This could be Hubble's law, or something related to it. I think there's a possibly Doppler RED SHIFT in the optical spectra of stars etc as observed on the earth. It seems that they are accelerating away from the earth, and that the further away they are the faster they are moving.
It seems that this has been connected to the idea of "The Big Bang" theory of the origin of the universe which seems to have superceded Professor Sir Fred Hoye's Steady State theory of the universe.
There's some Special Relativity in this lot, too.
Answer:
A)
B)
Explanation:
<u>Given:</u>
Length of the room 
Width of the room 
A) Let A be the area of the room

B)We will calculate uncertainty in each dimension
%uncertainty in length
%uncertainty in width =
The uncertainty in area will be sum of uncertainty in length and width
%uncertainty in Area= %uncertainty in length + %uncertainty in width
%uncertainty in Area
%uncertainty in Area=0.0106
Uncertainty in Area
There Area is
To solve this problem we will apply the first law of thermodynamics and we will make a balance between the heat transferred, its internal energy and the total work. Recall that for gases the definition of work can be expressed in terms of its pressure and volume. Let's start

Here,
dU = Internal Energy
dW = Work
But internal energy is unchanged, then


Where
= Change in Volume
P = Pressure
Finally, the expression of the heat transferred can be expressed in terms of pressure and volume, so it would end up becoming

Replacing,


Therefore the correct answer is B.
Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
Potential energy behind dams