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:
d. 944 mm^3
Step-by-step explanation:
The area of a circle is given by ...
A = πr² . . . . . where r is the radius, half the diameter
The area of a circle with diameter 9 mm is ...
A = π(4.5 mm)² = 20.25π mm²
The area of the semi-circular end of the prism is half this value, or ...
semicircle area = (1/2)(20.25π mm²) = 10.125π mm² ≈ 31.809 mm²
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The area of the rectangular portion of the end of the prism is the product of its width and height:
A = wh = (9 mm)(6 mm) = 54 mm²
Then the base area of the prism is ...
base area = rectangle area + semicircle area
= (54 mm²) +(31.809 mm²) = 85.809 mm²
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This base area multiplied by the 11 mm length of the prism gives its volume:
V = Bh = (85.809 mm²)(11 mm) ≈ 944 mm³
The volume of the composite figure is about 944 mm³.
Answer:
Step-by-step explanation:
5n + 5 = 45
Coefficient of n = 5
Constant = 5 & 45
Answer: 
Step-by-step explanation:
See attached image, it's direct formula substitution.
