Answer:
2.2 (rounded)
Step-by-step explanation:
1 m = 100 cm
1 m^3 = 1^6 cm^3
so 1 cm^3 = 10^-6 m^3
and 1 L = 1000 cm^3 = 10^-3 m^3
Answer:
y
=
−
3/
2x
−
15
Step-by-step explanation:
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:
Although the traditional tinker bank option is more convenient and suitable for her parents’ wishes, as it being the primary place for her revenue and learning tool it might be best for her to stick with her gut. As she has great accessibility to her digital phone, in her respect it could be far more convenient than a physical building to attend where using her phone is easier to figure out and like second nature to a young person today. Additionally, she can try to navigate the challenge of depositing cash into a non tangible account however if there is no accessibility to this aspect it might be best for her to utilize a traditional like system.
Step-by-step explanation: