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:
So you can cancel out the 16. See, the -16 is negative, and if we add +16 to it, then it cancels out and is 0. That way, you have x=1.
Step-by-step explanation:
-16x+x=-15
+16x +16
0+x=1
x=1
Answer:
the answer is c
Step-by-step explanation:
i just took a test with this question on it.
Answer:
It compares 3 meals to 1 day.
Step-by-step explanation:
"Unit rate" refers to a rate that has 1 unit in the denominator.
(3 meals)/(1 day) . . . . "three meals per day"
has one unit (1 day) in the denominator, so that is a unit rate.
