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3 years ago

10

In a car engine, gasoline is burned to create mechanical energy. Which of the following statements is true? Some energy is lost

as heat. The total amount of energy in the system increases. The total amount of energy in the system decreases. Nuclear energy is transformed into mechanical energy.

2 answers:

earnstyle [38]3 years ago

7 0

1 and 3 ???

not nuclear ... chemical (outside the nucleus)

energy is v v v often lost as heat.

Nitella [24]3 years ago

7 0

Answer:

Some energy is lost as heat.

Explanation:

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The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the ca

Xelga [282]

The question is incomplete. The complete question is :

A dielectric-filled parallel-plate capacitor has plate area A = 10.0 cm2 , plate separation d = 10.0 mm and dielectric constant k = 3.00. The capacitor is connected to a battery that creates a constant voltage V = 15.0 V . Throughout the problem, use ϵ0 = 8.85×10−12 C2/N⋅m2 .

Find the energy U1 of the dielectric-filled capacitor. I got U1=2.99*10^-10 J which I know is correct. Now I need these:

1. 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.

2. The capacitor is now disconnected from the battery, and the dielectric plate is slowly removed the rest of the way out of the capacitor. Find the new energy of the capacitor, U3.

Solution :

Given :

$A = 10 \ cm^2$

$=0.0010 \ m^2$

d = 10 mm

= 0.010 m

Then, Capacitance,

$C=\frac{k \epsilon_0 A}{d}$

$C=\frac{8.85 \times 10^{12} \times 3 \times 0.0010}{0.010}$

$C=2.655 \times 10^{12} \ F$

$U_1 = \frac{1}{2}CV^2$

$U_1 = \frac{1}{2} \times 2.655 \times 10^{-12} \times (15V)^2$

$U_1=2.987 \times 10^{-10}\ J$

Now,

$C_k=\frac{1}{2} \frac{k \epsilon_0}{d} \times \frac{A}{2}$

And

$C_{air}=\frac{1}{2} \frac{\epsilon_0}{d} \times \frac{A}{2}$

In parallel combination,

$C_{eq}= C_k + C_{air}$

$C_{eq} = \frac{1}{2} \frac{\epsilon_0 A}{d}(1+k)$

$C_{eq} = \frac{1}{2} \times \frac{8.85 \times 10^{-12} \times 0.0010}{0.01} \times (1+3)$

$C_{eq} = 1.77 \times 10^{-12}\ F$

Then energy,

$U_2 =\frac{1}{2} C_{eq} V^2$

$U_2=\frac{1}{2} \times 1.77 \times 10^{-12} \times (15V)^2$

$U_2=1.99 \times 10^{-10} \ J$

b). Now the charge on the $\text{capacitor}$ is :

$Q=C_{eq} V$

$Q = 1.77 \times 10^{-12} \times 15 V$

$Q = 26.55 \times 10^{-12} \ C$

Now when the capacitor gets disconnected from battery and the $\text{dielectric}$ is slowly $\text{removed the rest}$ of the way out of the $\text{capacitor}$ is :

$C_3=\frac{A \epsilon_0}{d}$

$C_3 = \frac{0.0010 \times 8.85 \times 10^{-12}}{0.01}$

$C_3=0.885 \times 10^{-12} \ F$

$C_3 = 0.885 \times 10^{-12} \ F$

Without the dielectric,

$U_3=\frac{1}{2} \frac{Q^2}{C}$

$U_3=\frac{1}{2} \times \frac{(25.55 \times 10^{-12})^2}{0.885 \times 10^{-12}}$

$U_3=3.98 \times 10^{-10} \ J$

4 0

3 years ago