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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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Determine the thrust produced if 1.5 x 10^3 kg of gas exits the combustion chamber each second, with a speed of 4.00 x 10^3 m/s.
ozzi

Answer:

The thrust is 6\times 10^6\ N

Explanation:

Given that,

Mass of gas, m=1.5\times 10^3\ kg

The rate at which the gas is expelling, \dfrac{dv}{dt}=4\times 10^{3}\ m/s

We need to find the thrust produced by the gas.

We know that force is equal to the rate of change of momentum. So,

F=\dfrac{p}{t}

Also, p = mv

F=\dfrac{mv}{t}

So,

F=1.5\times 10^3\times 4\times 10^3\\\\F=6\times 10^6\ N

So, the thrust is 6\times 10^6\ N

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3 years ago
If you were capable of converting mass to energy with 100%, efficiency, how much mass would you need to produce 3.5x10^12 Joules
Alexeev081 [22]

Answer:

a) 3.9 x 10⁻⁵ kg

Explanation:

The amount of mass required to produce the energy can be given by Einstein's formula:

E = mc^2\\\\m = \frac{E}{c^2}

where,

m = mass required = ?

E = Energy produced = 3.5 x 10¹² J

c = speed of light = 3 x 10⁸ m/s

Therefore,

m = \frac{3.5\ x\ 10^{12}\ J}{(3\ x\ 10^8\ m/s)^2} \\\\m = 3.9\ x\ 10^{-5}\ kg

Hence, the correct option is:

<u>a) 3.9 x 10⁻⁵ kg</u>

7 0
3 years ago
A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
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Answer

given,

mass of the ball = 3 kg

swing in vertical circle with radius = 2 m

   work done by the gravity = ?          

   work done by the tension = ?            

Work done by the gravity = - m g Δh            

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Work done by the gravity =- 3 \times 9.8 \times 4

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work done by gravity is equal to -117.6 J            

Work done by tension will be equal to zero.        

Zero because tension is always perpendicular to velocity

work done by tension is equal to 0 J                          

7 0
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