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3 years ago

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Resistors of 150 Ω and 100 Ω are connected in parallel. What is their equivalent resistance?

1 answer:

aliina [53]3 years ago

6 0

<h3>your question : </h3>

Resistors of 150 Ω and 100 Ω are connected in parallel. What is their equivalent resistance?

Answer:

the equivalent resistance will be 60 ohms

Explanation:

see the solution in attached picture

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For the following gear train, if the blue gear is moving at 50 rpm, what are the speeds of the other gears?

Flauer [41]

Answer:

6

Explanation:

6 teddy bears

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3 years ago

Consider air entering a heated duct at P1 = 1 atm and T1 = 288 K. Ignore the effect of friction. Calculate the amount of heat pe

Ne4ueva [31]

Answer:

The solution for the given problem is done below.

Explanation:

M1 = 2.0

$\frac{p1}{p*}$ = 0.3636

$\frac{T1}{T*}$ = 0.5289

$\frac{T01}{T0*}$ = 0.7934

Isentropic Flow Chart: M1 = 2.0 , $\frac{T01}{T1}$ = 1.8

T1 = $\frac{1}{0.7934}$ (1.8)(288K) = 653.4 K.

In order to choke the flow at the exit (M2=1), the above T0* must be stagnation temperature at the exit.

At the inlet,

T02= $\frac{T02}{T1}T1$ = (1.8)(288K) = 518.4 K.

Q= Cp(T02-T01) = $\frac{1.4(287 J / (Kg.K)}{1.4-1}(653.4-518.4)K$ = 135.7* $10^{3}$ J/Kg.

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3 years ago

Read 2 more answers

Why was information revolution different or unique <br>

Xelga [282]

Answer:

It made information easily accessible and ensured individuals became more vast in subject topics of interest.

Explanation:

Information revolution is different and unique and it came with the advent of computers and the internet. A lot of information is stored there which is too large and complex for the human brain.This helped people to access information without much stress as informations about almost every subject is on the Internet.

Individuals can check the informations up and become more vast in interested topics.

8 0

3 years ago

A ductile hot-rolled steel bar has a minimum yield strength in tension and compression of 350 MPa. Using the distortion-energy a

Ratling [72]

Answer:

Explanation:

From the given question:

Using the distortion energy theory to determine the factors of safety FOS can be expressed by the relation:

$\dfrac{Syt}{FOS}= \sqrt{ \sigma x^2+\sigma y^2-\sigma x \sigma y+3 \tau_{xy^2}}$

where; syt = strength in tension and compression = 350 MPa

The maximum shear stress theory can be expressed as:

$\tau_{max} = \dfrac{Syt}{2FOS}$

where;

$\tau_{max} =\sqrt{ (\dfrac{\sigma x-\sigma y}{2})^2+ \tau _{xy^2$

a. Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{94^2+0^2-94*0+3 (-75)^2}}$

$\dfrac{350}{FOS}=160.35$

${FOS}=\dfrac{350}{160.35}$

FOS = 2.183

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{94-0}{2})^2+ (-75)^2$

$\dfrac{350}{2 FOS} =88.51$

$\dfrac{350}{ FOS} =2 \times 88.51$

${ FOS} =\dfrac{350}{2 \times 88.51}$

FOS = 1.977

b. σx = 110 MPa, σy = 100 MPa

Using distortion - energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 110^2+100^2-110*100+3(0)^2}$

$\dfrac{350}{FOS}= \sqrt{ 12100+10000-11000$

$\dfrac{350}{FOS}=105.3565$

$FOS=\dfrac{350}{105.3565}$

FOS =3.322

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{110-100}{2})^2+ (0)^2$

$\dfrac{350}{2 FOS} ={ (\dfrac{110-100}{2})^2$

$\dfrac{350}{2 FOS} =25$

FOS = 350/2×25

FOS = 350/50

FOS = 70

c. σx = 90 MPa, σy = 20 MPa, τxy =−20 MPa

Using distortion- energy theory formula:

$\dfrac{350}{FOS}= \sqrt{ 90^2+20^2-90*20+3(-20)^2}$

$\dfrac{350}{FOS}= \sqrt{ 8100+400-1800+1200}$

$\dfrac{350}{FOS}= 88.88$

FOS = 350/88.88

FOS = 3.939

USing the maximum-shear stress theory;

$\dfrac{350}{2 FOS} =\sqrt{ (\dfrac{90-20}{2})^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ (35)^2+ (-20)^2$

$\dfrac{350}{2 FOS} =\sqrt{ 1225+ 400$

$\dfrac{350}{2 FOS} =40.31$

$FOS} =\dfrac{350}{2*40.31}$

FOS = 4.341

7 0

3 years ago

¿Qué aditivo se debe incorporar a la masa de hormigón para aumentar su resistencia frente a los ciclos alternados de hielo-deshi

tamaranim1 [39]

Answer:

Los aditivos que deben incorporarse a la masa de concreto para aumentar su resistencia a los ciclos alternos de congelación y descongelación son;

1. Agentes de arrastre de aire (AEA) o

2. Materiales poliméricos súper absorbentes

Explanation:

La resistencia alterna de los ciclos de congelación y descongelación en el concreto puede aumentarse mediante la adición de agentes de arrastre de aire.(AEA) que es un surfactante, crea burbujas de aire muy pequeñas en el concreto resultante para mejorar la durabilidad y resistencia del cemento al ciclo repetido de congelación y descongelación o materiales poliméricos súper absorbentes

Ejemplos de agentes de arrastre de aire son;

Sulfonatos alcalinos

Acidos de resinas sulfonadas

Sales de ácidos grasos

Ejemplos de materiales poliméricos superabsorbentes son;

SAP0.26CT

SAP0.39PT.

6 0

4 years ago