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
The first one is the correct answer due to 4 is the initial height of the plaint and 0.75 are inches that being added every week
There are 12 inches in a foot. To find how many inches are in 5 feet, multiply by 12:
5*12 = 60 inches
Add the 8 additional inches:
60 + 8 = 68 inches
The answer is 68.
Answer:
One solution: (2.5,0)
Step-by-step explanation:
We can use substitution for the solution. Substitute y=2x-5 to -8x-4y=-20 so that you end up with -8x-8x+20=-20. Next you want to add like terms which will be -16x+20=-20, next you want isolate x by subtracting 20 from both sides and leaves you with -16x=-40. Divide -16 on both sides to fully isolate x and will leave you with x=2.5. Now substiture in 2.5 for x in y=2x-5 to get y=2(2.5)-5 which will then lead to y=0.