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:
19.9875 feet
Step-by-step explanation:
The formula is given as:
Shadow of the student/Height of the student = Shadow of the telephone pole/Height of the telephone pole.
1 inch = 0.0833 feet
Shadow of the student = 5ft
Height of the student = 5 feet 4 inches
4 inches to feet
= 4 × 0.0833 feet
= 0.33 feet
Hence: Height of the student = 5 + 0.33 = 5.33 feet
Shadow of the telephone pole = 18 feet 9 inches long
9 inches to feet
= 9 × 0.0833 feet
= 0.75 feet
Hence: Shadow of the telephone pole = 18 + 0.75 = 18.75 feet
Height of the telephone pole= x
Therefore:
5/5.33 = 18.75/x
Cross Multiply
5x = 5.33 × 18.75
x = 5.33 × 18.75/5
x = 19.9875 feet
The height of the telephone pole = 19.9875 feet
Answer:
it will be 9738.54
Step-by-step explanation:
Answer:
The circumference of the circle
Step-by-step explanation:
