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3 years ago

When a balloon is deflating, why does air leave the balloon?

Physics

1 answer:

Gwar [14]3 years ago

6 0

Answer:

<em>When a balloon deflates air moves out of the balloon </em><em>because the pressure inside the balloon is higher than the pressure outside the balloon.</em>

Explanation:

An inflated balloon has a high pressure region on its inside. Gases always move from a region of high pressure to a region of low pressure. When a balloon is inflated its membrane stretches making it even more porous.

The gas molecules inside the balloon easily diffuse out through this membrane. The diffusion rate may differ depending on the type of gas filled inside the balloon and the material of the balloon. For example helium balloon deflates faster than common air balloon.

This is because helium is a light element and can escape easier than gases like nitrogen and oxygen through the porous membrane of the balloon.

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Conditions of equilibrium of parallel coplanar forces<br><br><br>

IRISSAK [1]

Answer:

Bruh............

Explanation:

8 0

3 years ago

brainly an andean condor with a wingspan and a mass soars along a horizontal path. model its wings as a rectangle with a width.

Anna35 [415]

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is 101,204 Pa.

<h3>Difference in pressure between the top and bottom of the wingspan</h3>

The difference in pressure between the top and bottom of the wingspan is calculated as follows;

ΔP = P(top) - P(bottom)

<h3>Area of the wingspan</h3>

A = bh

A = 2.7 m x 0.27 m

A = 0.729 m²

<h3>Weight of the Andean condor</h3>

W = mg

W = 9 x 9.8

W = 88.2 N

<h3>Pressure at the top surface of condor's wings</h3>

The pressure at the top surface of condor's wings is due to atmospheric pressure

P(top) = 14.7 Psi = 101,325 Pa

<h3>Pressure at the bottom surface of condor's wings</h3>

The pressure at the bottom surface is due to weight of andean condor.

P = W/A

P(bottom) = 88.2/0.729

P(bottom) = 120.99 Pa

The difference between the pressure at the top surfaces of the condor's wings and the pressure at the bottom surfaces is calculated as;

ΔP = P(top) - P(bottom)

ΔP = 101,325 Pa - 120.99 Pa

ΔP = 101,204 Pa

The complete question is below;

An Andean condor with a wingspan of 270 cm and a mass of 9.00 kg soars along a horizontal path. Model its wings as a rectangle with a width of 27.0 cm.

Learn more about pressure here: brainly.com/question/25736513

5 0

2 years ago

Find the potential inside and outside a uniformly charged solid sphere whose radius is R and whose total charge is q. Use infini

amid [387]

Answer:

Recall that the electric field outside a uniformly charged solid sphere is exactly the same as if the charge were all at a point in the centre of the sphere:

$E_{outside} =\frac{1}{4\pi(e_{0})}\frac{Q}{r^{2} } r^{'}$

lnside the sphere, the electric field also acts like a point charge, but only for the proportion of the charge further inside than the point r:

$E_{inside} =\frac{1}{4\pi(e_{0})}\frac{Q}{R^{2} } \frac{r}{R} r^{'}$

To find the potential, we integrate the electric field on a path from infinity (where of course, we take the direct path so that we can write the it as a 1 D integral):

$V(r>R)=\int\limits^r_\infty {\frac{1}{4\pi(e_{0)} }\frac{Q}{r^2} } \, dr=\frac{q}{4\pi(e_{0)} } \frac{1}{r} \\V(r$