Answer:
ρ_air = 0.15544 kg/m^3
Step-by-step explanation:
Solution:-
- The deflated ball ( no air ) initially weighs:
m1 = 0.615 kg
- The air is pumped into the ball and weight again. The new reading of the ball's weight is:
m2 = 0.624 kg
- The amount of air ( mass of air ) pumped into the ball can be determined from simple arithmetic between inflated and deflated weights of the ball.
m_air = Δm = m2 - m1
m_air = 0.624 - 0.615
m_air = 0.009 kg
- We are to assume that the inflated ball takes a shape of a perfect sphere with radius r = 0.24 m. The volume of the inflated ( air filled ) ball can be determined using the volume of sphere formula:
V_air = 4*π*r^3 / 3
V_air = 4*π*0.24^3 / 3
V_air = 0.05790 m^3
- The density of air ( ρ_air ) is the ratio of mass of air and the volume occupied by air. Expressed as follows:
ρ_air = m_air / V_air
ρ_air = 0.009 / 0.05790
Answer: ρ_air = 0.15544 kg/m^3
Answer:
Step-by-step explanation:
We require 2 equations with the repeating decimal after the decimal point in both equations.
The bar above 136 indicates that 136 is being repeated
let x = 0.136136...... → (1) Multiply both sides by 1000
1000x = 136.136.... → (2)
Subtract (1) from (2) to eliminate the decimal part
999x = 136 ( divide both sides by 999 )
x =
Answer:
hold on i'm abt to answer it
Step-by-step explanation:
Answer:
2 solutions
-4 and 4 are possible answers
Step-by-step explanation:
