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hram777 [196]
3 years ago
12

If a range hood removes contaminants at a flow rate of F liters of air per​ second, then the percent P of contaminants that are

also removed from the surrounding air can be modeled by Upper P equals 1.06 Upper F plus 22.18P=1.06F+22.18​, where F is between and inclusive of 1010 and 7070. What are the minimum and maximum values for P in this​ model?
Physics
1 answer:
podryga [215]3 years ago
3 0

Explanation:

The given value of P is as follows.

           P = 1.06F + 22.18,           10 \leq F \leq 70

or,       P' = 1.06

As p' is defined and non-zero. Hence, only critical points are boundary points.

For F = 10, the value of P will be calculated as follows.

         P = 1.06 \times 10 + 22.18

            = 32.78

For F = 70, the value of P will be calculated as follows.

        P = 1.06 \times 70 + 22.18

           = 96.38

Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.

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To answer this problem, we will use the equations of motions.

Part (a):
For the ball to start falling back to the ground, it has to reach its highest position where its final velocity will be zero.
The equation that we will use here is:
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v is the final velocity = 0 m/sec
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t is the time that we want to find.
Substitute in the equation to get the time as follows:
v = u + at
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t = 160/9.8 = 16.3265 sec
Therefore, the ball would take 16.3265 seconds before it starts falling back to the ground

Part (b):
First, we will get the total distance traveled by the ball as follows:
s = 0.5 (u+v)*t
s = 0.5(160+0)*16.3265
s = 1306.12 meters
The equation that we will use to solve this part is:
v^2 = u^2 + 2as where
v is the final velocity we want to calculate
u is the initial velocity of falling = 0 m/sec (ball starting falling when it reached the highest position, So, the final velocity in part a became the initial velocity here)
a is acceleration due to gravity = 9.8 m/sec^2 (positive as ball is moving downwards)
s is the distance covered = 1306.12 meters
Substitute in the above equation to get the final velocity as follows:
v^2 = u^2 + 2as
v^2 = (0)^2 + 2(9.8)(1306.12)
v^2 = 25599.952 m^2/sec^2
v = 159.99985 m/sec
Therefore, the velocity of the ball would be 159.99985 m/sec when it hits the ground.
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