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:
$93.50
Step-by-step explanation:
81.30 x 1.15 = 93.50
Answer:
y = 1/2x + 5
Step-by-step explanation:
Given :
Slope, m = 1/2
(x1, y1) = (6, 8)
y - y1 = m(x - x1)
y - 8 = 1/2(x - 6)
y - 8 = 1/2x - 3
y - 8 + 8 = 1/2x - 3 + 8
y = 1/2x + 5
Stamdard form of equation :
y = mx + c
Hence, equation is :
y = 1/2x + 5
Answer:
Step-by-step explanation:
add 36 and 4 and multiply that by 6
Answer:
x = -1/2
Step-by-step explanation:
