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marysya [2.9K]
2 years ago
12

If a electrician measures the current of a circuit to be 100 Amps and the circuit is only designed to have a current of 80 Amps,

which of the following is the best explanation of what has happened to the circuit ?
A. a There are too many batteries in the circuit.
B. There are too many resistors in the circuit.
C. A battery in the circuit is missing.
D. A resistor in the circuit is missing.
Physics
1 answer:
Talja [164]2 years ago
6 0

There are too many batteries in the circuit, which increased the emf of the circuit and in turn increased the current in the circuit.

<h3>Current in the circuit</h3>

The current in the circuit is determined by applying Ohms law as shown below;

V = IR

where;

  • I is the current
  • V is the voltage
  • R is the resistance

I = V/R

At a constant resistance of the circuit, the current in a circuit will increase when the emf of the circuit increases.

Thus, we can conclude that, there are too many batteries in the circuit, which increased the emf of the circuit and in turn increased the current in the circuit.

Learn more about current in circuit here: brainly.com/question/13195529

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Answer:

The intensity will be 1/9 as much.

Explanation:

The intensity of the light or any source is inversely related to the square of the distance.

I\alpha \frac{1}{r^{2} }

Now according to the question the distance is increased by three times than,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{r_{2}^{2} }

Therefore,

\frac{I_{2} }{I_{1} }=\frac{r_{1}^{2} }{(3r_{1})^{2} }\\\frac{I_{2} }{I_{1} }=\frac{1}{9} \\{I_{2}=\frac{1}{9}{I_{1} }

Therefore the intensity will become 1/9 times to the initial intensity.

3 0
3 years ago
Radiant heat makes it impossible to stand close to a hot lava flow. Calculate the rate of heat loss by radiation from 1.00 m^2 o
VARVARA [1.3K]

The rate of heat loss by radiation is equal to <u>-207.5kW</u>

Why?

To calculate the heat loss rate (or heat transfer rate) by radiation, from the given situation, we can use the following formula:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

Where,

E, is the emissivity of the body.

A, is the area of the body.

T, are the temperatures.

S, is the Stefan-Boltzmann constant, which is equal to:

5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }

Now, before substitute the given information, we must remember that the given formula works with absolute temperatures (Kelvin), so,  we need to convert the given values of temperature from Celsius degrees to Kelvin.

We know that:

K=Celsius+273.15

So, converting we have:

T_{1}=1110\°C+273.15=1383.15K\\\\T_{2}=36.2\°C+273.15=309.35K

Therefore, substituting the given information and calculating, we have:

HeatLossRate=E*S*A*((T_{cold})^{4} -(T_{hot})^{4} )

HeatLossRate=1*5.67x10^{-8}\frac{W}{m^{2}*K^{-4} }*1m^{2} *((309.35K)^{4} -(1383.15})^{4} )\\\\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(95697.42K^{4} -3.66x10^{12}K^{4})\\ \\HeatLossRate=5.67x10^{-8}\frac{W}{K^{-4} }*(-3.66x10^{12} K^{4})=-207522W=-207.5kW

Hence, we have that the rate of heat loss is equal to -207.5kW.

8 0
3 years ago
What is the resistance of a voltage of 65 V and a current of 2.2 A? Include units.
kozerog [31]

Answer:

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Explanation:

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= 65 / 2.2

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Substituting the data of the problem, we find
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So, Natalie gave 220.7 J of energy to the rock.
4 0
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