You would need to distribute the 3 on the left side and the -6 on the right side. p=7/3
The second answer is 4 times as small as the first
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:
The place value to which the number is rounded is unit place.
Step-by-step explanation:
Here, the number 3.695 is rounded to the number 4 and we have to name the place value to which the number is rounded.
Here, the place value to which the number is rounded is unit place.
And the name of the rounding is the rounding to the nearest whole number i.e. there will be no fractional part in the rounded number.
Another example is the number 0.917 is rounded to the nearest whole number to give 1. (Answer)
