This is a blueprint drawing of the stage area at Millennium Park. The length of one square on the grid is equal to 5 feet. Accor
1 answer:
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Answer:
Explanation:
We can assume that the general formula for the drag force is given by:
And we can see that is proportional to the area. On this case we can calculate the area with the product of the width and the height. And we can express the grad force like this:
Where w is the width and h the height.
The last formula is without consider the area of the carrier, but if we use the area for the carrier we got:
If we want to find the additional power added with the carrier we just need to take the difference between the multiplication of drag force by the velocity (assuming equal velocities for both cases) of the two cases, and we got:
We can assume the same drag coeeficient and we got:
1.7 ft =0.518 m
60 mph = 26.822 m/s
In order to find the drag coeffcient we ned to estimate the Reynolds number first like this:
And the value for the kinematic vicosity was obtained from the table of physical properties of the air under standard conditions.
Now we can find the aspect ratio like this:
And we can estimate the calue of from a figure.
And we can calculate the power difference like this:
Answer:
the work done during this cooling is −28.7 kJ
Explanation:
Given data
mass (m) = 1 kg
r = 287 J/kg-K
pressure ( p) = 50 kPa
temperature (T) = 100°C = ( 100 +273 ) = 373 K
to find out
the work done during this cooling
Solution
we know the first law of thermodynamics
pv = mRT ....................1
here put value of p, m R and T and get volume v(a) when it initial stage in equation 1
50 v(a) = 1 × 0.287 × 373
v(a) = 107.051 / 50
v(a) = 2.1410 m³ .......................2
now we find out volume when temperature is 0°C
so put put value of p, m R and T and get volume v(b) when temperature is cooled in equation 1
50 v(b) = 1 × 0.287 × 273
v(a) = 78.351 / 50
v(a) = 1.5670 m³ .......................3
by equation 2 and 3 we find out work done to integrate the p with respect to v i.e.
work done =
integrate it and we get
work done = p ( v(b) - v(a) ) ................4
put the value p and v(a) and v(b) in equation 4 and we get
work done = p ( v(b) - v(a) )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 ( 1.5670 - 2.1410 )
work done = 50 (−0.574)
work done = −28.7 kJ
here we can see work done is negative so its mean work done opposite in direction of inside air
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