Answer:
The temperature required is near about 3 million kelvin
Explanation:
The brilliance of the star results from the nuclear reaction that take place in the core of the star and radiate a huge amount of thermal energy resulting from the fusion of hydrogen into helium.
For this reaction to take place, the temperature of the star's core must be near about 3 million kelvin.
The hydrogen atoms collide and starts and the energy from the collision results in the heating of the gas cloud. As the temperature comes to near about 
, the nuclear fusion reaction takes place in the core of the gas cloud.
The huge amount of thermal energy from the nuclear reaction gives the gas cloud a brilliance resulting in a protostar.
v₀ = initial speed of the object = 8 meter/second
v = final speed of the object = 16 meter/second
t = time taken to increase the speed = 10 seconds
d = distance traveled by the object in the given time duration = ?
using the kinematics equation
d = (v + v₀) t/2
inserting the above values in the above equation
d = (16 + 8) (10)/2
d = 120 meter
Answer:Yes, water indeed expands when it changes form from liquid to solid. And this is because water has a property called “hydrogen bonds”, and these bonds occur between each water molecule. But when water is in a liquid form these hydrogen bonds break more easily and occur less frequently. When the temperature drops the kinetic energy also drops, which in turn makes hydrogen bonds form more frequently. So the water molecules form a lattice, which is less dense than regular liquid water.
Explanation:
"v0" means that there are no friction forces at that speed
<span>mgsinΘ = (mv0²/r)cosΘ → the variable m cancels </span>
<span>sinΘ/cosΘ = tanΘ = v0² / gr
</span><span>Θ = arctan(v0² / gr) </span>
<span>When v > v0, friction points downslope: </span>
<span>mgsinΘ + µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ + µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = ((v²/r)cosΘ - gsinΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
<span>When v > v0, friction points upslope: </span>
<span>mgsinΘ - µ(mgcosΘ + (mv²/r)sinΘ) = (mv²/r)cosΘ → m cancels: </span>
<span>gsinΘ - µ(gcosΘ + (v²/r)sinΘ) = (v²/r)cosΘ </span>
<span>µ = (gsinΘ - (v²/r)cosΘ) / (gcosΘ + (v²/r)sinΘ) </span>
<span>where Θ is defined above. </span>
When you are going down you pick up more speed
