The vibration is thermal energy ("heat" energy which every object possesses).
The second one is kinetic energy ("motion" energy of a massive object)
There would be 6 electrons placed on the third energy level.
Answer:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
Explanation:
Rutherford bombarded aluminum foil with beam of light known as alpha particles. The mass of this alpha particle is equivalent to helium atom.
When this alpha particles were made to strike the aluminum foil, some passed through the foil, some were reflected and speed others changed.
The ones reflected encountered heavier particle known as the nucleus, preventing them from passing through it. The whole observations indicated that atom is not is uniformly charged sphere as proposed by J.J Thomson.
Rutherford proposed new model known as the Planetary model of atom, which described atom as containing a nucleus which is revolved by electron, just like planets revolve round the sun. And this nucleus contains opposite charge to electron which is proton, to balance the motion.
1) 0.0011 rad/s
2) 7667 m/s
Explanation:
1)
The angular velocity of an object in circular motion is equal to the rate of change of its angular position. Mathematically:
where
is the angular displacement of the object
t is the time elapsed
is the angular velocity
In this problem, the Hubble telescope completes an entire orbit in 95 minutes. The angle covered in one entire orbit is
rad
And the time taken is
Therefore, the angular velocity of the telescope is
2)
For an object in circular motion, the relationship between angular velocity and linear velocity is given by the equation
where
v is the linear velocity
is the angular velocity
r is the radius of the circular orbit
In this problem:
is the angular velocity of the Hubble telescope
The telescope is at an altitude of
h = 600 km
over the Earth's surface, which has a radius of
R = 6370 km
So the actual radius of the Hubble's orbit is
Therefore, the linear velocity of the telescope is:
Answer:
0.2289
Explanation:
Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then
P= 700*9.81*2.5=17167.5 W= 17.1675 kW
To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%
