Answer:
Explanation:
Answer: Let ke = 1/2 IW^2 = 1/2 kMr^2 W^2 be Earth's rotational KE. W = 2pi/24 radians per hour rotation speed and k = 2/5 for a solid sphere M is Earth mass, r = 6.4E6 m.
Then ke = 1/2 2/5 6E24 (6.4E6)^2 (2pi/(24*3600))^2 = ? Joules. You can do the math, note W is converted to radians per second for unit consistency.
Let KE = 1/2 KMR^2 w^2 be Earth's orbital KE. w = 2pi/(365*24) radians per hour K = 1 for a point mass. Note I used 365 days, a more precise number is 365.25 days per year, which is why we have Leap Years.
Find KE/ke = 1/2 KMR^2 w^2//1/2 kMr^2 W^2 = (K/k)(w/W)^2 (R/r)^2 = (5/2) (365)^2 (1.5E11/6.4E6)^2 = 7.81E9 ANS
Answer:
225-Ib person weighs 1000 N
De broglie wavelength,
, where h is the Planck's constant, m is the mass and v is the velocity.

Mass of hydrogen atom, 
v = 440 m/s
Substituting
Wavelength 

So the de broglie wavelength (in picometers) of a hydrogen atom traveling at 440 m/s is 902 pm
Frequency:
Frequency is defined as number of events occurring per second.
- Mathematically written as

- The dimensions
are
![[f]=\frac{1}{[T]}](https://tex.z-dn.net/?f=%5Bf%5D%3D%5Cfrac%7B1%7D%7B%5BT%5D%7D)
as
![[T]=[S]](https://tex.z-dn.net/?f=%5BT%5D%3D%5BS%5D)
so,
![[f]=[S^{-1}]](https://tex.z-dn.net/?f=%5Bf%5D%3D%5BS%5E%7B-1%7D%5D)