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iogann1982 [59]
3 years ago
11

*

Physics
2 answers:
Mashutka [201]3 years ago
8 0

i think the answer is true i hope u get it correct

Colt1911 [192]3 years ago
8 0
The answer is true it’s the aphelion es the strongest
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A scientific hypothesis must be testable and capable of being proven false.
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A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o
Liono4ka [1.6K]
Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
g= \frac{GM}{r^2} 
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, d, and its volume (the volume of a sphere of radius r):
M=dV=d ( \frac{4}{3} \pi r^3)=  \frac{4}{3} \pi d r^3 

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
g' =  \frac{GM'}{r'^2} (1)
so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius: 
r'=2r (2)
and that its density is 2/3 of Earth's density:
d'= \frac{2}{3} d
so the mass M' of the new planet is, with respect to the Earth's mass:
M' = d'V' = \frac{4}{3} \pi d' (r')^3 =  \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) =  \frac{16}{3} M (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
g'= \frac{G( \frac{16}{3}M) }{(2r)^2}=  \frac{GM}{r^2}  \frac{4}{3} =  \frac{4}{3}g
And since g=9.81 m/s^2, we find
g'=  \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2
8 0
4 years ago
1. Order the materials from smallest refractive index to largest refractive index.
kakasveta [241]

Air = 1

Water = 1.33

Glass = 1.52

Quartz = 1.46

Diamond = 2.42

5 0
4 years ago
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alekssr [168]
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