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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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Answer: have a subject and a verb

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What is the resolution of an analog-to-digital converter with a word length of 12 bits and an analogue signal input range of 100
Elena L [17]

Answer:

The resolution of an analog-to-digital converter is 24.41 mV

Explanation:

Resolution of an analog-to-digital  = (analogue signal input range)/2ⁿ

where;

n is the number or length of bit, and in this question it is given as 12

Also, the analogue signal input range is 100V

Resolution of an analog-to-digital  = 100V/2¹²

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To calibrate your calorimeter cup, you first put 45 mL of cold water in the cup, and measure its temperature to be 24.7 °C. You
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Answer : The heat change of the cold water in Joules is, 1.6\times 10^3J

Explanation :

First we have to calculate the mass of cold water.

As we know that the density of water is 1 g/mL. The volume of cold water is 45 mL.

Density=\frac{Mass}{Volume}

Mass=Density\times Volume=1g/mL\times 45mL=45g

Now we have to calculate the heat change of cold water.

Formula used :

Q=m\times c\times (T_2-T_1)

where,

Q = heat change of cold water = ?

m = mass of cold water = 45 g

c = specific heat of water = 4.184J/g^oC

T_1 = initial temperature of cold water = 24.7^oC

T_2 = final temperature  = 33.4^oC

Now put all the given value in the above formula, we get:

Q=45g\times 4.184J/g^oC\times (33.4-24.7)^oC

Q=1638.036J=1.6\times 10^3J

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Answer:

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