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
The minimum frequency is
while the maximum frequency is
Using the relationship between frequency f of a wave, wavelength
and the speed of the wave v, we can find what wavelength these frequencies correspond to:
So, the wavelengths of the radio waves of the problem are within the range 188-545 m.
In terms of Snapchat streaks...
This is a number of days two users have been sending a picture or video to each other.
In the general term of a streak...
An irregular line left from smearing or a continuous series of like events.
It all depends on the context.
Answer:
The width of the slit is 0.167 mm
Explanation:
Wavelength of light, 
Distance from screen to slit, D = 88.5 cm = 0.885 m
The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m
We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :
where
a = width of the slit
a = 0.000167 m
a = 0.167 mm
So, the width of the slit is 0.167 mm. Hence, this is the required solution.
Answer:
19.6 cm.
Explanation:
From the question given above, the following data were obtained:
Focal length (f) = 13.8 cm
Magnification (M) = +2.37
Object distance (u) =.?
Next, we shall determine the image distance. This can be obtained as follow:
Magnification (M) = +2.37
Object distance (u) = u
Image distance (v) =?
M = v / u
2.37 = v / u
Cross multiply
v = 2.37 × u
v = 2.37u
Finally, we shall determine the object distance. This can be obtained as follow:
Focal length (f) = 13.8 cm
Image distance (v) = 2.37u
Object distance (u) =.?
1/v + 1/u = 1/f
vu / v + u = f
2.37u × u / 2.37u + u = 13.8
2.37u² / 3.37u = 13.8
Cross multiply
2.37u² = 3.37u × 13.8
2.37u² = 46.506u
Divide both side by u
2.37u² / u = 46.506u / u
2.37u = 46.506
Divide both side by 2.37
u = 46.506 / 2.37
u = 19.6 cm
Thus, the lens should be held at a distance of 19.6 cm.
