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DanielleElmas [232]
3 years ago
9

Technician A says that 18 gauge AWG wire can carry more current flow that 12 gauge AWG wire. Technician B says that metric wire

is sized by its cross-sectional area. Who is correct? Select one: a. Both Technicians A and B O b. Neither Technician A nor B C. Technician A d. Technicain B​
Engineering
1 answer:
denpristay [2]3 years ago
6 0

Answer:

Technician B

Explanation:

Both AWG and metric are sized by cross-sectional area.

Technician A is wrong:  12 gauge wire is larger diameter rated for 20 amps in free air.  18 awg is smaller diameter and typically used for speaker wiring, Class II or low voltage and sub-circuits within appliances.

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8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of
san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb=  225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material  K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

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
3 years ago
One of the flaws in the engineers' reasoning for galloping gertie's design was that they attributed prior failures of suspension
Jlenok [28]
False i think it would be
3 0
3 years ago
Steam flowing through a long, thin walled pipe maintains the pipe wall at a uniform temperature of 500 K. The pipe is covered wi
hjlf

Answer: I would love to learn this

Explanation:

3 0
3 years ago
A Venturi meter (see below) uses the principles of the Bernoulli equation to measure the velocity of a flow in a reactor system.
LenaWriter [7]

Answer:

Q=0.000604 m³/s

Explanation:

Given that

d₁=5 cm

d₂=1 cm

P= 30 KPa

Density of water ,ρ=1000 kg/m³

As we know that volume flow rate Q given as

Q=A_1A_2\sqrt{\dfrac{\dfrac{2\Delta P}{\rho}}{A_1^2-A_2^2}}

A_1=\dfrac{\pi}{4}\times 0.05^2\ m^2

A₁=0.0019 m²

A_2\dfrac{\pi}{4}\times 0.01^2\ m^2

A₂=0.000078 m²

Q=0.0019 \times 0.000078 \sqrt{\dfrac{\dfrac{2\times 30\times 1000}{1000}}{0.0019^2-0.000078^2}}\ m^3/s

Q=0.000604 m³/s

7 0
3 years ago
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