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3 years ago

Type the correct answer in the box. Spell all words correctly.

Engineering

1 answer:

AVprozaik [17]3 years ago

3 0

Answer:

Mechanical engineer? Thats my guess I didnt have alot of options sorry if I am wrong

Explanation:

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If the density of states function in the conduction band of a particular semiconductor is a constant equal to K, derive the expr

s2008m [1.1K]

Answer:

full details of the answer is attached

5 0

2 years ago

A 1 m3 rigid tank initially contains air whose density is 1.18kg/m3. The tank is connected to a high pressure supply line throug

Elanso [62]

To solve this problem it is necessary to apply the concepts related to density in relation to mass and volume for each of the states presented.

Density can be defined as

$\rho = \frac{m}{V}$

Where

m = Mass

V = Volume

For state one we know that

$\rho_1 = \frac{m_1}{V}$

$m_1 = \rho_1 V$

$m_1 = 1.18*1$

$m_1 = 1.18Kg$

For state two we have to

$\rho_2 = \frac{m_2}{V}$

$m_2 = \rho_2 V$

$m_1 = 7.2*1$

$m_1 = 7.2Kg$

Therefore the total change of mass would be

$\Delta m = m_2-m_1$

$\Delta m = 7.2-1.18$

$\Delta m = 6.02Kg$

Therefore the mass of air that has entered to the tank is 6.02Kg

5 0

2 years ago

Find the number of Btu conducted through a wall in 8 hours. The wall is 8 feet high by 24 feet long and has a total R-value of 1

dedylja [7]

Answer:

ΔQ = 4930.37 BTu

Explanation:

given data

height h = 8ft

Δt = 8 hours

length L = 24 feet

R value = 16.2 hr⋅°F⋅ft² /Btu

inside temperature t1 = 68°F

outside temperature t2 = 16°F

to find out

number of Btu conducted

solution

we get here number of Btu conducted by this expression that s

$\frac{\Delta Q}{\Delta t} =\frac{-A}{R} (t2 -t1)$ ......................1

here A is area that is = h × L = 8 × 24 = 1492 ft²

put here value we get

$\frac{\Delta Q}{8} =\frac{-192}{16.2} (16-68)$

solve it we get

ΔQ = 4930.37 BTu

7 0

3 years ago

Air with a mass flow rate of 2.3 kg/s enters a horizontal nozzle operating at steady state at 450 K, 350 kPa, and velocity of 3

viktelen [127]

Answer:

Given that

Mass flow rate ,m=2.3 kg/s

T₁=450 K

P₁=350 KPa

C₁=3 m/s

T₂=300 K

C₂=460 m/s

Cp=1.011 KJ/kg.k

For ideal gas

P V = m R T

P = ρ RT

$\rho_1=\dfrac{P_1}{RT_1}$

$\rho_1=\dfrac{350}{0.287\times 450}$

ρ₁=2.71 kg/m³

mass flow rate

m= ρ₁A₁C₁

2.3 = 2.71 x A₁ x 3

A₁=0.28 m²

Now from first law for open system

$h_1+\dfrac{C_1^2}{200}+Q=h_2+\dfrac{C_2^2}{2000}$

For ideal gas

Δh = CpΔT

by putting the values

$1.011\times 450+\dfrac{3^2}{200}+Q=1.011\times 300+\dfrac{460^2}{2000}$

$Q=1.011\times 300+\dfrac{460^2}{2000}-\dfrac{3^2}{200}-1.011\times 450$

Q= - 45.49 KJ/kg

Q =- m x 45.49 KW

Q= - 104.67 KW

Negative sign indicates that heat transfer from air to surrounding

4 0

2 years ago

How do we find percentage error in measuring voltage across a resistor

Black_prince [1.1K]

Answer:

use the percentage error relation

Explanation:

The percentage error in anything is computed from ...

%error = ((measured value)/(accurate value) -1) × 100%

The difficulty with voltage measurements is that the "accurate value" may be hard to determine. It can be computed from the nominal values of circuit components, but there is no guarantee that the components actually have those values.

Likewise, the measuring device may have errors. It may or may not be calibrated against some standard, but even measurement standards have some range of possible error.

6 0

2 years ago

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