The dielectric plate is now slowly pulled out of the capacitor, which remains connected to the battery. Find the energy U2 of the capacitor at the moment when the capacitor is half-filled with the dielectric.
Answer:
U_eq = 1.99 * 10^(-10) J
Explanation:
Given:
Plate Area = 10 cm^2
d = 0.01 m
k_dielectric = 3
k_air = 1
V = 15 V
e_o = 8.85 * 10 ^-12 C^2 / N .m
Equations used:
U = 0.5 C*V^2 .... Eq 1
C = e_o * k*A /d .... Eq 2
U_i = 0.5 e_o * k_i*A_i*V^2 /d ... Eq 3
For plate to be half filled by di-electric and half filled by air A_1 = A_2 = 0.5 A:
U_electric = 0.5 e_o * k_1*A*V^2 /2*d
U_air = 0.5 e_o * k_2*A*V^2 /2*d
The total Energy is:
U_eq = U_electric + U_air
U_eq = 0.5 e_o * k_1*A*V^2 /2*d + 0.5 e_o * k_2*A*V^2 /2*d
U_eq = (k_1 + k_2) * e_o * A*V^2 / 4*d
Plug the given values:
U_eq = (3 + 1) * (8.82 * 10^ -12 )* (0.001)*15^2 / 4*0.01
U_eq = 1.99 * 10^(-10) J
Answer:
1kg
Explanation:
this box is the smallest and weighs the least. Hope this helps :]
Answer:
Δ = 84 Ω, = (40 ± 8) 10¹ Ω
Explanation:
The formula for parallel equivalent resistance is
1 / = ∑ 1 / Ri
In our case we use a resistance of each
R₁ = 500 ± 50 Ω
R₂ = 2000 ± 5%
This percentage equals
0.05 = ΔR₂ / R₂
ΔR₂ = 0.05 R₂
ΔR₂ = 0.05 2000 = 100 Ω
We write the resistance
R₂ = 2000 ± 100 Ω
We apply the initial formula
1 / = 1 / R₁ + 1 / R₂
1 / = 1/500 + 1/2000 = 0.0025
= 400 Ω
Let's look for the error (uncertainly) of Re
= R₁R₂ / (R₁ + R₂)
R’= R₁ + R₂
= R₁R₂ / R’
Let's look for the uncertainty of this equation
Δ / = ΔR₁ / R₁ + ΔR₂ / R₂ + ΔR’/ R’
The uncertainty of a sum is
ΔR’= ΔR₁ + ΔR₂
We substitute the values
Δ / 400 = 50/500 + 100/2000 + (50 +100) / (500 + 2000)
Δ / 400 = 0.1 + 0.05 + 0.06
Δ = 0.21 400
Δ = 84 Ω
Let's write the resistance value with the correct significant figures
= (40 ± 8) 10¹ Ω
Answer:
The angular velocity is 15.37 rad/s
Solution:
As per the question:
Horizontal distance, x = 30.1 m
Distance of the ball from the rotation axis is its radius, R = 1.15 m
Now,
To calculate the angular velocity:
Linear velocity, v =
v =
v =
v =
Now,
The angular velocity can be calculated as:
Thus
b. It is not possible to convert work entirely into heat.
c. The second law of thermodynamics is a consequence of the first law of thermodynamics.
d. It is impossible to transfer heat from a cooler to a hotter body.
e. All of these statements are false.
Answer:
e. All of these statements are false.
Explanation:
As we know that heat transfer take place from high temperature to low temperature.
It is possible to convert all work into heat but it is not possible to convert all heat in to work some heat will be reject to the surrounding.
The first law of thermodynamics is the energy conservation law.
Second law of thermodynamics states that it is impossible to construct a device which convert all energy into work without rejecting the heat to the surrounding.
By using heat pump ,heat can transfer from cooler body to the hotter body.
Therefore all the answer is False.
Answer:
At a higher velocity.