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:
Confusion
Step-by-step explanation:
what the heck are you asking?
Answer:
4.80 mm to 2 d.p
Step-by-step explanation:
w = f(r)
f'(r) = (dw/dr) = (0.0218 mm/mm)
So, for a small difference in rainfall of 220 mm, what is the corresponding small difference in width of leaves in the two forests given.
One definition of a derivative or a rate of change is that it is the ratio of very small differences in the dependent variable to very small differences in the independent variable.
Mathematically,
(dw/dr) = (Δw/Δr) for very small Δw and Δr.
0.0218 = (Δw/220)
Δw = 0.0218 × 220 = 4.796 mm = 4.80 mm to 2 d.p
Hope this Helps!!!
Subtract the known trinomial from the total sum:
6x2 - 5x + 4 - 4x2 + 3x - 2 = 2X^2 -8x +6
I hope this helps :)
