The faster a string vibrates, the higher frequency sound it produces,
and the higher the pitch is that we hear.
This question involves the concepts of work done, pressure, and temperature.
The work done per mol of the air is "-724.71 J/mol".
Using the given formula for work done:

where,
W = work done per mol = ?
R = universal gas constant = 8.314 J/mol.k
T = absolute temperature = 30°C + 273 = 303 k
P₁ = initial pressure = 30 KPa
P₂ = final pressure = 40 KPa
Therefore,

<u>W = -724.71 J/mol</u>
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Power is the rate work done given by dividing work done by unit time. It is measured in watts equivalent to J/s.
In this case the force by the student is mg = 490 N (taking g as 9.8m/s²)
Work done is given by force × distance,
Therefore, Power =(force × distance)/ time, but velocity/speed =distance/time
Thus, Power = force × speed/velocity
= 490 N × 1.25
= 612.5 J/S (Watts)
Hence, power will be 612.5 Watts.
To solve the problem it is necessary to apply the concepts related to the conservation of energy through the heat transferred and the work done, as well as through the calculation of entropy due to heat and temperatra.
By definition we know that the change in entropy is given by

Where,
Q = Heat transfer
T = Temperature
On the other hand we know that by conserving energy the work done in a system is equal to the change in heat transferred, that is

According to the data given we have to,




PART A) The total change in entropy, would be given by the changes that exist in the source and sink, that is



On the other hand,



The total change of entropy would be,



Since
the heat engine is not reversible.
PART B)
Work done by heat engine is given by



Therefore the work in the system is 100000Btu
Answer:
Behaviour and uses of electromagnetic waves
Radio waves. Radio waves are used for communication such as television and radio. ...
Microwaves. Microwaves are used for cooking food and for satellite communications. ...
Infrared. ...
Visible light. ...
Ultraviolet radiation.