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jeka57 [31]
3 years ago
12

Draw a free body diagram of a person as their board accelerates them up into the sky. Be sure that, if a force is bigger than an

other, you show this clearly on your diagram.
Physics
1 answer:
trasher [3.6K]3 years ago
7 0

Answer:

I hope it helps.

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Cabbage juice can be used as a pH indicator. Using the above chart, choose all of the statements that are correct
GREYUIT [131]

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Google it muh homie

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The diagrams below represent different ways heat can be transferred on Earth. The arrows in the diagrams show the direction of h
Vadim26 [7]

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c

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heat radiates off of thing most of the time

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HELP When the forces are applied in the same direction, how do you determine net force?
aleksklad [387]

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Explanation:

If two forces act on an object in the same direction, the net force is equal to the sum of the two forces.

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DOMICILE (Choose the word that names a necessary part of the above word.)
8_murik_8 [283]
A house, address, area.

I hope this helps!
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3 years ago
The intensity of sunlight reaching the earth is 1360 w/m2. Assuming all the sunligh is absorbed, what is the radiation pressure
krok68 [10]

(a) 1.15\cdot 10^9 N

For an electromagnetic wave incident on a surface, the radiation pressure is given by (assuming all the radiation is absorbed)

p=\frac{I}{c}

where

I is the intensity

c is the speed of light

In this problem, I=1360 W/m^2; substituting this value, we find the radiation pressure:

p=\frac{1360 W/m^2}{3\cdot 10^8 m/s}=4.5\cdot 10^{-6} Pa

the force exerted on the Earth depends on the surface considered. Assuming that the sunlight hits half of the Earth's surface (the half illuminated by the Sun), we have to consider the area of a hemisphere, which is

A=2pi R^2

where

R=6.37\cdot 10^6 m

is the Earth's radius. Substituting,

A=2\pi (6.37\cdot 10^6 m)^2=2.55\cdot 10^{14}m^2

And so the force exerted by the sunlight is

F=pA=(4.5\cdot 10^{-6} Pa)(2.55\cdot 10^{14} m^2)=1.15\cdot 10^9 N

(b) 3.2\cdot 10^{-14}

The gravitational force exerted by the Sun on the Earth is

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=1.99\cdot 10^{30}kg is the Sun's mass

m=5.98\cdot 10^{24}kg is the Earth's mass

r=1.49\cdot 10^{11} m is the distance between the Sun and the Earth

Substituting,

F=(6.67\cdot 10^{-11} )\frac{(1.99\cdot 10^{30}kg)(5.98\cdot 10^{24} kg)}{(1.49\cdot 10^{11} m)^2}=3.58\cdot 10^{22}N

And so, the radiation pressure force on Earth as a fraction of the sun's gravitational force on Earth is

\frac{1.15\cdot 10^9 N}{3.58\cdot 10^{22}N}=3.2\cdot 10^{-14}

6 0
4 years ago
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