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Ugo [173]
1 year ago
6

The Sun radiates energy at a rate of about 4×1026W. Earth is about 150×106km from the Sun.

Physics
1 answer:
sergiy2304 [10]1 year ago
8 0

The average intensity received on the Earth's surface is 12.6*10^{10} W/m^{2}

​Intensity = Rate  of  energy radiated/4\pi R^{2}

where  R is the distance of the earth from the sun.

I=4*10^{26}/4\pi *[150*106*10^{3}]^{2}

I=0.12597*10^{12}

I=12.6*10^{10} W/m^{2}

What is the meaning of the intensity of sunlight?

Sun intensity refers to the amount of incoming solar energy, or radiation, that reaches the Earth's surface. The angle at which the rays from the sun hit the Earth determines this intensity.

What is the average intensity received on the Earth's surface?

The average intensity received on the Earth's surface is given by:

I=rate of energy radiated/4\pi R^{2}

Thus, the average intensity received on the Earth's surface is 12.6*10^{10} W/m^{2}

To know more about the intensity received from the sun:

brainly.com/question/29579262

#SPJ1

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6. A 145-g baseball moving 30.5 m/s strikes a stationary 5.75-kg brick resting on small rollers so it moves without significant
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Answer:

Explanation:

a )

momentum of baseball before collision

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= .145 x 30.5

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momentum of brick after collision

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= 6.325 kg m/s

Applying conservation of momentum

4.4225 + 0 = .145 x v + 6.325 , v is velocity of baseball after collision.

v = - 13.12 m / s

b )

kinetic energy of baseball  before collision = 1/2 mv²

= .5 x .145 x 30.5²

= 67.44 J

Total kinetic energy before collision = 67.44 J

c )

kinetic energy of baseball after collision = 1/2 x .145 x 13.12²

= 12.48 J .

 kinetic energy of brick after collision

= .5 x 5.75 x 1.1²

= 3.48 J

Total kinetic energy after collision

= 15.96 J

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4 years ago
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The average kinetic energy of the atoms that make up a substance which is changing from a liquid to a solid is _____________.
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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the
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Answer:

(a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

Explanation:

Given that,

Activity R_{0}=10\ mCi

Time t_{1}=4\ hours

Activity R= 8 mCi

(a). We need to calculate the decay constant

Using formula of activity

R=R_{0}e^{-\lambda t}

\lambda=\dfrac{1}{t}ln(\dfrac{R_{0}}{R})

Put the value into the formula

\lambda=\dfrac{1}{4\times3600}ln(\dfrac{10}{8})

\lambda=0.0000154\ s^{-1}

\lambda=1.55\times10^{-5}\ s^{-1}

We need to calculate the half life

Using formula of half life

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{\lambda}

Put the value into the formula

T_{\dfrac{1}{2}}=\dfrac{ln(2)}{1.55\times10^{-5}}

T_{\dfrac{1}{2}}=44.719\times10^{3}\ s

T_{\dfrac{1}{2}}=11.3\ hr

(b). We need to calculate the value of N₀

Using formula of N_{0}

N_{0}=\dfrac{3.70\times10^{6}}{\lambda}

Put the value into the formula

N_{0}=\dfrac{3.70\times10^{6}}{1.55\times10^{-5}}

N_{0}=2.38\times10^{11}\ nuclei

(c). We need to calculate the sample's activity

Using formula of activity

R=R_{0}e^{-\lambda\times t}

Put the value intyo the formula

R=10e^{-(1.55\times10^{-5}\times30\times3600)}

R=1.87\ mCi

Hence, (a). The decay constant is 1.55\times10^{-5}\ s^{-1}

The half life is 11.3 hr.

(b). The value of N₀ is 2.38\times10^{11}\ nuclei

(c). The sample's activity is 1.87 mCi.

4 0
3 years ago
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