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2 years ago

I need help with #24 ASAP .. I have to get it done today.. I need help with #24 right now

Physics

1 answer:

Savatey [412]2 years ago

8 0

Answer:

Explanation: the temperature range is at 1,400

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A device that converts mechanical energy into electricity is a

IgorLugansk [536]

Answer:

An electric generator is a device that converts a form of energy into electricity.

Explanation:

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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

Which type of drug would be useful in dilating the pupils for an examination of the retina? Which type of drug would be useful i

givi [52]

Answer:

muscarinic receptor inhibitor

Explanation:

6 0

3 years ago

Read 2 more answers

Jak możemy służyć innym ludziom. Pilne

Troyanec [42]

Could u put it in english u can use goggle translation if u do not know how to

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3 years ago

The speed of sound in air is 10 times faster than the speed of a wave on a certain string. The density of the string is 0.002kg/

nata0808 [166]

Answer:

The tension on the string is 2.353 N.

Explanation:

Given;

the speed of sound in air, v₀ = 343 m/s

then, the speed of sound on the string, v = 343 / 10 = 34.3 m/s

mass per unit length, m/l = μ = 0.002 kg/m

The speed of sound on the string is given as;

$v = \sqrt{\frac{T}{\mu} } \\\\v^2 = \frac{T}{\mu} \\\\T = v^2 \mu$

where;

T is the tension on the string

T = (34.3)²(0.002)

T = 2.353 N

Therefore, the tension on the string is 2.353 N.

3 0

3 years ago