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3 years ago

At 10 seconds, you used 55 Joules. Can you tell me the power?

Physics

2 answers:

Kipish [7]3 years ago

7 0

Answer:

5.5

Explanation:

power =energy÷time

power=55÷10

power=5.5

SashulF [63]3 years ago

5 0

5.5 I think, I’m just decent at this stuff

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When two or more identical capacitors are connected in series across a potential difference?

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When two or more identical capacitors (or resistors) are connected
in series across a potential difference, the potential difference divides
equally among them.

For example, if you have nine identical capacitors (or resistors) all
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4 years ago

A police car parked on the side of the highway emits a 1200 Hz sound that bounces off a vehicle farther down the highway and ret

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f' = frequency observed by the police car after sound reflected from the vehicle and comes back to police car = 1250 Hz

f = frequency emitted by the police car = 1200 Hz

V = speed of sound = 340 m/s

v = speed of vehicle = ?

frequency observed by the police car is given as

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VARVARA [1.3K]

The rate of heat loss by radiation is equal to -207.5kW

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.

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3 years ago