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3 years ago

Write a comparison of the current in three bulbs of increasing resistance connected in a series arrangement.

Physics

1 answer:

Igoryamba3 years ago

8 0

Answer:

are you still looking for an answer

Explanation:

because if you are i could give it to you

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How do you do this?Or what are the formulus

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You got the formulas on the sheet on the top :) So just use those, exchanging v (as in velocity, expressed in m/s) and the d (in meters) and t (in seconds). Hope you will manage it.

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3 years ago

The hot glowing surfaces of stars emit energy in the form of electromagnetic radiation. It is a good approximation to assume tha

belka [17]

Given that,

Energy $H=2.7\times10^{31}\ W$

Surface temperature = 11000 K

Emissivity e =1

(a). We need to calculate the radius of the star

Using formula of energy

$H=Ae\sigma T^4$

$A=\dfrac{H}{e\sigma T^4}$

$4\pi R^2=\dfrac{H}{e\sigma T^4}$

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.7\times10^{31}}{1\times5.67\times10^{-8}\times(11000)^4\times 4\pi}}$

$R=5.0\times10^{10}\ m$

(b). Given that,

Radiates energy $H=2.1\times10^{23}\ W$

Temperature T = 10000 K

We need to calculate the radius of the star

Using formula of radius

$R^2=\dfrac{H}{e\sigma T^4\times4\pi}$

Put the value into the formula

$R=\sqrt{\dfrac{2.1\times10^{23}}{1\times5.67\times10^{-8}\times(10000)^4\times4\pi}}$

$R=5.42\times10^{6}\ m$

Hence, (a). The radius of the star is $5.0\times10^{10}\ m$

(b). The radius of the star is $5.42\times10^{6}\ m$

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3 years ago

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Sonja [21]

Answer:

need more context

Explanation:

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