Answer:
Rectangular path
Solution:
As per the question:
Length, a = 4 km
Height, h = 2 km
In order to minimize the cost let us denote the side of the square bottom be 'a'
Thus the area of the bottom of the square, A = 
Let the height of the bin be 'h'
Therefore the total area, 
The cost is:
C = 2sh
Volume of the box, V =
(1)
Total cost,
(2)
From eqn (1):

Using the above value in eqn (1):


Differentiating the above eqn w.r.t 'a':

For the required solution equating the above eqn to zero:


a = 4
Also

The path in order to minimize the cost must be a rectangle.
Answer:
P = 147,75 W
Explanation:
A man whose mass is 59.1kg climbs up 30 steps of a stair each step is 25 cm high
Height at 30 steps , h=30×2.5= 7.5 m
Change in potential energy , =mgh=59.1×10×7.5 = 4432.5 J
So, Work done by the man , W= 4432.5J
Power used , P= 
P = 4432.5 /30
P = 147,75 W
Solve any question of Work, Energy and Power with:-
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The answer is 17 m because you have to add the 15 m and the 8 m together to get the answer so it will be like this 17x17 = 15x15 + 8x8 got it?
To develop this problem we will start from the definition of entropy as a function of total heat, temperature. This definition is mathematically described as

Here,
Q = Total Heat
T = Temperature
The total change of entropy from a cold object to a hot object is given by the relationship,

From this relationship we can realize that the change in entropy by the second law of thermodynamics will be positive. Therefore the temperature in the hot body will be higher than that of the cold body, this implies that this term will be smaller than the first, and in other words it would imply that the magnitude of the entropy 'of the hot body' will always be less than the entropy 'cold body'
Change in entropy
is smaller than 
Therefore the correct answer is C. Will always have a smaller magnitude than the change in entropy of the cold object
It is determined by the nature of the green light. Because lasers create light at almost a single frequency, green laser light would appear as a thin line of pure green. Other sources of "green" light emit light at a variety of frequencies, including yellow and blue, resulting in a strong green band in the center that fades into blue-green and yellow-green at the borders.
For example, here’s a graph of the spectrum of a green LED, showing the color range: Attachment #1
and here’s a graph of the transmission spectra of several standard photographic filters, including green: Attachment #2
Learn more about the color spectrum:
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