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3 years ago

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Electricity The power P, in watts, that a circular solar cell produces and radius of the cell in centimeters are related by the

square root equation
$r = \sqrt{ \frac{p }{0.02\pi} }$
about how much power is produced by a cell with a radius of 12 cm?

1 answer:

gavmur [86]3 years ago

5 0

Answer:

9.05 W

Explanation:

The given formula is $r=\sqrt {\frac {P}{0.02\pi}}$ where r is in centimeters and P is in Watts

Making Power, P the subject from the above formula

$P=0.02\pi r^{2}$

Substituting r with 12 cm then

$P=0.02\pi 12^{2}=
9.047786842 W\approx 9.05 W$

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What force is keeping the Earth in orbit around the Sun? * HELP FAST PLEASE.

sveta [45]

Answer:

magnetic force due to gravity

Explanation:

nowno that is because that's what the geography teacher taught us back in don't remember but it's true

6 0

2 years ago

create a poem that incorporates those ten words. Feel free to make it as silly as you like! MINIMUM of 6 lines with a MINIMUM of

Luba_88 [7]

I could make a poem for you if you actually gave the words...... what 10 words do i need to incorporate???☹︎

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3 years ago

A positive kaon (K+) has a rest mass of 494 MeV/c² , whereas a proton has a rest mass of 938 MeV/c². If a kaon has a total energ

vitfil [10]

Answer:

<em>0.85c </em>

Explanation:

Rest mass of Kaon $M_{0K}$ = 494 MeV/c²

Rest mass of proton $M_{0P}$ = 938 MeV/c²

The rest energy is gotten by multiplying the rest mass by the square of the speed of light c²

for the kaon, rest energy $E_{0K}$ = 494c² MeV

for the proton, rest energy $E_{0P}$ = 938c² MeV

Recall that the rest energy, and the total energy are related by..

$E$ = γ $E_{0}$

which can be written in this case as

$E_{K}$ = γ $E_{0K}$ ...... equ 1

where $E$ = total energy of the kaon, and

$E_{0}$ = rest energy of the kaon

γ = relativistic factor = $\frac{1}{\sqrt{1 - \beta ^{2} } }$

where $\beta = \frac{v}{c}$

But, it is stated that the total energy of the kaon is equal to the rest mass of the proton or its equivalent rest energy, therefore...

$E_{K}$ = $E_{0P}$ ......equ 2

where $E_{K}$ is the total energy of the kaon, and

$E_{0P}$ is the rest energy of the proton.

From $E_{K}$ = $E_{0P}$ = 938c²

equ 1 becomes

938c² = γ494c²

γ = 938c²/494c² = 1.89

γ = $\frac{1}{\sqrt{1 - \beta ^{2} } }$ = 1.89

1.89 $\sqrt{1 - \beta ^{2} }$ = 1

squaring both sides, we get

3.57( 1 - $\beta^{2}$ ) = 1

3.57 - 3.57 $\beta^{2}$ = 1

2.57 = 3.57 $\beta^{2}$

$\beta^{2}$ = 2.57/3.57 = 0.72

$\beta = \sqrt{0.72}$ = 0.85

but, $\beta = \frac{v}{c}$

v/c = 0.85

v = <em>0.85c </em>

7 0

3 years ago

If the mass of an object increases by a factor 2, kinetic energy?

emmainna [20.7K]

Answer:

A) Increases by a factor of 2

Explanation:

Kinetic energy can be defined as an energy possessed by an object or body due to its motion.

Mathematically, kinetic energy is given by the formula;

$K.E = \frac{1}{2}MV^{2}$

Where;

K.E represents kinetic energy measured in Joules.

M represents mass measured in kilograms.

V represents velocity measured in metres per seconds square.

Given that mass, m = 2m

Substituting into the equation, we have;

K.E = ½mv²

K.E = ½*2mv²

Cross-multiplying, we have;

2K.E = 2mv²

Hence, if the mass of an object increases by a factor 2, kinetic energy is increased by a factor of 2.

5 0

3 years ago

Radiation from the Sun The intensity of the radiation from the Sun measured on Earth is 1360 W/m2 and frequency is f = 60 MHz. T

Zina [86]

a) Total power output: $3.845\cdot 10^{26} W$

b) The relative percentage change of power output is 1.67%

c) The intensity of the radiation on Mars is $540 W/m^2$

Explanation:

a)

The intensity of electromagnetic radiation is given by

$I=\frac{P}{A}$

where

P is the power output

A is the surface area considered

In this problem, we have

$I=1360 W/m^2$ is the intensity of the solar radiation at the Earth

The area to be considered is area of a sphere of radius

$r=1.5\cdot 10^{11} m$ (distance Earth-Sun)

Therefore

$A=4\pi r^2 = 4 \pi (1.5\cdot 10^{11})^2=2.8\cdot 10^{23}m^2$

And now, using the first equation, we can find the total power output of the Sun:

$P=IA=(1360)(2.8\cdot 10^{23})=3.845\cdot 10^{26} W$

b)

The energy of the solar radiation is directly proportional to its frequency, given the relationship

$E=hf$

where E is the energy, h is the Planck's constant, f is the frequency.

Also, the power output of the Sun is directly proportional to the energy,

$P=\frac{E}{t}$

where t is the time.

This means that the power output is proportional to the frequency:

$P\propto f$

Here the frequency increases by 1 MHz: the original frequency was

$f_0 = 60 MHz$

so the relative percentage change in frequency is

$\frac{\Delta f}{f_0}\cdot 100 = \frac{1}{60}\cdot 100 =1.67\%$

And therefore, the power also increases by 1.67 %.

c)

In this second case, we have to calculate the new power output of the Sun:

$P' = P + \frac{1.67}{100}P =1.167P=1.0167(3.845\cdot 10^{26})=3.910\cdot 10^{26} W$

Now we want to calculate the intensity of the radiation measured on Mars. Mars is 60% farther from the Sun than the Earth, so its distance from the Sun is

$r'=(1+0.60)r=1.60r=1.60(1.5\cdot 10^{11})=2.4\cdot 10^{11}m$

Now we can find the radiation intensity with the equation

$I=\frac{P}{A}$

Where the area is

$A=4\pi r'^2 = 4\pi(2.4\cdot 10^{11})^2=7.24\cdot 10^{23} m^2$

And substituting,

$I=\frac{3.910\cdot 10^{26}}{7.24\cdot 10^{23}}=540 W/m^2$

Learn more about electromagnetic radiation:

brainly.com/question/9184100

brainly.com/question/12450147

#LearnwithBrainly

4 0

3 years ago