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2 years ago

Please help me with this question

Engineering

1 answer:

Nat2105 [25]2 years ago

5 0

Answer:

The answer is c and the teacher helped me

Explanation:

i had help from the tescger and the assignment is done

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When asked about favorite Thanksgiving leftovers, 9% of the people said turkey and 7100 said mashed potatoes. Which food is more

irakobra [83]

Answer:

how many people were asked though

Explanation:

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3 years ago

A steel bar 100 mm (4.0 in.) long and having a square cross section 20 mm (0.8 in.) on an edge is pulled intension with a load o

grigory [225]

Answer:

The elastic modulus of the steel is 139062.5 N/in^2

Explanation:

Elastic modulus = stress ÷ strain

Load = 89,000 N

Area of square cross section of the steel bar = (0.8 in)^2 = 0.64 in^2

Stress = load/area = 89,000/0.64 = 139.0625 N/in^2

Length of steel bar = 4 in

Extension = 4×10^-3 in

Strain = extension/length = 4×10^-3/4 = 1×10^-3

Elastic modulus = 139.0625 N/in^2 ÷ 1×10^-3 = 139062.5 N/in^2

7 0

3 years ago

Economics is the study of how individuals and societies make choices under the condition of

k0ka [10]

Answer:

ig scarcity

Explanation:

Hope it Helps!!!

3 0

2 years ago

A one-dimensional plane wall of thickness 2L=80 mm experiences uniform thermal generation of q= 1000 W/m^3 and is convectively c

Eduardwww [97]

Answer:

$h=1.99998\ W/m^2.C$

$k=33.333\ W/m.C$

Explanation:

Considering the one dimensional and steady state:

From Heat Conduction equation considering the above assumption:

$\frac{\partial^2T}{\partial x^2}+\frac{\dot e_{gen}}{k}=0$ Eq (1)

Where:

k is thermal Conductivity

$\dot e_{gen}$ is uniform thermal generation

$T(x) = a(L^2-x^2)+b$

$\frac{\partial\ T(x)}{\partial x}=\frac{\partial\ a(L^2-x^2)+b}{\partial x}=-2ax\\\frac{\partial^2\ T(x)}{\partial x^2}=\frac{\partial^2\ -2ax}{\partial x^2}=-2a$

Putt in Eq (1):

$-2a+\frac{\dot e_{gen}}{k}=0\\ k=\frac{\dot e_{gen}}{2a}\\ k=\frac{1000}{2*15}\\ k=33.333\ W/m.C$

Energy balance is given by:

$Q_{convection}=Q_{conduction}$

$h(T_L-T_{inf})=-k(\frac{dT}{dx}) _L$ Eq (2)

$T(x) = a(L^2-x^2)+b$

Putting x=L

$T(L) = a(L^2-L^2)+b\\T(L)=b\\T(L)=40^oC$

$\frac{dT}{dx}=\frac{d(a(L^2-x^2)+b}{dx}=-2ax\\Put\ x\ =\ L\\\frac{dT}{dx}=-2aL\\(\frac{dT}{dx})_L=-2*15*0.04=-1.2$

From Eq (2)

$h=\frac{-k*-1.2}{(40-20)} \\h=\frac{-33.333*-1.2}{(40-20)}\\h=1.99998\ W/m^2.C$

7 0

3 years ago

Consider a half-wave rectifier circuit with a triangular-wave input of LaTeX: 6V6 V peak-to-peak amplitude and zero average valu

Varvara68 [4.7K]

The question is not complete. We are supposed to find the average value of v_o.

Answer:

v_o,avg = 0.441V

Explanation:

Let t1 and t2 be the start and stop times of the output waveforms. Thus, from the diagram i attached, using similar triangles, we have;

3/(T/4) = 0.7/t1

So, 12/T = 0.7/t1

So, t1 = 0.7T/12

t1 = 0.0583 T

Also, from symmetry of triangles,

t2 = T/2 - t1

So, t2 = T/2 - 0.0583 T

t2 = 0.4417T

Average of voltage output is;

v_o,avg = (1/T) x Area under small triangle

v_o,avg = (1/T) x (3 - 0.7) x (T/4 - t1)

v_o,avg = (1/T) x (2.3) x (T/4 - 0.0583 T)

v_o,avg = (1/T) x 2.3 x 0.1917T

T will cancel out to give;

v_o,avg = 0.441V

8 0

3 years ago