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tatiyna
2 years ago
6

I need the answer with steps

Physics
1 answer:
Elis [28]2 years ago
6 0

Answer: What Question

Explanation:

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A device that uses electricity and magnetism to create motion is called a _________motor,magnet,generator . In a reverse process
lawyer [7]
A device that uses electricity and magnetism to create motion is called a "Motor" (which converts electric energy into mechanical energy) & <span>In a reverse process, a device that uses motion and magnetism can be used to create "Electromagnetism".

In short, 1st Blank = Motor
2nd Blank = Electromagnetism 

Hope this helps!</span>
5 0
3 years ago
Read 2 more answers
Explain what the ionsphere is and how it interacts with some radio waves
mrs_skeptik [129]

Answer:

not sure

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

6 0
3 years ago
During a medieval siege of a castle, the attacking army uses a trebuchet to heavy stones at the castle the trebuchet launches th
vlada-n [284]

Answer:

Explanation:

The question relates to time of flight of a projectile .

Time of flight = 2 u sinθ / g

u is speed of projectile , θ is angle of projectile

= 2 x 48.5 sin42 / 9.8

= 6.6 seconds  .

Maximum height attained

= u² sin²θ / g

= 48.5² sin²42 / 9.8

= 107.47 m .

7 0
3 years ago
Help can’t find the answer no where
katen-ka-za [31]

i'm stuck on that question also

5 0
3 years ago
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