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mars1129 [50]
4 years ago
7

Some metal conductor has a resistance of 12 ohms. When plugged into a circuit, the amount of heat 144J is released in 12s. What

is the voltage at the ends of the conductors?​
Physics
1 answer:
worty [1.4K]4 years ago
5 0

Answer:

12V

Explanation:

Given parameters:

Resistance  = 12ohms

Heat released = 144J

Time taken = 12s

Unknown:

Voltage at the ends of the conductors = ?

Solution:

To solve this problem, we must first determine the power expended by the circuit.

Power is the rate at which work is done;

                P  = \frac{work done }{time}

                    = \frac{144}{12}

                    = 12W

Electrical power is given as;

             P  = IV  = \frac{V^{2} }{R}

            Since V is the unknown, we  solve for it;

          12  = \frac{V^{2} }{12}

           V  = 12V

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Answer:

In electromagnetic waves.

Explanation:

Energy is transferred through vibrations of electric and magnetic fields. In sound waves, energy is transferred through vibration of air particles or particles of a solid through which the sound travels. In water waves, energy is transferred through the vibration of the water particles.

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Newton's Law of Gravitation says that the magnitude F of the force exerted by a body of mass m on a body of mass M is F = GmM r2
Nookie1986 [14]
<h2>Answers:</h2>

<h2>(a) </h2>

According to Newton's Law of Gravitation, the Gravity Force is:

F=\frac{GMm}{{r}^{2}}     (1)

This expression can also be written as:

F=GMm{r}^{-2}    (2)

If we derive this force F respect to the distance r between the two masses:

\frac{dF}{dr}dFdr=\frac{d}{dr}(GMm{r}^{-2})dr     (3)

Taking into account GMm are constants:

\frac{dF}{dr}dFdr=-2GMm{r}^{-3}     (4)

Or

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

<h2> (b) dF/dr represents the rate of change of the force with respect to the distance between the bodies.  </h2><h2 />

In other words, this means how much does the Gravity Force changes with the distance between the two bodies.

More precisely this change is inversely proportional to the distance elevated to the cubic exponent.

As the distance increases, the Force decreases.

<h2>(c) The minus sign indicates that the bodies are being forced in the negative direction.  </h2>

This is because Gravity is an attractive force, as well as, a central conservative force.

This means it does not depend on time, and both bodies are mutually attracted to each other.

<h2>(d) </h2>

In the first answer we already found the decrease rate of the Gravity force respect to the distance, being its unit N/km:

\frac{dF}{dr}dFdr=-2\frac{GMm}{{r}^{3}}     (5)

We have a force that decreases with a rate 1 \frac{dF_{1}}{dr}dFdr=4N/km when r=20000km:

4N/km=-2\frac{GMm}{{(20000km)}^{3}}     (6)

Isolating -2GMm:

-2GMm=(4N/km)({(20000km)}^{3})     (7)

In addition, we have another force that decreases with a rate 2 \frac{dF_{2}}{dr}dFdr=X when r=10000km:

XN/km=-2\frac{GMm}{{(10000km)}^{3}}     (8)

Isolating -2GMm:

-2GMm=X({(10000km)}^{3})     (9)

Making (7)=(9):

(4N/km)({(20000km)}^{3})=X({(10000km)}^{3}       (10)

Then isolating X:

X=\frac{4N/km)({(20000km)}^{3}}{{(10000km)}^{3}}  

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7 0
4 years ago
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Solve these problems on charges by finding out Q1 and Q2. Q1Q2= 4x10^6, Q1 + Q2 =8.00×10^-6​
inna [77]

Answer:

Answer:

Q_1 = 7Q

1

=7

Q_2 = 10Q

2

=10

Q_3 = 13.5Q

3

=13.5

Step-by-step explanation:

Given

5, 7, 7, 8, 10, 11, 12, 15, 17.

Required

Determine Q1, Q2 and Q3

The number of data is 9

Calculating Q1:

Q1 is calculated as:

Q_1 = \frac{1}{4}(N + 1)Q

1

=

4

1

(N+1)

Substitute 9 for N

Q_1 = \frac{1}{4}(9 + 1)Q

1

=

4

1

(9+1)

Q_1 = \frac{1}{4}*10Q

1

=

4

1

∗10

Q_1 = 2.5th\ itemQ

1

=2.5th item

This means that the Q1 is the mean of the 2nd and 3rd data.

So:

Q_1 = \frac{1}{2}(7+7)Q

1

=

2

1

(7+7)

Q_1 = \frac{1}{2}*14Q

1

=

2

1

∗14

Q_1 = 7Q

1

=7

Calculating Q2:

Q2 is calculated as:

Q_2 = \frac{1}{2}(N + 1)Q

2

=

2

1

(N+1)

Substitute 9 for N

Q_2 = \frac{1}{2}(9 + 1)Q

2

=

2

1

(9+1)

Q_2 = \frac{1}{2}*10Q

2

=

2

1

∗10

Q_2 = 5th\ itemQ

2

=5th item

Q_2 = 10Q

2

=10

Calculating Q3:

Q3 is calculated as:

Q_3 = \frac{3}{4}(N + 1)Q

3

=

4

3

(N+1)

Substitute 9 for N

Q_3 = \frac{3}{4}(9 + 1)Q

3

=

4

3

(9+1)

Q_3 = \frac{3}{4}*10Q

3

=

4

3

∗10

Q_3 = 7.5th\ itemQ

3

=7.5th item

This means that the Q3 is the mean of the 7th and 8th data.

So:

Q_3 = \frac{1}{2}(12+15)Q

3

=

2

1

(12+15)

Q_3 = \frac{1}{2}*27Q

3

=

2

1

∗27

Q_3 = 13.5Q

3

=13.5

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