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vichka [17]
3 years ago
11

Find the minimum radius of a helium balloon that will lift her off the ground. the density of helium gas is 0.178 kg/m3, and the

density of air is 1.29 kg/m3. neglect the weight of the deflated balloon itself (but not the weight of the helium). assume that the shape of the balloon is a sphere. neglect the buoyant force of the air on the woman.
Physics
1 answer:
marysya [2.9K]3 years ago
4 0

Volume of balloon = \frac{4}{3} \pi r^{3}

Where r is the radius of balloon.

Here mass of woman = 68 kg

Mass of air displaced by a balloon with volume V = 1.29*V

Mass of helium inside balloon = 0.178*V

Total mass to be lifted by balloon = 68 +0.178*V

Buoyant force = 1.29V-0.178V=1.112V

So we have 1.112 V =  68+ 0.178*V

   0.934 V = 68

     V = 72.81 m^3

    \frac{4}{3} \pi r^{3}[/tex]= 72.81

     r = 2.59 m

So radius of helium balloon = 2.59 m

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The membrane that surrounds a certain type of living cell has a surface area of 5.3 x 10-9 m2 and a thickness of 1.1 x 10-8 m. A
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Answer:

2.1\times 10^{-12} c

Explanation:

We are given that

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Thickness of membrane=1.1\times 10^{-8} m

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Substitute the values then we get

Capacitance between parallel plate capacitor=\frac{5.9\times 8.85\times 10^{-12}\times 5.3\times 10^{-9}}{1.1\times 10^{-8}}

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since centripetal acceleration is always towards the center of the circle

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also we know that

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A 2.0 m × 4.0 m flat carpet acquires a uniformly distributed charge of −10 μC after you and your friends walk across it several
mamaluj [8]

Answer:

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    The length is  l = 2.0 \ m

    The width is  w = 4.0 \ m

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Generally the electric field on the carpet is mathematically represented as

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=>      q_d  =  \frac{5.0 *10^{-9} * 9.8}{70621.5}

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