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nadezda [96]
3 years ago
6

To play basketball with her friends, Evangeline needs to pump air in her ball, which is completely deflated. Before inflating it

, the ball weighs
0.615
0.6150, point, 615 kilograms. Afterwards, it weighs
0.624
0.6240, point, 624 kilograms. The diameter of the ball is
0.24
0.240, point, 24 meters.
Assuming the inflated ball is perfectly spherical, what is the air density within it? Can someone tell me answer please ?
Mathematics
2 answers:
Pani-rosa [81]3 years ago
6 0

Answer:

1.24

Step-by-step explanation:

Nadya [2.5K]3 years ago
4 0

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                      

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