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Lina20 [59]
3 years ago
10

Is the housekeeping satisfactory? ​

Engineering
2 answers:
inessss [21]3 years ago
6 0

Answer:

yes

Explanation:

Nat2105 [25]3 years ago
3 0

Answer:

yess

Explanation:

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What is the primary function of NCEES?
charle [14.2K]

Answer:

It is a non profit organization that dedicates to licensing professional engineers and surveyors

Explanation:

6 0
3 years ago
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

  = 0.0214 * 3.738

  = 0.0792 ohm.

Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
4 years ago
Match the following parts of a crane
trasher [3.6K]
It is auxillary sorry i couldn’t help it happens to the best of us
5 0
3 years ago
Now that we have a second enemy, you will need to make some changes to the script that is attached to your backdrop. Look at tha
JulsSmile [24]

Answer:

<u><em>≡</em></u>

Explanation:

8 0
3 years ago
a cubical box 20-cm on a side is contructed from 1.2 cm thick concrete panels. A 100-W light bulb is sealed inside the box. What
Flura [38]

Answer:

Temperature on the inside ofthe box

Explanation:

The power of the light bulb is the rate of heat conduction of the bulb, dq/dt = 100 W

The thickness of the wall, L = 1.2 cm = 0.012m

Length of the cube's side, x = 20cm = 0.2 m

The area of the cubical box, A = 6x²

A = 6 * 0.2² = 6 * 0.04

A = 0.24 m²

Temperature of the surrounding, T_0 = 20^0 C = 273 + 20 = 293 K

Temperature of the inside of the box, T_{in} = ?

Coefficient of thermal conductivity, k = 0.8 W/m-K

The formula for the rate of heat conduction is given by:

dq/dt = \frac{kA(T_{in} - T_0)}{L} \\\\100 = \frac{0.8*0.24(T_{in} - 293)}{0.012}\\\\T_{in} - 293 = \frac{100 * 0.012}{0.8*0.24} \\\\T_{in} - 293 = 6.25\\\\T_{in} = 293 + 6.25\\\\T_{in} = 299.25 K\\\\T_{in} = 299.25 - 273\\\\T_{in} = 26.25^0 C

5 0
4 years ago
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