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son4ous [18]
3 years ago
7

What determines the length of a year for a planet

Physics
2 answers:
nlexa [21]3 years ago
8 0
The turning of the planet determines it's time
yawa3891 [41]3 years ago
7 0
How long it takes to revolve around the sun
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3 A 100 g steel ball falls from a height of 1.8 m on to a metal plate and rebounds to a height of 1.25 m.​
BigorU [14]

Given values:

Mass of the steel ball, m = 100 g = 0.1 kg

Height of the steel ball, h1 = 1.8 m

Rebound height, h2 = 1.25 m

a.  PE= mgh

0.1 x 9.8 x 1.8 =

1.764 Joules

b. KE = PE ->

1.764 Joules

c. KE= 1/2 mv square

so v = square root 2ke/m

square root 2 x 1.764/ 0.1

= 5.93 m/s

d. KE=PE=mgh square

0.1 x 9.8 x 1.21 =

1.186 joules

velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s

6 0
3 years ago
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The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to the virt
atroni [7]
Because you see yourself the opposite way in a mirror. So yes your “seeing” yourself but not how everyone else sees you.
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Tần số của dao động duy trì
AfilCa [17]
Amman oak smoke skins kaons
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A solenoidal coil with 25 turns of wire is wound tightly around another coil with 310 turns. the inner solenoid is 25.0 cm long
Ymorist [56]
(a) Image 1
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3 years ago
Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×10^7 light-years from Earth. If the lifetime of a human
kupik [55]

Answer:

0.9999986*c

Explanation:

The ship would travel 2.54*10^7 light years, which means that at a speed close to the speed of light the trip would take 2.54*10^7 years from the point of view of an observer on Earth. However from the point of view of a passenger of that ship it will take only 70 years if the speed is close enough to the speed of light.

\Delta t = \Delta t' * \sqrt{1 - (\frac{v}{c})^2}

Where

Δt is the travel time as seen by a passenger

Δt' is the travel time as seen by someone on Earth

v is the speed of the ship

c is the speed of light in vacuum

We can replace the fraction v/c with x

\Delta t = \Delta t' * \sqrt{1 - x^2}

\sqrt{1 - x^2} = \frac{\Delta t}{\Delta t'}

1 - x^2 = (\frac{\Delta t}{\Delta t'})^2

x^2 = 1 - (\frac{\Delta t}{\Delta t'})^2

x = \sqrt{1 - (\frac{\Delta t}{\Delta t'})^2}

x = \sqrt{1 - (\frac{70}{2.54*10^7})^2} = 0.9999986

It would need to travel at 0.9999986*c

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3 years ago
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