Answer: that’s chemistry not physics
Explanation:
Answer:
If the same volume of air is inhaled and exhaled, the air we breathe out normally weighs more than the air we breathe in.
Since the output from the body normally exceeds the input, breathing leads to weight loss.
Explanation:
If equal volumes of gas is inhaled and exhaled, the exhaled gas is heavier.
The inhaled gas contains Oxygen and majorly Nitrogen.
The exhaled gas contains CO₂, H₂O and a very large fraction of the unused inhaled air that goes into the lungs.
So, basically, the body exchanges O₂ with CO₂ and H₂O (and some other unwanted gases in the body) in a composition that CO₂, the heavier gas of the ones mentioned here, is prominent.
So, because the mass leaving the body is more than the mass entering, breathing leads to a loss of weight. This is one of the reasons why we need food for sustenance. Breathing alone will wear one out.
Answer:
Temperature of the air in the balloon = 272°C
Explanation:
Given:
Volume of balloon = 500 m³
Air temperature = 15° C = 273 + 15 = 288 K
Total weight = 290 kg
Density of air = 1.23 kg/m³
Find:
Temperature of the air in the balloon
Computation:
Density of hot air = Density of air - [Total weight / Volume of balloon]
Density of hot air = 1.23 - [290 - 500]
Density of hot air = 0.65 kg/m³
[Density of hot air][Temperature of the air in the balloon] = [Density of air][Air temperature ]
Temperature of the air in the balloon = [(1.23)(288)]/(0.65)
Temperature of the air in the balloon = 544.98
Temperature of the air in the balloon = 545 K
Temperature of the air in the balloon = 545 - 273 = 272°C
Answer:
5.65 times
Explanation:
60 db sound is equal to 60 phons sound when frequency is kept at 1000Hz.
But when the frequency of sound is changed to 100 Hz , according to equal loudness curves , the loudness level on phon scale will be 35 phons.
A decrease of 10 phon on phon- scale makes sound 2 times less loud
Therefore a decrease of 25 phons will make loudness less intense by a factor equal to 2²°⁵ or 5.65 less intense . Therefore intensity at 100 Hz
must be increased 5.65 times so that its intensity matches intensity of 60 dB sound at 1000 Hz frequency.
