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3 years ago

A lamp draws a current of 0.50 A when it is connected to a 120 V source? What is the resistance of the lamp?

2 answers:

8 0

Given,
Current (I) = 0.50A
Voltage (V) = 120 volts
Resistance (R) =?
We know that:-
Voltage (V) = Current (I) x Resistance (R)
→Resistance (R) = Voltage (V) / Current (I)
= 120/0.50
= 24Ω
∴ Resistance (R) = 24Ω

Ronch [10]3 years ago

3 0

For this case we have that by definition, Ohm's Law states that:

$V = I * R$

Where:

V: It's the voltage

I: It's the current

R: It's the resistance

According to the data we have to:

$I = 0.50A\\V = 120V$

Clearing the resistance formula we have:

$R = \frac {V} {I}\\R = \frac {120} {0.50}\\R = 240$

So, the resistance is 240 ohms!

Answer:

240Ω

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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?

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Where,

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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:

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

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

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3 years ago

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Maslowich

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2 years ago

Through which one of the following mediums is the velocity of a sound wave the greatest?

tino4ka555 [31]

Answer: C. Steel

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3 years ago

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A basketball player grabbing a rebound jumps 76.0 cm vertically. How much total time (ascent and descent) does the player spend

Viefleur [7K]

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s this is the time to fall from the top; it would take the same time to travel

upward the final 0.13 m, so the total time spent in the upper 0.15 m is 2x0.175

= 0.35s

b) there are a couple of ways of finding thetime it takes to travel the bottom 0.13m first way: we can use d=1/2gt^2 twice

to solve this problem the time it takes to fall the final 0.13 m is: time it

takes to fall 0.76 m - time it takes to fall 0.63 m t = sqrt[2d/g] = 0.399 s to

fall 0.76 m, and this equation yields it takes 0.359 s to fall 0.63 m, so it

takes 0.04 s to fall the final 0.13 m. The total time spent in the lower 0.13 m

is then twice this, or 0.08s

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3 years ago

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