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Pachacha [2.7K]

2 years ago

14

An empty rubber balloon has a mass of 12.5 g. The balloon is filled with helium at a density of 0.181 kg/m3. At this density the

balloon has a radius of 0.498 m. If the filled balloon is fastened to a vertical line, what is the tension in the line? The density of air is 1.29 kg/m3.

1 answer:

Taya2010 [7]2 years ago

8 0

Answer:

5.5 N

Explanation:

mass of balloon (m) = 12.5 g = 0.0125 kg

density of helium = 0.181 kg/m^{3}

radius of the baloon (r) = 0.498 m

density of air = 1.29 kg/m^{3}

acceleration due to gravity (g) = 1.29 m/s^{2}

find the tension in the line

the tension in the line is the sum of all forces acting on the line

Tension =buoyant force + force by helium + force of weight of rubber

force = mass x acceleration

from density = \frac{mass}{volume} , mass = density x volume

buoyant force = density x volume x acceleration

where density is the density of air for the buoyant force

buoyant force = 1.29 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 6.54 N

force by helium = density x volume x acceleration

force by helium = 0.181 x (\frac{4]{3} x π x 0.498^{3}) x 9.8 = 0.917 N

force of its weight = mass of rubber x acceleration

force of its weight = 0.0125 x 9.8 = 0.1225 N

Tension = buoyant force + force by helium + force of weight of rubber

the force of weight of rubber and of helium act downwards, so they

carry a negative sign.

Tension = 6.54 - 0.917 - 0.1225 = 5.5 N

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strojnjashka [21]

Answer:

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Explanation:

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From the data given we have to:

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3 years ago

kherson [118]

The friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 × 10^8 respectively. Also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 × 10^8 respectively.

<h3>How to determine the friction factor</h3>

Using the formula

μ = viscosity = 0. 06 Pas

d = diameter = 120mm = 0. 12m

V = velocity = 1m/s and 3m/s

ρ = density = 0.9

a. Velocity = 1m/s

friction factor = 0. 52 × $\frac{0. 06}{0. 12* 1* 0. 9}$

friction factor = 0. 52 × $\frac{0. 06}{0. 108}$

friction factor = 0. 52 × 0. 55

friction factor $= 0. 289$

b. When V = 3mls

Friction factor = 0. 52 × $\frac{0. 06}{0. 12 * 3* 0. 9}$

Friction factor = 0. 52 × $\frac{0. 06}{0. 324}$

Friction factor = 0. 52 × 0. 185

Friction factor $= 0.096$

Loss When V = 1m/s

Head loss/ length = friction factor × 1/ 2g × velocity^2/ diameter

Head loss = 0. 289 × $\frac{1}{2*6. 6743 * 10^-11}$ × $\frac{1^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 1. 80 × 10^8

Head loss When V = 3m/s

Head loss = $0. 096$ × $\frac{1}{1. 334 *10^-10}$ × $\frac{3^2}{0. 120}$ × $\frac{1}{100}$

Head loss = 5. 3× 10^8

Thus, the friction factor and head loss when velocity is 1m/s is 0.289 and 1.80 ×10^8 respectively also, the friction factor and head loss when velocity is 3m/s is 0.096 and 5.3 ×10^8 respectively.

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