One cubic centimeter (1.0 cm3) of water has a mass of 1.0 × 10-3 kg

  1. One cubic centimeter (1.0 cm3) of water has a mass of 1.0 × 10-3 kg. (a) Determine the mass of 1.0 m3 of water. (b) Assuming that biological substances are 98% water, estimate the masses of a cell with a diameter of 1.0 μ m, a human kidney, and a fly. Take a kidney to be roughly a sphere with a radius of 4.0 cm and a fly to be roughly a cylinder 4.0 mm long and 2.0 mm in diameter.
  2. A billionaire offers to give you $1 billion if you can count out that sum with only $1 bills. Should you accept her offer? Assume that you can count at an average rate of one bill every second, and be sure to allow for the fact that you need about 8 hours a day for sleeping and eating.

The Correct Answer and Explanation is :

Solution to Question 1

(a) Mass of 1.0 m³ of water

We are given that 1.0 cm³ of water has a mass of 1.0 × 10⁻³ kg.
Since 1.0 m³ = 1.0 × 10⁶ cm³, the mass of 1.0 m³ of water is:

[
\text{Mass} = (1.0 \times 10^6 \text{ cm}^3) \times (1.0 \times 10^{-3} \text{ kg/cm}^3)
]

[
= 1.0 \times 10^3 \text{ kg} = 1000 \text{ kg}
]

(b) Estimating the masses of a cell, kidney, and fly

Since biological substances are 98% water, their mass is approximately the same as water’s density (1.0 × 10³ kg/m³).

Cell with a diameter of 1.0 μm
  • Radius (r) = 0.5 μm = 0.5 × 10⁻⁶ m
  • Volume of a sphere: [
    V = \frac{4}{3} \pi r^3
    ] [
    V = \frac{4}{3} \pi (0.5 \times 10^{-6})^3
    ] [
    \approx 5.24 \times 10^{-19} \text{ m}^3
    ]
  • Mass = Volume × Density = [
    (5.24 \times 10^{-19}) \times (10^3)
    ] [
    = 5.24 \times 10^{-16} \text{ kg}
    ]
Kidney with radius 4.0 cm
  • r = 4.0 cm = 0.04 m
  • Volume: [
    V = \frac{4}{3} \pi (0.04)^3 = 2.68 \times 10^{-4} \text{ m}^3
    ]
  • Mass = ( 2.68 \times 10^{-4} \times 10^3 = 0.268 ) kg (268 g)
Fly (cylinder: length = 4.0 mm, diameter = 2.0 mm)
  • Radius = 1.0 mm = 1.0 × 10⁻³ m, Height = 4.0 mm = 4.0 × 10⁻³ m
  • Volume of a cylinder: [
    V = \pi r^2 h
    ] [
    = \pi (1.0 \times 10^{-3})^2 (4.0 \times 10^{-3})
    ] [
    = 1.26 \times 10^{-8} \text{ m}^3
    ]
  • Mass = ( 1.26 \times 10^{-8} \times 10^3 = 1.26 \times 10^{-5} ) kg (12.6 mg)

Solution to Question 2: Should You Accept the Offer?

You must count $1 billion ($1 × 10⁹) dollar bills. If you count at a rate of 1 bill per second, the total time required is:

[
\text{Time} = 10^9 \text{ seconds}
]

Convert to days:

[
10^9 \text{ sec} \times \frac{1 \text{ min}}{60 \text{ sec}} \times \frac{1 \text{ hour}}{60 \text{ min}} \times \frac{1 \text{ day}}{24 \text{ hours}}
]

[
= 1.16 \times 10^4 \text{ days} = 31,709 \text{ years}
]

Since you need 8 hours per day for rest, you are only counting for 16 hours per day, so the actual time doubles:

[
\approx 47,564 \text{ years}
]

Conclusion:

You should not accept the offer, because it would take nearly 48,000 years to complete the counting, which is far beyond a human lifespan.


Now, I’ll generate an image to visually represent the calculations.

I’ve generated an illustration depicting the challenge of counting one billion $1 bills, emphasizing the impossibility of the task humorously. Let me know if you need any modifications!

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