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
Hey There!
In order to simplify this expression, you would have to combine the like terms.
9+1 = 10
Nothing else could be done, therefore the answer would be.
7v + 10
Hope this helped!
All we really need to do here is to convert 2 cm to meters. There are 100 cm in 1 meter, so the appropriate cm to meters conversion factor is
1 meter
-----------
100 cm
Thus,
2 cm 1 m
-------- * ----------- = (1/50) m = 0.02 m or 2.0*10^(-2) m (answer)
1 100 cm
Answer:
b. the base is 9 and the height is 6
Step-by-step explanation:
Answer:
The file will help
Step-by-step explanation:
