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muminat
2 years ago
7

A heating element for a cooking appliance is stretched too far during installation. What action can be performed? A. Dispose of

the element and purchase a new one. B. Chill the element in a freezer to shrink it. C. Cut the element to shorten it. D. Recoil the element over a rod.​
Engineering
1 answer:
Liula [17]2 years ago
3 0

Because the heating element for a cooking appliance is stretched too far during installation, Dispose of the element and purchase a new one.

<h3>What is meant by heating element?</h3>

The Definition of heating element is known to be the aspect of an electric heating appliance where the electrical energy is said to be changed to heat.

Note that Because the heating element for a cooking appliance is stretched too far during installation, Dispose of the element and purchase a new one.

Learn more about  heating element from

brainly.com/question/4428189

#SPJ1

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A well-insulated tank in a vapor power plant operates at steady state. Saturated liquid water enters at inlet 1 at a rate of 125
kompoz [17]

Answer:

a. The mass flow rate (in lbm/s) is 135lbm/s

b. The temperature (in o F) is 200.8°F

Explanation:

We assume that potential energy and kinetic energy are negligible and the control volume operates at a steady state.

Given

a. The mass flow rate (in lbm/s) is 135lbm/s

b.

m1 = Rate at inlet 1 = 125lbm/s

m2 = Rate at inlet 2 = 10lbm/s

The mass flow rate (in lbm/s) is calculated as m1 + m2

Mass flow rate = 125lbm/s + 10lbm/s

Mass flow rate = 135lbm/s

Hence, the mass flow rate (in lbm/s) is 135lbm/s

b. To calculate the temperature.

First we need to determine the enthalpy h1 at 14.7psia

Using table A-3E (thermodynamics)

h1 = 180.15 Btu/Ibm

h2 at 14.7psia and 60°F = 28.08 Btu/Ibm

Calculating h3 using the following formula

h3 = (h1m1 + h2m2) / M3

h3 = (180.15 * 125 + 28.08 * 10)/135

h3 = 168.8855555555555

h3 = 168.89 Btu/Ibm

To get the final temperature; we make use of table A-2E of thermodynamics.

Because h3 < h1, it means the liquid is at a compressed state.

The corresponding temperature at h3 = 168.89 is 200.8°F

The temperature (in o F) is 200.8°F

6 0
3 years ago
Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 k
In-s [12.5K]

Consider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pmConsider an area-source box model for air pollution above a peninsula of land. The length of the box is 15 km, its width is 80 km, and a radiation inversion restricts mixing to 15 m. Wind is blowing clean air into the long dimension of the box at 0.5 m/s. On average, there are 250,000 vehicles on the road, each being driven 40 km in 2 hours and each emitting 4 g/km of CO.

Required:

a. Estimate the steady-state concentration of CO in the air. Should the city be designated as "nonattainment" (i.e., steady-state concentration is over the NAAQS standard)?

b. Find the average rate of CO emissions during this two-hour period.

c. If the windspeed is zero, use the formula to derive relationship between CO and time and use it to find the CO over the peninsula at 6pm

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3 0
3 years ago
An aluminium alloy tube has a length of 750 mm at a temperature of 223°C. What will be its length at 23°C if its coefficient of
uranmaximum [27]

Answer:

Final length= 746.175 mm

Explanation:

Given that Length of aluminium at 223 C is 750 mm.As we know that when temperature of material increases or decreases then dimensions of material also increases or decreases respectively with temperature.

Here temperature of aluminium decreases so the final length of aluminium decreases .

As we know that

\Delta L=L\alpha\Delta T

Now by putting the values

\Delta L=750\times \25.5\times 10^{-6}\times 200

ΔL=3.82 mm

So final length =750-3.82 mm

Final length= 746.175 mm

3 0
3 years ago
Is a jeep cherokee faster than a bmw 325i
soldier1979 [14.2K]

Answer:

Yes

Explanation:

8 0
3 years ago
Read 2 more answers
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

Electron concentration n(x)=10^{16}\exp (\frac{-x}{18})

                                        \frac{dn}{dx}=(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})

Electron diffusion coefficient is Dn = 25cm^2/s \approx 2.5*10^{-3}

Electron mobility is \mu n = 960 cm^2/V-s \approx0.096m/V

Electron current density Jn = -40 A/cm^2 \approx -40*10^{4}A/m^2

Generally the equation for the semiconductor is mathematically given by

Jn=qb_n\frac{dn}{dx}+nq \mu E

Therefore

-40*10^{4}=1.6*10^{-19} *(2.5*10^{-3})*(10^{16} *(\frac{-1}{16})\exp\frac{-x}{16})+(10^{16}\exp (\frac{-x}{18}))*1.6*10^{-19}*0.096* E

E=\frac{-2.5*10^-^7 exp(\frac{-x}{18})+40*10^{4}}{1.536*10^-4exp(\frac{-x}{18} )}

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

7 0
3 years ago
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