Why do protons repel protons but attract electrons ??

Find the voltage gain vO/vS and current gain iO/iX for the circuit for g = 0.0025 S. Then, for vS = 4 V, find the power supplied

Lelechka [254]

Answer:

Incomplete question, no circuit diagram.

Check attachment for further explanation and circuit diagram

Explanation:

We want to find Vo/Vs and Io/Ix and also power delivered to the 2 kΩ resistor

Given that,

g= 0.0025 S (i.e conductance )

Vs= 4 V

R1 = 1 kΩ = 1000 Ω

R2 = 3 kΩ = 3000 Ω

R3 = 10 kΩ = 10000 Ω

R4 = 500Ω = 0.5 kΩ

R5 = 2 kΩ = 2000 Ω

a. At Loop 1: let use voltage divider rule to get Vx

Then, Vx = R2/(R1+R2) • Vs

Vx=3/(1+3) •Vs

Vx=¾ Vs.

The small signal current is given as

Is=g•Vx, since Vx=¾Vs

Then, Is= 0.025×¾Vs

Is=3/1600 Vs. Equation 1

Note: Ressistor R4 and R5 are in series, then the equivalent resistance of R4 and R5 is given as

Req= R4+R5

Req=2+0.5=2.5 kΩ

So, using current divider rule between R3 and the equivalent resistance of R4 and R5.

Therefore, Io= R3/(R3+Req) • Is

Io= R3/(R3+Req) • Is equation 2

Note : using ohms law on resistor R5,

V=iR. , R5=2 kΩ

Vo=IoR5

Vo=2Io

Io=Vo/2. Equation 3

Substitute equation 1 and 3 into 2

Io= R3/(R3+Req) • Is

½Vo = 10/(10+2.5) • 3/1600 Vs

½Vo = 10/12.5 • 3/1600 Vs

½Vo = 3/2000 Vs

Vo/Vs = 3/2000 × 2

Vo/Vs = 1 / 1000

The voltage output gain is

Vo/Vs = 1 / 1000

b. From equation 2

Io= R3/(R3+Req) • Is

Also, applying ohms law to resistor R2,

Vx = Ix• R2, R2=3kΩ

Vx = 3•Ix

Given that, Is= g•Vx

Is=0.0025(3•Ix)

Is= 3/400 Ix

Then, Io= R3/(R3+Req) • Is

Io= 10/(10+2.5) • 3/400 Ix

Io= 10/12.5 • 3/400 Ix

Io= 3/500 Ix

Io/Ix= 3/500

The current gain is

Io/Ix= 3/500

c. Output power

Power is given as

P=I²R

Then, output power at Resistor 5 is

Po = Io²•R5

R5=2000 Ω

From loop 1: using KVL, sum of voltage in a loop is zero

-Vs+1000Ix+3000Ix=0

4000Ix=Vs

Since Vs=4

Then, 4000Ix=4

Ix =4/4000

Ix = 1/1000 A

Since, Io/Ix = 3/500

Then, Io = 3/500 • Ix

Io=3/500 × 1/1000

Io= 6×10^-6 A

Therefore,

Po=Io²•R5. ,R5=2000

Po= (6×10^-6l² × 2000

Po=7.2×10^-8 W

Po=72×10^-9 W

Po=72 nW

The output power at resistor R5 is

72 nW

6 0

3 years ago

The 20 kg at angle of 53⁰ in an inclined plane is realsed from rest the coefficient of friction bn the block and the inclined pl

Scorpion4ik [409]

Given :-

A 20kg block at an angle 53⁰ in an inclined plane is released from rest .
$\mu_s = 0.3 \ \& \ \mu_k = 0.2$

To Find :-

Would the block move ?
If it moves what is its speed after it has descended a distance of 5m down the plane .

Solution :-

For figure refer to attachment .

So the block will move if the angle of the inclined plane is greater than the angle of repose . We can find it as ,

$\longrightarrow \theta_{repose}= tan^{-1}(\mu_s)$

Substitute ,

$\longrightarrow \theta_{repose}= tan^{-1}( 0.2)$

Solve ,

$\longrightarrow\underline{\underline{\theta_{repose}= 16.6^o }}$

Hence ,

$\longrightarrow\theta_{plane}>\theta_{repose}$

Hence the block will slide down .

Now assuming that block is released from the reset , it's initial velocity will be 0m/s .

And the net force will be ,

$\longrightarrow F_n = mgsin53^o - \mu_k N$

Substitute, N = mgcos53⁰ ( see attachment)

$\longrightarrow ma_n = mgsin53^o - \mu_k mgcos53^o$

Take m as common,

$\longrightarrow\cancel{m }(a_n) = \cancel{m}( gsin53^o - \mu gcos53^o)$

Simplify ,

$\longrightarrow a_n = gsin53^o - \mu_k g cos53^o$

Substitute the values of sin , cos and g ,

$\longrightarrow a_n = 10( 0.79 - 0.2 (0.6))$

Simplify ,

$\longrightarrow a_n = 10 ( 0.79 - 0.12 ) \\\\ \longrightarrow a_n = 10 (0.67)\\\\ \longrightarrow \underline{\underline{a_n = 6.7 m/s^2}}$

Now using the Third equation of motion namely,

$\longrightarrow2as = v^2-u^2$

Substituting the respective values,

$\longrightarrow2(6.7)(5) = v^2-(0)^2$

Simplify and solve for v ,

$\longrightarrow v^2 = 67 m/s\\\\\longrightarrow v =\sqrt{67} m/s \\\\\longrightarrow\underline{\underline{ v = 8.18 m/s }}$

Hence the velocity after covering 5m is 8.18 m/s .

7 0

3 years ago

PLEASE HELPPPPPP <333

True [87]

Answer:

the answer is b

Explanation:

gravity pulls you down so on a scale you will weigh more. less gravity you will weigh less.

8 0

3 years ago

PLEASE HELP TOPIC - CONVECTION

STALIN [3.7K]

Answer:

SO ANSWER AGRE AS FOLLOWS:

Explanation:

A. Wind is the movement of air, caused by the uneven heating of the Earth by the sun and the Earth's own rotation. Winds range from light breezes to natural hazards such as hurricanes and tornadoes.

B.At night, the roles reverse. The air over the ocean is now warmer than the air over the land. The land loses heat quickly after the sun goes down and the air above it cools too. ... This causes the low surface pressure to shift to over the ocean during the night and the high surface pressure to move over the land.

<h2>HOPE THIS IS CORRECT </h2>

5 0

3 years ago

a body of mass 5kg falls from height of 10m above the ground what kinetic energy of the body before it strike the ground

Snowcat [4.5K]

Gravitational Potential Energy (GPE) before fall = Kinetic energy on impact
GPE = mgh
GPE = 5kg x 9.8m/s^2 x 10m
GPE = 490 J
Kinetic Energy = 490 J

(This is assuming that gravitation field strength (g) is 9.8m/s^2, sometimes you may round this to 10m/s^2, therefore making the final result:
Kinetic energy = 500 J)

4 0

4 years ago