The force is reasonable for making fusion possible in the Sun is heat energy.
<h3>What is nuclear fission and fusion?</h3>
When the slow moving neutrons are bombarded with the heavy radioactive nuclei, the product is the more number of neutrons are produced with the large amount of energy. This multiplying process is called nuclear fusion.
The amount of energy produced in such a reaction can be calculated using the equivalence of mass and energy relationship.
E = mc²
The same happens in nuclear fusion where large amount of energy is needed to make more heavy nuclei.
Thus, fusion requires heat energy to continue the reaction.
Learn more about nuclear fusion and fission.
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Energy to lift something =
(mass of the object) x (gravity) x (height of the lift).
BUT ...
This simple formula only works if you use the right units.
Mass . . . kilograms
Gravity . . . meters/second²
Height . . . meters
For this question . . .
Mass = 55 megagram = 5.5 x 10⁷ grams = 5.5 x 10⁴ kilograms
Gravity (on Earth) = 9.8 m/second²
Height = 500 cm = 5.0 meters
So we have ...
Energy = (5.5 x 10⁴ kilogram) x (9.8 m/s²) x (5 m)
= 2,696,925 joules .
That's quite a large amount of energy ... equivalent to
straining at the rate of 1 horsepower for almost exactly an
hour, or burning a 100 watt light bulb for about 7-1/2 hours.
The reason is the large mass that's being lifted.
On Earth, that much mass weighs about 61 tons.
Answer: Enceladus
Explanation:
Enceladus is a small, icy body with an undergound ocean beneath its crust. Cassini discovered that geyser-like jets spew water vapor and ice particles. It is also the sixth largest moon in Saturn and just about a tenth of the largest moon in Saturn; Titan. It is often regarded as one of the most reflective body in the solar system as a result of its icy surface.
Answer:
a)Distance traveled during the first second = 4.905 m.
b)Final velocity at which the object hits the ground = 38.36 m/s
c)Distance traveled during the last second of motion before hitting the ground = 33.45 m
Explanation:
a) We have equation of motion
S = ut + 0.5at²
Here u = 0, and a = g
S = 0.5gt²
Distance traveled during the first second ( t =1 )
S = 0.5 x 9.81 x 1² = 4.905 m
Distance traveled during the first second = 4.905 m.
b) We have equation of motion
v² = u² + 2as
Here u = 0, s= 75 m and a = g
v² = 0² + 2 x g x 75 = 150 x 9.81
v = 38.36 m/s
Final velocity at which the object hits the ground = 38.36 m/s
c) We have S = 0.5gt²
75 = 0.5 x 9.81 x t²
t = 3.91 s
We need to find distance traveled last second
That is
S = 0.5 x 9.81 x 3.91² - 0.5 x 9.81 x 2.91² = 33.45 m
Distance traveled during the last second of motion before hitting the ground = 33.45 m
Potential energy is highest at the top of the loop, and kinetic energy is highest at the bottom of the loop.
