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jek_recluse [69]
2 years ago
12

What is the change in weight of a hollow cylinder of height 8 cm and radius 3 cm when the air is pumped out of it

Physics
1 answer:
Trava [24]2 years ago
8 0

This question involves the concepts of density, weight, and volume.

The change in weight of the cylinder will be "2.72 x 10⁻³ N".

The change in weight of the cylinder will be equal to the weight of the air inside the cylinder which is being removed from it:

\Delta W = W_a

where,

  • \Delta W = change in weight of cylinder = ?
  • W_a = Weight of air inside cylinder = \rho Vg
  • \rho = density of air = 1.225 kg/m³
  • V = Volume of air inside the cylinder = πr²h = π(0.03 m)²(0.08 m) = 2.26 x 10⁻⁴ m³
  • g = acceleration due to gravity = 9.81 m/s²

Therefore,

\Delta W = W_a = \rho Vg\\\\\Delta W = (1.225\ kg/m^3)(2.26\ x\ 10^{-4}\ m^3)(9.81\ m/s^2)

<u>ΔW = 2.72 x 10⁻³ N</u>

Learn more about density here:

brainly.com/question/952755

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Answer:

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Explanation:

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initial \:  velocity \:  (u) = 0 \\  \\ Final  \: Velocity \:  (v) = 72 km /h   \\  \\ Time \:  (t) = 5 \:  hours \\  \\ Acceleration \:  (a) =?  \\  \\  \because \: a =  \frac{v - u}{t}  \\  \\  \therefore \: a =  \frac{72 - 0}{5}  \\  \\ a =  \frac{72}{5}  \\  \\ a = 14.5 \: km {h}^{ - 2}  \\  \\ a =  \frac{14.5 \times 1000}{3600\times 3600} \: m {s}^{ - 2}   \\  \\ a = 0.00111882716 \\  \\ a \approx \: 0.001 \: m {s}^{ - 2}

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