Answer:
B.the same as Lily's.
Explanation:
When a disc is rotating on its axis uniformly with some angular velocity , every particle on it rotates with the same angular velocity , because every particle on it completes a rotation simultaneously . So in one rotation they take same time . Hence particles whether on the outer edge or near the axis , all have same angular velocity .
They but differ due to their different linear velocity . Linear velocity of particle near outer edge have greater magnitude. .
Answer:

Explanation:
Given data:
particle mass = 0.923 g
particle charge is 4.52 micro C
speed of particle 45.7 m/s
In this particular case, coulomb attraction will cause centrifugal force and taken as +ve and Q is taken as -ve

solving for Q WE GET



where

Answer:
The separation between the two lowest levels = 
The values of n where the energy of molecule reaches 1/2 kT at 300K = 
The separation at this level = 1.8 *
J
Explanation:
Knowing the formula
En = 
Mass of oxygen molecule
m (O2) = 32 amu * 
So the energy diference between the two lowest levels:
E2 - E1 = 
E2 - E1 = 
Now we should find n where the energy of molecule reaches 1/2 kT
En =
= 
= 


by the end is necessary to calculate the separation of the level
En - En-1 = 
= 1.8 *
J
Answer:
The 15 ⁰C measured at this altitude is above the standard temperature for the altitude.
Explanation:
The standard temperature at sea level is 15 degrees Celsius. It decreases about 2 degrees C (or 3.5 degrees F) per 1,000 feet of altitude above sea level.
235 meters is equal to 771 feet.
Using the formula below, we can estimate temperature loss due to this change in altitude, that is 771 feet above sea level.
temperature loss = (3.5 x Change in altitude)/1000ft
temperature loss = (3.5 x 771ft)/1000ft = 2.7⁰F, (32 -2.7 = 29.3 ⁰F)
this is equivalent to 1.5⁰C temperature loss.
Thus, the standard temperature of the engineering quadrangle at 235 meters above sea level is 13.5 ⁰C.
Therefore, the 15 ⁰C measured at this altitude is above the standard temperature for the altitude.