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
the answer is 76%
Step-by-step explanation:
I did all my equations and showed my work
1) divide 26.32 by 8 = 3.29
2) 3.29 x 20= 65.80
3) he would spend 65.80 on 20 gallons of gas
Answer:
512(cm)3
Step-by-step explanation:
Answer:
Smartphones can imrove student lives by giving students the ability to contact their teachers to ask questions or comment on a specific work. they can also allow the student to look up answers and learn from what others think. Smartphones can also add distractions in class and allow an easy way to cheat on a test or a quiz. Overall Smartphones can be beneficial and harmful to a students life.
Step-by-step explanation:
