Core
Home of atoms of hydrogen also the lightest element in the universe.
Radiative Zone
Outside the inner Core it radiates energy through the process of photon emission.
Convection Layer
Outer most Layer of the Core, it extends form a depth of 200,000 kilometres to the visible surface. Energy is created by Convection. This is where light is produced.
Photosphere
Surrounds the stars and is where light and heat radiate.
Chromosphere
Reddish gas layer outside of the photosphere I think it also works with the Corona.
Corona
Aura of Plasma that surrounds the Sun and other stars, it extends millions of kilometres and easily seen during a total eclipse.
When sphere A and B are brought in contact and separated, charge on each sphere becomes [2x10^-6 + (-4x10^-6)]/ 2 = -1x10^-6 C.
That is, charge is equally separated and is the average of charges on both spheres. The reason behind equal charge on both spheres after separation is, when they are kept in contact, their potential difference becomes same.
There are no appropriate units for power on the list you provided
For this case, let's
assume that the pot spends exactly half of its time going up, and half going
down, i.e. it is visible upward for 0.245 s and downward for 0.245 s. Let us take
the bottom of the window to be zero on a vertical axis pointing upward. All calculations
will be made in reference to this coordinate system. <span>
An initial condition has been supplied by the problem:
s=1.80m when t=0.245s
<span>This means that it takes the pot 0.245 seconds to travel
upward 1.8m. Knowing that the gravitational acceleration acts downward
constantly at 9.81m/s^2, and based on this information we can use the formula:
s=(v)(t)+(1/2)(a)(t^2)
to solve for v, the initial velocity of the pot as it enters
the cat's view through the window. Substituting and solving (note that
gravitational acceleration is negative since this is opposite our coordinate
orientation):
(1.8m)=(v)(0.245s)+(1/2)(-9.81m/s^2)(0.245s)^2
v=8.549m/s
<span>Now we know the initial velocity of the pot right when it
enters the view of the window. We know that at the apex of its flight, the
pot's velocity will be v=0, and using this piece of information we can use the
kinematic equation:
(v final)=(v initial)+(a)(t)
to solve for the time it will take for the pot to reach the
apex of its flight. Because (v final)=0, this equation will look like
0=(v)+(a)(t)
Substituting and solving for t:
0=(8.549m/s)+(-9.81m/s^2)(t)
t=0.8714s
<span>Using this information and the kinematic equation we can find
the total height of the pot’s flight:
s=(v)(t)+(1/2)(a)(t^2) </span></span></span></span>
s=8.549m/s (0.8714s)-0.5(9.81m/s^2)(0.8714s)^2
s=3.725m<span>
This distance is measured from the bottom of the window, and
so we will need to subtract 1.80m from it to find the distance from the top of
the window:
3.725m – 1.8m=1.925m</span>
Answer:
<span>1.925m</span>
Answer:
0.80 m
Explanation:
elastic potential energy formula
elastic potential energy = 0.5 × spring constant × (extension) 2
