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elena-s [515]
4 years ago
13

If the forces acting upon an object are balanced, then the object

Physics
2 answers:
tatuchka [14]4 years ago
8 0
<h2>The Correct is C. Might be moving at a constant velocity. </h2><h2>Good Luck</h2>
poizon [28]4 years ago
4 0
C) Might be moving at a constant velocity


Good luck! (:
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Conductive deafness occurs Because of damage to what part of the body
aliya0001 [1]
Conductive deafness refers to a malfunction of the middle ear. Those 3 ossicles (malleus, incus, stapes) are responsible for converting the sound wave and conducting it to the inner ear.
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3 years ago
Suppose a log's mass is 5kg. After burning, the mass of the ash is 1 kg. Explain what could have happened to the other 4kg.
snow_lady [41]
Mass never just disappears. The other 4kg had to go somewhere. It could have left the scene of the fire in the form of smoke particles and hot gases.
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4 years ago
1.) Identify two types of heat?
inn [45]

Answer:

Explanation:

Thermal Energy and Heat

Shards of ice fly from the sculptor’s chisel. As the crowd

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another? To answer this question, you need to think about the

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5 0
4 years ago
PLEASE HELP!!! SCIENCE QUESTION!
Tatiana [17]
C. has to be right, also its the only one that makes sense.
Hope this helps!
5 0
3 years ago
Read 2 more answers
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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