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Gala2k [10]
3 years ago
7

Why does a hot-air balloon rise?

Physics
1 answer:
prohojiy [21]3 years ago
4 0

Answer: The weight of the air displaced by the balloon is less than the volume of the balloon.

Explanation:

A hot air balloon is a cloth wrap that contains several thousand cubic meters of air inside (a large volume of air). The burner heats the liquid propane to a gaseous state to generate a huge flame, which can reach more than 3 meters, thus heating the air mass inside the balloon. In this way,<u> its density is modified with respect to the air that surrounds it</u>, because the hot air is lighter than the outside air (less dense), causing the balloon to rise and float.

Now, if we know that the density of a body d is directly proportional to its mass m and inversely proportional to its volume V:

d=\frac{m}{V}

We can deduce that <u>by increasing the volume of the body, its density will decrease.</u>

This is proof of <em><u>Archimedes' Principle</u></em>:  

<em>A body totally or partially immersed in a fluid at rest, experiences a vertical upward thrust equal to the mass weight of the body volume that is displaced.</em>

In this case the fluid is the air outside. So, the warm air inside the balloon, being less dense, will weigh less than the outside air and therefore will receive an upward pushing force or thrust that will make the balloon ascend.

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Explanation:

As we know, Effort×effort distance = load × load distance

So when effort distance is increases,

The effort decreases

So when the spanner’s handle is long

A tight knot can easily be opened by less effrot

I hope it helped

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A planet orbits a sun in a clockwise elliptical orbit as shown in the diagram below
STatiana [176]

Answer:

Greatest gravitational energy is at "C".

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3 years ago
What momentum does a 70 kg person running 10m/s (a fast sprint) have ?
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8 0
3 years ago
If 1495 j of heat is needed to raise the temperature of a 351 g sample of a metal from 55.0°c to 66.0°c, what is the specific he
forsale [732]
The amount of heat needed to increase the temperature of a substance by \Delta T is given by
Q= mC_s \Delta T
where m is the mass of the substance, Cs is its specific heat capacity and \Delta T is the increase of temperature.

If we re-arrange the formula, we get
C_s =  \frac{Q}{m \Delta T}
And if we plug the data of the problem into the equation, we can find the specific heat capacity of the substance:
C_s =  \frac{1495 J}{(351 g)(66.0^{\circ}C-55.0^{\circ}C)}=0.39 J/g^{\circ}C
6 0
3 years ago
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