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
One possible way:
-86+(4*12)=-38
-86 is the number where the diver is. 86 feet below the surface.
4*12 is the total number of feet he has ascended in 12 minutes.
Because he is ascending, so you add 4*12 (or 48) to -86.
The answer is -38, meaning he's 38 feet below the surface.
The diver is now 38 feet below the surface.
Hope this helps.
Answer:
120 min
or
2 hrs
Step-by-step explanation:
60 min=1 hr
60 min/5=12 min
12(4)=48 min
24 min=1/5 of room
24(5)=120 min
120 min=2 hrs
Yes because they both have the same degrees, but one is slightly rotated
Answer:
x-2
Step-by-step explanation:
