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kykrilka [37]
3 years ago
11

Why does the hood of a car heat up after the car has been running for a while? A) According to the fourth law of thermodynamics,

the temperature of the other parts of the car increases due to the coolant used for the engine. B) According to the first law of thermodynamics, the hood of the car heats up using heat from the surroundings in-order to achieve thermal equilibrium with the engine. C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car. D) According to the third law of thermodynamics, the increase in the velocity of the car changes the entropy of the tires. To balance this change, the temperature of the other parts is increased.
Physics
2 answers:
Svetlanka [38]3 years ago
6 0

Answer:

C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.  

Explanation:

A) According to the fourth law of thermodynamics, the temperature of the other parts of the car increases due to the coolant used for the engine.  

B) According to the first law of thermodynamics, the hood of the car heats up using heat from the surroundings in-order to achieve thermal equilibrium with the engine.  

C) According to the second law of thermodynamics, not all energy from the burnt fuel is used to do work on the piston. It also produces heat which warms other parts of the car.  

D) According to the third law of thermodynamics, the increase in the velocity of the car changes the entropy of the tires. To balance this change, the temperature of the other parts is increased.

gtnhenbr [62]3 years ago
3 0

Answer:

c

Explanation:

kuz i smrt

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Answer:

i_2=3.61\ A

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

q, q_1, q_2 = charge of the capacitor in any time t, t_1, t_2

q_o = initial charge of the capacitor

\omega=angular frequency of the circuit

i, i_1, i_2 = current through the circuit in any time t, t_1, t_2

The charge in an LC circuit is given by

q(t) = q_0 \, cos (\omega t )

The current is the derivative of the charge

\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).

We are given

q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}

It means that

q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]

i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]

From eq 1:

\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}

From eq 2:

\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}

Squaring and adding the last two equations, and knowing that

sin^2x+cos^2x=1

\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1

Operating

\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2

Solving for q_o

\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}

Now we know the value of q_0, we repeat the procedure of eq 1 and eq 2, but now at the second time t_2, and solve for i_2

\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2

Solving for i_2

\displaystyle i_2=w\sqrt{q_o^2-q_2^2}

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

\displaystyle f=\frac{1}{4\pi}\ KHz

w=2\pi f=500\ rad/s

\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}

q_0=0.01166\ c

Finally

\displaystyle i_2=500\sqrt{0.01166^2-.006^2}

i_2=5\ A

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