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3 years ago

What were the two different alleles for height in the pea plants that Mendel studied?

1 answer:

8 0

You might want to go to this website, http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P

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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

$P_r = 0.63$

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

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

6 0

4 years ago

at a major league baseball game, you are sitting a distance 52 meters from home plate. how much time passes between seeing josh

brilliants [131]

sound travels through air at around 340 - 350 m/s.

to cover 52 meters takes it, say, (52/345) = about 0.15 second

This would be the same time-lag even if it were not a major-league game,
except that you might not be able to get a seat 52 meters from home plate
at a little league or t-ball game.

8 0

4 years ago

Which shared property belongs in the region marked "B"?

NNADVOKAT [17]

The properti that you didnt share

4 0

3 years ago

Discuss the factors that are used to determine power.

True [87]

<span>The factors that are used to determine power are:
Voltage,current and the power factor.

</span><span>Power = Voltage x Current x K
Watts = Volts x Amps x Power Factor</span>

5 0

4 years ago

Read 2 more answers

Mathphys :( im sorry i annoy you

Vitek1552 [10]

Answer:

4. 7.59276

Explanation:

Add up the x components:

Aₓ + Bₓ + Cₓ = 5 − 1.6 + 2.4 = 5.8

Add up the y components:

Aᵧ + Bᵧ + Cᵧ = -2.4 + 3.3 + 4 = 4.9

Use Pythagorean theorem to find the magnitude:

√(x² + y²)

√(5.8² + 4.9²)

√57.65

7.59276

3 0

3 years ago