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Salsk061 [2.6K]
3 years ago
8

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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A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed
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we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

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Explanation:

A student throws a small rock straight upwards. The rock rises to its highest point and then falls back down. How does the speed of the rock at 2m on the way down compare with its speed at 2m on the way up?

It decreases in speed on its way down and increases in speed on its way down.

it decreases in speed on its way up because the the vertical motion is against the earths gravitational pull on an object to the earth's center

.It increases in speed on his way down because its under the influence of gravity

from newton's equation of motion we can check by

using V^2=u^2+2as

we assume that it starts with a velocity of 10m/s. At 2m height above ground level, its velocity decreases at 3m above ground level

for its way down the velocity at 3m on its way down is more than its velocity at 2m on its way down.

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2 years ago
In a hydroelectric power plant, water flows from an elevation of 400 ft to a turbine, where electric power is generated. For an
VashaNatasha [74]

Answer:

V=3.475ft^3/s=3.48ft^3/s

Explanation:

We have here values from SI and English Units. I will convert the units to English Units.

We hace for the power P,

P= 100kW \rightarrow 100kW*(\frac{737.56ft.lbf/s}{1kW})

P= 73.7*10^{3}ft.lbf/s

we have other values such h=400ft and  \gamma= 62.42lb/ft^3 (specific weight of the water), and 0.85 for \eta

We need to figure the flow rate of the water (V) out, that is,

V=\frac{P}{\gamma h \eta_0}

Where \eta_0 is the turbine efficiency, at which is,

\eta_0 = \frac{P}{\gamma Vh}

Replacing,

V=\frac{73.7*10^{3}}{62.42*400*0.85}

V=3.475ft^3/s

With this value (the target of this question) we can also calculate the mass flow rate of the waters,

through the density and the flow rate,

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converting the slugs to lbm, 1slug = 32.174lbm, we have that the mass flow rate of the water is,

m= 217lbm/s

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2 years ago
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