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2 years ago

12

When a steady current flows through a resistor, the resistor heats up. We say that "electrical energy is dissipated" by the resi

stor, that is, converted into heat. But if energy is dissipated, where did it come from? Did it come from the voltage source through the wires?

1 answer:

drek231 [11]2 years ago

7 0

Answer:

The dissipated energy in a resistor comes from the potential electric energy of a current source. This is known as the Joule effect. In fact, this energy is carried by the wires until the resistor.

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particle of mass 59 g and charge 51 µC is released from rest when it is 32 cm from a second particle of charge −14 µC. Determine

AleksAgata [21]

Answer:

The initial acceleration of the 59g particle is $1062.7\frac{m}{s^{2}}$

Explanation:

Newton's second laws relates acceleration (a), net force(F) and mass (m) in the next way:

$F=ma$ (1)

We already know the mass of the particle so we should find the electric force on it to use on (1), the magnitude of the electric force between two charged objects by Columb's law is:

$F=k\frac{\mid q_{1}q_{2}\mid}{r^{2}}$

with q1 and q2 the charge of the particles, r the distance between them and k the constant $k=9.0\times10^{9}\,\frac{Nm^{2}}{C^{2}}$ . So:

$F=(9.0\times10^{9})\frac{\mid (51\times10^{-6})(-14\times10^{-6})\mid}{0.32^{2}}$

$F=62.7 N$

Using that value on (1) and solving for a

$a=\frac{F}{m}=\frac{62.7}{0.059}=1062.7\frac{m}{s^{2}}$

4 0

3 years ago

What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a

viktelen [127]

Complete Question

A certain refrigerator, operating between temperatures of -8.00°C and +23.2°C, can be approximated as a Carnot refrigerator.

What is the refrigerator's coefficient of performance? COP

(b) What If? What would be the coefficient of performance if the refrigerator (operating between the same temperatures) was instead used as a heat pump? COP

Answer:

a

$COP = 8.49$

b

$COP_1 = 9.49$

Explanation:

From the question we are told that

The lower operation temperature of refrigerator is $T_1 = -8.00^oC = 265 \ K$

The upper operation temperature of the refrigerator is $T_2 = 23.2 ^oC = 296.2 \ K$

Generally the refrigerators coefficient of performance is mathematically represented as

$COP = \frac{T_1}{T_2 - T_1 }$

=> $COP = \frac{265}{296.2 - 265 }$

=> $COP = 8.49$

Generally if a refrigerator (operating between the same temperatures) was instead used as a heat pump , the coefficient of performance is mathematically represented as

$COP_1 = \frac{T_2}{ T_2 - T_1}$

=> $COP_1 = \frac{296.2}{ 296.2 - 265 }$

=> $COP_1 = 9.49$

8 0

3 years ago

Near the end of a marathon race, the first two runners are separated by a distance of 45.6 m. The front runner has a velocity of

morpeh [17]

Answer:17.08 s

Explanation:

Given

distance between First and second Runner is 45.6 m

speed of first runner $(v_1)$ =3.1 m/s

speed of second runner $(v_2)$ =4.65 m/s

Distance between first runner and finish line is 250 m

Second runner need to run a distance of 250+45.6=295.6 m

Time required by second runner $t=\frac{295.6}{4.65}=63.56 s$

time required by first runner to reach finish line $=\frac{250}{3.1}=80.64 s$

Thus second runner reach the finish line 80.64-63.56=17.08 s earlier

3 0

3 years ago

Without friction, what is the mass of an ball accelerating at 1.8 m/sec2 to which an

Juli2301 [7.4K]

Answer:

<h2>23.33 kg </h2>

Explanation:

The mass of the object can be found by using the formula

$m = \frac{f}{a} \\$

f is the force

a is the acceleration

From the question we have

$m = \frac{42}{1.8} = 23.3333... \\$

We have the final answer as

<h3>23.33 kg</h3>

Hope this helps you

4 0

2 years ago

Read 2 more answers

If 3.61 m3 of a gas initially at STP is placed under a pressure of 2.67 atm , the temperature of the gas rises to 37.9 ∘C. What

Pavel [41]

Answer: $1.54m^3$

Explanation:

Combined gas law is the combination of Boyle's law, Charles's law and Gay-Lussac's law.

The combined gas equation is,

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

where,

$P_1$ = initial pressure of gas at STP = 1 atm

$P_2$ = final pressure of gas = 2.67 atm

$V_1$ = initial volume of gas = $3.61m^3$

$V_2$ = final volume of gas = ?

$T_1$ = initial temperature of gas at STP = $0^oC=273+0=273K$

$T_2$ = final temperature of gas = $37.9^oC=273+37.9=310.9K$

Now put all the given values in the above equation, we get:

$\frac{1atm\times 3.61m^3}{273K}=\frac{2.67\times V_2}{310.9K}$

$V_2=1.54m^3$

Thus the final volume will be $1.54m^3$

7 0

3 years ago