All categories

3 years ago

What is engineering?

Engineering

1 answer:

ANTONII [103]3 years ago

8 0

Answer:

the branch of science and technology concerned with the design, building, and use of engines, machines, and structures. Anything that involves engines, wires, etc. basically

You might be interested in

Given V1 = 8 vpp, R1 = 6 kΩ, V2 = 5 vpp, R2 = 3 kΩ, V3 = 10 vpp, R3 = 3 kΩ, and Rf = 14 kΩ, find Vout. Vout = vpp. (Round your a

iragen [17]

Answer:

Vout= 93.3V

Explanation:

For this question, consider circuit in the attachment 1.

This is the circuit of an inverting amplifier. In an inverting amplifier

Vout/Vin= -Rf/Rin

To calculate the Vout, we must find Rin and Vin. For this we must solve the input circuit (attachment 2) using Thevinine theorem. Thevnine theorem states that all voltage sources in a circuit can be replaced by an equivalent voltage source Veq and and all resistances can be replaced by an equivalent resistance Req. To find out Req all voltage sources must be short circuited (attachment 3)

1/Req= 1/R1+1/R2+1/R3

1/Req=1/6+1/3+1/3

Req=6/5

To find out Veq consider circuit in attachment 4. We will solve this circuit using nodal analysis. In nodal analysis, we use the concept that sum of currents entering a node is equal to the sum of currents leaving a node. So,

I1= I2+I3

(10-Veq)/6= (Veq-5)/3+(Veq-10)/3

Veq=8V

Now the input circuit can be simplified as shown in attachment 5. Solve for Vout using equation

Vout/Veq= -Rf/Req

Vout/8= -14/(6/5)

Vout= - 93.3

It is at an angle of 180° from Veq

7 0

3 years ago

Me ayudas plis noce

alekssr [168]

Answer:

Explanation:

7 0

3 years ago

State three characteristic of lines of magnetic flux

stepladder [879]

Answer:

1. The magnetic flux line form a closed loop.

2. The magnetic flux line repel each other.

3. The magnetic flux line never intersect.

5 0

3 years ago

Read 2 more answers

Consider a very long rectangular fin attached to a flat surface such that the temperature at the end of the fin is essentially t

kodGreya [7K]

Answer:

$\frac{T-20}{130-20}= e^{-14.28*0.05}$

And if we solve for T we got:

$T= 20 + 110e^{-14.28*0.05} = 73.86 C$

The answer for this case would be $T = 73.86 C$ at 5cm from the base of the fin.

Explanation:

Data given

For this case we have the following data given:

$h = 20 \frac{W}{m^2 K}$ represent the heat transfer coefficient.

$p$ represent the perimeter for this case and would be given by:

$p = 2*0.05m +2*0.001m= 0.102m$

$k = 200 \frac{W}{m C}$ represent the thermal conductivity

$w = 5cm =0.05 m$ represent the width

$h = 1mm =0.001m$ represent the thickness

$A= wh= 0.05m *0.001m = 0.00005 m^2$

Solution to the problem

For this case we assume that we have steady conditions, the temperature of the fins varies just in one direction, the heat transfer coefficient not changes with the time and the thermal properties of the fin not change.

We can determine the temperature if the fin at x=5 cm=0.05 m from the base with the following formula:

$\frac{T-T_{\infty}}{T_b -T_{\infty}} = e^{-mx}$