Answer:
a) ΔV₁ = 21.9 V, b) U₀ = 99.2 10⁻¹² J, c) U_f = 249.9 10⁻¹² J, d) W = 150 10⁻¹² J
Explanation:
Let's find the capacitance of the capacitor
C =
C = 8.85 10⁻¹² (8.00 10⁻⁴) /2.70 10⁻³
C = 2.62 10⁻¹² F
for the initial data let's look for the accumulated charge on the plates
C =
Q₀ = C ΔV
Q₀ = 2.62 10⁻¹² 8.70
Q₀ = 22.8 10⁻¹² C
a) we look for the capacity for the new distance
C₁ = 8.85 10⁻¹² (8.00 10⁻⁴) /6⁴.80 10⁻³
C₁ = 1.04 10⁻¹² F
C₁ = Q₀ / ΔV₁
ΔV₁ = Q₀ / C₁
ΔV₁ = 22.8 10⁻¹² /1.04 10⁻¹²
ΔV₁ = 21.9 V
b) initial stored energy
U₀ =
U₀ = (22.8 10⁻¹²)²/(2 2.62 10⁻¹²)
U₀ = 99.2 10⁻¹² J
c) final stored energy
U_f = (22.8 10⁻¹²) ² /(2 1.04 10⁻⁻¹²)
U_f = 249.9 10⁻¹² J
d) the work of separating the plates
as energy is conserved work must be equal to energy change
W = U_f - U₀
W = (249.2 - 99.2) 10⁻¹²
W = 150 10⁻¹² J
note that as the energy increases the work must be supplied to the system
Answer:
6926.4J
Explanation:
Given parameters:
Mass of iron = 200g
Initial temperature = 100°C
Final temperature = 22°C
Unknown:
Amount of heat transferred to the water = ?
Solution:
The quantity of heat transferred to the water is a function of mass and temperature of the iron;
H = m c Ф
m is the mass of the iron
Ф is the change in temperature
C is the specific heat capacity of iron = 0.444 J/g°C
Now;
insert the parameters and solve;
H = 200 x 0.444 x (100-22)
H = 6926.4J
Violet would refract the most as it has the shortest wavelength amongst the visible spectrum.
Answer: 5.68g/ML.
Explanation; Divide the mass of the unknown substance and the volume of the unknown substance and you will have your answer.
Hope this helps! ^^ if you can pls mark me brainiest!
Answer:
a)
, b)
, c) 
Explanation:
a) The net torque is:

Let assume a constant angular acceleration, which is:



The moment of inertia of the wheel is:



b) The deceleration of the wheel is due to the friction force. The deceleration is:



The magnitude of the torque due to friction:


c) The total angular displacement is:



The total number of revolutions of the wheel is:


