Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
No. the answer to the question if can an argon atom undergo vibrational motion is no. it can not even spin either. the argon atom, or the argon is a chemical element that is the third most abundant gas in the earth's atmosphere. it is ore than twice as abundance as water vapor. Thank you for this question.
Answer:
They speed up as temperature increases.
Explanation:
As temperatures increases within a molecule the particles will speed up.
That depends on a few things that you haven't told us about the setup.
So I'm going to assume one of them, and then give you the answer
in terms of another one:
-- Assume a Class-I lever . . . the fulcrum is between the load and the effort.
-- Then the effort needed to lift the load is
(the weight of the load) x (13 / the distance between the fulcrum and the effort)
