According to the what the universe's total amount of energy does not change

When the input voltages of a difference amplifier are 5.1 v and 6.4 v, the output voltage is 64.7 v. The inputs are changed to 4

daser333 [38]

Answer:

a) Differential mode gain = 48

b) Common mode gain = 0.4

c) CMRR = 120

Explanation:

The output of a difference amplifier is related to the input by the equation:

$V_{0} = A_{1} V_{1} + A_{2} V_{2} \\$

When V₁ = 6.4 V, V₂ = 5.1 V and V₀ = 64.7 V, the equation becomes

6.4 A₁ + 5.1 A₂ = 64.7.....................(1)

When V₁ = 5.6 V, V₂ = 4.9 V and V₀ = 35.7 V, the equation becomes

5.6 A₁ + 4.9 A₂ = 35.7.....................(2)

Multiply equation (1) by 5.6 and (2) by 6.4

35.84 A₁ + 28.56A₂ = 362.32.....................(3)

35.84 A₁ + 31.36 A₂ = 228.48....................................(4)

Subtract equation (3) from (4)

2.8 A₂ = -133.84

A₂ = -133.84/2.8

A₂ = -47.8

Put the value of A₂ into equation (1)

6.4 A₁ + 5.1 (-47.8) = 64.7

6.4 A₁ = 64.7 + 243.78

A₁ = 308.48/6.4

A₁ = 48.2

a) Common mode gain = A₁ + A₂ = 48.2 + (-47.8)

Common mode gain = 0.4

b) Differential mode gain = (A₁ -A₂)/2

Differential mode gain = (48.2 - (-47.8))/2

Differential mode gain = 96/2

Differential mode gain = 48

c) Common Mode Rejection Ratio (CMRR)

$CMRR = |\frac{Differential Mode Gain}{Common Mode Gain} |$

$CMRR = |\frac{48}{0.4} |\\CMRR = 120$

4 0

3 years ago

Two north magnets will do this

DerKrebs [107]

Answer:

Repel one another

Explanation

I am assuming you meant "two north poles will do this" Or something similar.

because you can't isolate poles, so a magnet wouldn't be just north

ANYWAYS i say repel because like terms repel, opposites attract.

5 0

3 years ago

Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach

Svetlanka [38]

Answer:

$\Delta V = 1.8 \times 10^7 V$

Explanation:

GIVEN

diameter = 15 fm = $15 \times 10^{-15}$ m

we use here energy conservation

$K_{i}+U_{i} =K_{f}+U_{f}$

there will be some initial kinetic energy but after collision kinetic energy will zero

$K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}$

on solving these equations we get kinetic energy initial

$KE_{i} = 5.65\times 10 ^{-12} \times \frac {1 eV}{1.6 \times 10^{-19}}$

$KE_{i} = 35.33$ J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e

and gains kinetic energy K =e∆V ..........(ii)

by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the ${238}_U$ nucleus after being accelerated through a potential difference ∆V

equating (i) and second equation we get

e∆V = 35.33 Me V

$\Delta V = \frac{35.33}{2} MV\\\Delta V = 1.8 \times 10^7 V$

7 0

3 years ago

PLEASE HELP ASAP!! CORRECT ANSWER ONLY PLEASE!!

Liula [17]

The answer is Trend Line.

5 0

4 years ago

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What are the rows of the periodic table of the elements also called?

Ostrovityanka [42]

Answer:

periods

Explanation:

7 0

3 years ago

Read 2 more answers