This approach is called the dimensional analysis which involves only the units of measurement without their magnitudes. You simply have to do the operations by using variables. Cancel out like items that may appear both in the numerator and denominator side. The solution is as follows:
F = mv²/r = [kg][m/s]²/[m] = [kg][m²⁻¹][1/s²] = [kg·m/s²]
Answer:
Ecu/Eag = 0.46
Explanation:
E = PI/A
Ecu = Pcu × I/A
Pcu = 1.72×10^-8 ohm-meter
r = 0.8 mm = 0.8/1000 = 8×10^-4 m
A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π
Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1
Eag = Pag × I/A
Pag = 1.47×10^-8 ohm-meter
r = 0.5 mm = 0.5/1000 = 5×10^-4 m
A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π
Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π
Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46
The volume of the room is the product of its dimensions:

Now, from the equation

where d is the density, m is the mass and V is the volume, we deduce

So, multiply the density and the volume to get the mass of air in the room.
is the best estimate of the density of the air on the planet.
Given:
The mass of the conical flask with stopper is 457.23 grams and the volume is
.
Mass of conical flask and a stopper after removing the air is 456.43 g
To find:
The density of the air on the planet.
Solution;
Mass of the conical flask and stopper with air on the planet= 457.23 g
Mass of conical flask with a stopper and without air on the planet = 456.43 g
Mass of the air in the conical flask on the planet =m

The volume of the conical flask = 
The volume of the air in the conical flask = 

The density of the air on the planet = d

is the best estimate of the density of the air on the planet.
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