8
B(max) = 9*10^-12 T
c) and lastly,
1T = 1 (V.s/m^2)
Answer:
A) 1.67 x 10 ⁻⁶ m/s
B)5.59 x 
%
Explanation:
A)
Given:
d = 5.0 km,
mₐ = 2.5 x 
kg
u₁ = 4.0 x 10⁴ m/s
= 5.98 x 10 ²⁴ kg
Solve using kinetic conserved energy
mₐ x u₁ + x u₂ = uₓ x (mₐ + )
(2.5 x ) (4.0 x 10⁴ )+ (5.98 x 10 ²⁴ )(0) = uₓ x (2.5 x + 5.98 x 10 ²⁴ )
uₓ = ( 2.5 x x 4.0 x 10⁴ ) / (2.5 x + 5.98 x 10 ²⁴ )
uₓ = 1.67 x 10 ⁻⁶ m/s
B) Assuming earth radius as a R = 1.5 x 10 ¹¹ m
t = 365 days x 24 hr / 1 day x 60 minute / 1 hr x 60s / 1 minute = 31536000 s
t = 31536000 s
D = 2 π R = 2 π( 1.5 x 10 ¹¹ )
D = 9.4247 x 10 ¹¹ m
u₂ = D / t = 9.4247 x 10 ¹¹ / 31536000
u₂ = 29885.775 m/s
% = ( 1.67 x 10 ⁻⁶ m/s ) / (29885.775 m/s) x 100
% = 5.59 x %
