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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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A 750 watt hairdryer is used for 60 seconds. how many joules of energy are used?
daser333 [38]
We know, Power = Energy / time
Here, P = 750 Watt
t = 60 sec

Substitute their values, 
750 = E / 60
E = 750 * 60
E = 45,000 J

In short, Your Answer would be 45,000 Joules

Hope this helps!
4 0
4 years ago
Read 2 more answers
Suppose 1.4 mol of an ideal gas is taken from a volume of 2.5 m3 to a volume of 1.0 m3 via an isothermal compression at 27°C. (a
zysi [14]

Answer:

Part a)

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

Explanation:

Part a)

Energy transfer during compression of gas is same as the work done on the gas

In isothermal process work done is given by the equation

W = nRT ln(\frac{V_2}{V_1})

now we know that

n = 1.4 moles

T = 27 degree C = 300 K

V_2 = 2.5 m^3

V_1 = 1 m^3

now we have

W = (1.4)(8.31)(300)(ln\frac{2.5}{1})

Q = 3198 J

Part b)

It is compression of gas so this is energy transferred to the gas

6 0
4 years ago
What are wind names according to direction
Vsevolod [243]

North, East, South, West (in order)

8 0
3 years ago
Two charges, each 9 µC, are on the x axis, one at the origin and the other at x = 8 m. Find the electric field on the x axis at
bearhunter [10]

a) Electric field at x = -2 m: 21,060 N/C to the left

b) Electric field at x = 2 m: 18,000 N/C to the right

c) Electric field at x = 6 m: 18,000 N/C to the left

d) Electric field at x = 10 m: 21,060 N/C to the right

e) Electric field is zero at x = 4 m

Explanation:

a)

The electric field produced by a single-point charge is given by

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

Here we have two charges of

q=9\mu C = 9\cdot 10^{-6} C

each. Therefore, the net electric field at any point in the space will be given by the vector sum of the two electric fields. The two charges are both positive, so the electric field points outward of the charge.

We call the charge at x = 0 as q_0 , and the charge at x = 8 m as q_8.

For a point located at x = -2 m, both the fields E_0 and E_8 produced by the two charges point to the left, so the net field is the sum of the two fields in the negative direction:

E=-\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=-kq(\frac{1}{(-2)^2}+\frac{1}{(8-(-2))^2})=-21060 N/C

b)

In this case, we are analyzing a point located at

x = 2 m

The field produced by the charge at x = 0 here points to the right, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(2)^2}-\frac{1}{(8-2)^2})=18000 N/C

And since the sign is positive, the direction is to the right.

c)

In this case, we are considering a point located at

x = 6 m

The field produced by the charge at x = 0 here points to the right again, while the field produced by the charge at x = 8 m here points to the left. Therefore, the net field is given by the difference between the two fields, as before; so:

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(6)^2}-\frac{1}{(8-6)^2})=-18000 N/C

And the negative sign indicates that the electric field in this case is towards the left.

d)

In this case, we are considering a point located at

x = 10 m

This point is located to the right of both charges: therefore, the field produced by the charge at x = 0 here points to the right, and the field produced by the charge at x = 8 m here points to the right as well. Therefore, the net field is given by the sum of the two fields:

E=\frac{kq_0}{(0-x)^2}+\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(10)^2}+\frac{1}{(8-(10))^2})=21060 N/C

And the positive sign means the field is to the right.

e)

We want to find the point with coordinate x such that the electric field at that location is zero. This point must be in between x = 0 and x = 8, because that is the only region where the two fields have opposite directions. Therefore, te net field must be

E=\frac{kq_0}{(0-x)^2}-\frac{kq_8}{(8-x)^2}=kq(\frac{1}{(-x)^2}-\frac{1}{(8-x)^2})=0

This means that we have to solve the equation

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0

Re-arranging it,

\frac{1}{x^2}-\frac{1}{(8-x)^2}=0\\\frac{(8-x)^2-x^2}{x^2(8-x)^2}=0

So

(8-x)^2-x^2=0\\64+x^2-16x-x^2=0\\64-16x=0\\64=16x\\x=4 m

So, the electric field is zero at x = 4 m, exactly halfway between the two charges (which is reasonable, because the two charges have same magnitude)

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

6 0
3 years ago
Read 2 more answers
Someone, please help!! cant understand
ch4aika [34]

Answer:

4400 N

Explanation: (c)

P= \frac{F}{A}

F= PA

P= pressure

A= area

P=800x10^{3\\ and A= 5.5 x 10^{-3}m^{2}

Force F= 800x10^{3}x0.0055= 4400 N

force exerted by the gas on the piston is 4400 N

3 0
3 years ago
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