Explanation:
Below is an attachment containing the solution.
The air pressure inside the balloon is: 0.1432 Pa
The formulas and procedures that we will use to solve this problem are:
Where:
- a = area of the sphere
- r = radius
- π = mathematical constant
- P = Pressure
- F = Force
- a = surface area
Information about the problem:
- r = 5.0 m
- F = 45 N
- 1 Pa = N/m²
- 1 N = kg * m/s²
- a=?
- P=?
Using the formula of the sphere area we get:
a = 4 * π * r²
a = 4 * 3.1416 * (5.0 m)²
a = 314.16 m²
Applying the pressure formula we get:
P = F/a
P = 45 N/314.16 m²
P = 0.1432 Pa
<h3>What is pressure?</h3>
It is a physical quantity that expresses the force applied on the area of a surface.
Learn more about pressure at: brainly.com/question/26269477
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Answer: 
Explanation:
Given
Cross-sectional area of wire 
Extension of wire 
Extension in a wire is given by

where, 

for same force, length and material

Divide (i) and (ii)

Answer:
i dont know
Explanation:
cuz i dont know what it wrote its too blurry
Answer:
the mass of the air in the classroom = 2322 kg
Explanation:
given:
A classroom is about 3 meters high, 20 meters wide and 30 meters long.
If the density of air is 1.29 kg/m3
find:
what is the mass of the air in the classroom?
density = mass / volume
where mass (m) = 1.29 kg/m³
volume = 3m x 20m x 30m = 1800 m³
plugin values into the formula
1.29 kg/m³ = <u> mass </u>
1800 m³
mass = 1.29 kg/m³ ( 1800 m³ )
mass = 2322 kg
therefore,
the mass of the air in the classroom = 2322 kg