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ValentinkaMS [17]
3 years ago
6

While traveling along a highway, a driver slows from 27 m/s to 13 m/s in 11 seconds. What is the automobile’s acceleration?

Physics
1 answer:
Travka [436]3 years ago
4 0

Answer:

a=-1.27\ m/s^2

Explanation:

Given that,

The initial speed of a driver, u = 27 m/s

Final speed of the driver, v = 13 m/s

Time, t = 11 s

We need to find the automobile's acceleration. We know that the acceleration of an object is equal to the change in speed divided by time taken.

a=\dfrac{13-27}{11}\\\\a=-1.27\ m/s^2

So, the magnitude of automobile's acceleration is 1.27\ m/s^2.

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A student creates interest in a visualization by contrasting the amount of light and dark. what color element is used?
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A student creates interest in a visualization by contrasting the amount of light and dark. What color element is used?Grayscale
3 0
3 years ago
Why can't 41 be represented in a binary table?
Airida [17]

Answer:

I think it is but I don't know for sure

Explanation:

41 101001

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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
Can you please solve this for me urgently want to make sure if my answers are correct?
MA_775_DIABLO [31]

Answer:

<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

therefore, for resultant:

d =  \sqrt{ {d _{y} }^{2} + d _{x}  {}^{2}  }

substitute:

d =  \sqrt{ {( - 5.702)}^{2} +  {( - 11.476)}^{2}  }  \\  \\ d =  \sqrt{164.211}  \\  \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}

6 0
3 years ago
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anastassius [24]

Answer: 0K

Explanation:

Absolute 0 (0K) is the point where nothing could be colder and no heat energy remains in a substance.

7 0
3 years ago
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