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:
15
Step-by-step explanation:
Calculation for What is a reasonable estimate of the number of gallons of gas Karl used
Estimated number of gallons used=619.5 miles/41 miles
Estimated number of gallons used=15
Therefore the reasonable estimate of the number of gallons of gas Karl used is 15
Answer:
5,-1,-7
Step-by-step explanation:
Answer:
3
Step-by-step explanation:
2 / (2/3) = 2 * 3/2 = 6/2 = 3
Y = 5
3(3y - 15) is also equal to 9y - 45, which means to find y, you have to divide 45 by 9, which is 5.