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:
what? if u mean by rounding, its 103
Answer:
P(x) = (x - 1)(x + 2)(x - 3)(3x + 1)
Step-by-step explanation:
Since P(1) = 0 and P(- 2) = 0, then
(x - 1) and (x + 2) are factors of P(x)
(x - 1)(x + 2) = x² + x - 2 ← is also a factor of P(x)
dividing 3
- 5x³ - 17x² + 13x + 6 by x² + x - 2 gives
P(x) = (x - 1)(x + 2)(3x² - 8x - 3) = (x - 1)(x + 2)(x - 3)(3x + 1)
You pass the free terms wich don't contain 'x' in the othe part of the equal with changed signe ans the same for the terms wich contain x.
