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larisa86 [58]
3 years ago
13

A meteorologist plans to release a weather balloon from ground level, to be used for high-altitude atmospheric measurements. The

balloon is spherical, with a radius of 2.20 m, and filled with hydrogen. The total mass of the balloon (including the hydrogen within it) and the instruments it carries is 21.0 kg. The density of air at ground level is 1.29 kg/m3. (a) What is the magnitude of the buoyant force (in N) acting on the balloon, just after it is released from ground level
Physics
1 answer:
Slav-nsk [51]3 years ago
8 0

Answer:

563.86 N

Explanation:

We know the buoyant force F = weight of air displaced by the balloon.

F = ρgV where ρ = density of air = 1.29 kg/m³, g = acceleration due to gravity = 9.8 m/s² and V = volume of balloon = 4πr/3 (since it is a sphere) where r = radius of balloon = 2.20 m

So, F = ρgV = ρg4πr³/3

substituting the values of the variables into the equation, we have

F =  1.29 kg/m³ × 9.8 m/s² × 4π × (2.20 m)³/3

= 1691.58 N/3

= 563.86 N

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Anestetic [448]

Answer:

The Gravitational Force is reduced 4 times

Explanation:

The equation of Gravitational force follows:

F = (G*m1*m2)/r^2

Assume that G*m1*m2 = 1 and r = 1:

F = 1/1^2 = 1 N

Multiply the radius by 2

F = 1/2^2 = 1/4 N

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7 0
3 years ago
A rectangular tank is filled to a depth of 10m with freshwater and open to air at atmospheric pressure.
charle [14.2K]

Answer:

<em>1.</em> <em>39068.07 N</em>

<em>2. 19534.036 N</em>

Explanation:

depth of water h = 10 m

atmospheric pressure Patm = 101325 Pa

density of water p = 1000 kg/m^3

acceleration due to gravity g = 9.81 m/s^2

pressure due to depth of water = pgh

P = 1000 x 9.81 x 10 = 98100 Pa

total pressure on the bottom of the tank is Patm + p = 101325 + 98100 = 199425 Pa

Left plug has diameter = 50 cm = 0.5 m

radius = 0.5/2 = 0.25 m

height = 1 cm = 0.01 m

<em>height below tank surface = 10 - 0.01 = 9.99</em>

pressure at this depth =  1000 x 9.81 x 9.99 = 98001.9 Pa

<em>total pressure = Patm + P = 101325 + 98001.9 = 199326.9 Pa</em>

surface area of plug = πr^{2} = 3.142 x 0.25^{2} = 0.196 m^{2}

<em>force required to lift left plug = pressure x area</em>

F =  199326.9 x 0.196 = <em>39068.07 N</em>

<em>The right side is a hemisphere with the same diameter, therefore surface area is half of the left plug</em>

A = 0.196/2 = 0.098 m^{2}

force F required to lift right plug =  199326.9  x 0.098 =<em> 19534.036 N</em>

6 0
3 years ago
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Answer:

it is a force

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a force is a push or a pull

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It has been hypothesized, and some studies have supported the conjecture,
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