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Tpy6a [65]
3 years ago
6

A group of Advanced Open Water Divers plans to make two dives. The first dive is on a reef in 90 feet of water for 20 minutes. T

he group then remains on the surface for 1 hour. The second dive is on a wreck in 60 feet of water, with a planned bottom time of 30 minutes. What will be the ending pressure group after the second dive
Physics
1 answer:
Kaylis [27]3 years ago
5 0

Answer:

The ending pressure is R.

Explanation:

This is Recreational scuba diving (R). According to standards, this diving has prescribed depth limit of 130 ft. This means that the limit must include a depth that is not greater than 130 ft while using only compressed air and not even requiring a decompression stop.

Now, from the question, we can see that that there was a prescribed limit of 60ft of water with a planned bottom time of 30 minutes.

Thus, the ending pressure is R.

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Help me please, i need it fast
s2008m [1.1K]

Answer:

2)  C.  (x - 3)² + (y + 2)² = 25

5)   x² +  y² - 8x - 16y + 54 = 0

6)   x² + y²  - 10x - 12y + 36  = 0

Explanation:

2)

center of circle = 3, -2

                            x1, y1

end point of circle = 7, 1

                                 x2, y2

 

the equation of a circle is Pythagorean theorem

x² + y² = r²    (where r is the radius of a circle)

distance between points  

(x2 - x1)² + (y2 - y1)² = r²

(7 - 3)² + (1 - (-2))² = r²

r² = 25

therefore the equation to the circle is

(x - 3)² + (y + 2)² = 25

=========================================

5)

write the general form of a circle with the center (4,8)

and containing the point (-1, 7)

distance between points  

(x2 - x1)² + (y2 - y1)² = r²

(-1 - 4)² + (7 - 8)² = r²

r² = 26

(x - 4)² + (y - 8)² = 26

(x - 4)(x - 4) +  (y - 8)(y - 8) = 26

x² - 8x + 16 + y² - 16y + 64 -26 = 0

x² +  y² - 8x - 16y + 54 = 0

=========================================

6)

find the general form of a circle with center (5,6)

and tangent to the y-axis.

           

center (5,6)

           h, k

radius = r²

r = 5

(x - h)² + (y - k)² = r²

(x - r)² + (y - k)² = r²

(x - 5)(x - 5) + (y - 6)(y - 6) = r²

x² - 10x + 25 + y² - 12y + 36 = 25

x² + y²  - 10x - 12y + 36  = 0

=========================================

8 0
2 years ago
3.1 Two waves A and B have frequencies 256 Hz and 1024 Hz respectively have amplitude in ratio 3:1 1.What is their ratio of freq
scoundrel [369]

Answer:

1. the one with the raito

2. the one that stubbed their toe

Explanation:

8 0
3 years ago
An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst
sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

T_1 = 500°C

T_2 = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to T_{avg} = 535.5K \approx 550K

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

P_r = 0.63

A)

Number for the first strips is equal to

R_e_x = \frac{u_o.L}{v}

R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4

Calculating heat transfer coefficient from the first strip

h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3

h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2

The rate of convection heat transfer from the first strip is

q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W

The rate of convection heat transfer from the fifth trip is equal to

q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)

h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2

Calculating h_o_-_4

h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2

The rate of convection heat transfer from the tenth strip is

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)

h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2

Calculating

h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W

The rate of convection heat transfer from 25th strip is equal to

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)

Calculating h_o_-_2_5

h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2

Calculating h_o_-_2_4

h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2

Calculating the rate of convection heat transfer from the tenth strip

q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W

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