Answer:
Step-by-step explanation:
the answer is 7.9
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:
12/5
Step-by-step explanation:
m= slope
m=Y2-Y1/X2-X1
Y2=5, Y1=-7, X2=3, X1=-2
Substitute into equation
m=5-(-7)/3-(-2)
m=5+7/3+2
m=12/5
slope is 12/5
I'm guessing 32.5 is the percentage.
32.5/100 = x/12.48
Cross-multiply:
100x = 405.6
Divide both sides by 100 to leave x alone.
x = 4.056
