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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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Answer:

(B) sound/kinetic

(E) Gravitational/kinetic

(F) mechanical/sound

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Calculate the work done on a Pressure / Volume (PV) isothermal expansion where 400 Pa and 0.08 volume in m3?
Ainat [17]
PV = 400 x 0.08 = 32 J

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The pressure, P, of a gas varies directly with its temperature, T, and inversely with its volume, V, according to the equation f
lbvjy [14]
The equation formula:
P V = n R T
1,245 * 2 l = n R * 300 K
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Unless indicated otherwise, assume the speed of sound in air to be v = 344 m/s. A pipe closed at both ends can have standing wav
uranmaximum [27]

Answer:

68.8 Hz

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Explanation:

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v = Velocity of sound in air = 344 m/s

Distance between nodes is given by

L=\dfrac{\lambda}{2}+m\dfrac{\lambda}{2}\\\Rightarrow \dfrac{\lambda(n+1)}{2}=L\\\Rightarrow \lambda=\dfrac{2L}{n+1}

Where n = 0, 1, 2, 3, ...

Making n+1 = n

\lambda=\dfrac{2L}{n}

where n = 1, 2, 3 .....

For fundamental frequency n = 1

\lambda=\dfrac{2\times 2.5}{1}\\\Rightarrow \lambda=5\ m

Frequency is given by

f=\dfrac{v}{\lambda}\\\Rightarrow f=\dfrac{344}{5}\\\Rightarrow f=68.8\ Hz

The fundamental frequency is 68.8 Hz

First overtone

2f=2\times 68.8=137.6\ Hz

Second overtone

3f=3\times 68.8=206.4\ Hz

The overtones are 137.6 Hz, 206.4 Hz

4 0
4 years ago
Define fundamental unit ? write its two example. ​
stealth61 [152]

the International System of Units, the fundamental units are: The meter (symbol: m), used to measure length. The kilogram (symbol: kg), used to measure mass. The second (symbol: s), used to measure time. The ampere (symbol: A), used to measure electric current.

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