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
We are given
area of sphere is 200.96cm^2
it means that we are given surface area
so, we can use surface area formula
where S is surface area
r is radius
so, firstly we need to find radius or r
we can solve for r
now, we can find volume
we can plug r
now, we can find volume
.............Answer
Answer:
The increment in the model is 106cm
Step-by-step explanation:
Given
Required
Determine the increment
To do this, we simply subtract the initial height of the building from the final height
<em>Hence, the increment in the model is 106cm</em>
One box is 24lb another box is 8 because i half of 32 is 16 and 1/2 of 166 is 8 so 16 + 8 = 24 leaving another 8 behind
<span>I : 2n + 1
II : 2n + 3
III : 2n +5
(2n + 1) + (2n + 3) +(2n+5)=63
6n+9=63
n=9
I : 2n+1=9*2+1=19
II : 2n+3=9*2+3=21
III : 2n+5=9*2+5=23
19+21+23=63
</span>
