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3 years ago

Errors in the output voltage of an actual integrated circuit operational amplifier can be caused by : Select one:

Engineering

1 answer:

natta225 [31]3 years ago

7 0

Answer:

Option B

Explanation:

An operational amplifier usually has a high open loop gain of around 10^5 which allows a wide range get of feed back levels in order to achieve the desired performance so therefore a low open loop gain reduces the range feed back level thereby reducing the performance which can cause errors in the output voltage.

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Explain the differences between 1- Energy 2- Power 3- Work 4- Heat Your answer should explain the mathematica and physical meani

vazorg [7]

Answer:

Explanation:

Energy : Energy can be defined as both qualitative and quantitative property that can be given to an object in the form of work or heat.

For example - water stored in a tank at a height,h from the ground has a potential energy of U = ρgh.

Mathematically, Energy of water stored in a tank is

U = ρ.g.h

where, ρ is water density

g is acceleration due to gravity

h is height of water tank from ground

Power : Power is defined as rate at which work can be done.

For example - When a force F is applied to an object to move it through a distance d in time t seconds is an example of power delivered.

Mathematically, Power = Work done / Time

$P = \frac{F\times d}{t}$

$P = \frac{W}{t}$

Work : When a force act on a body and the body moves, work is said to be done.

For example - Gas confined in a cylinder by a piston. The gas expands when it is heated, doing work on the piston.

Mathematically, Work,W = F X d

where, F is force

d is distance

Heat : Heat is a form of energy which is produced as a result of motion of atoms and particles.

For example - while increasing the temperature of water, heat is added to the water by means of gas burner or any other form.

Mathematically we can calculate heat release or heat absorb by

q = m.C.ΔT

where q is heat

m is mass

C is specific heat capacity

ΔT is change in temperature

4 0

3 years ago

Note that common skills are listed toward the top, and less common skills are listed toward the bottom.

Sladkaya [172]

Answer:

BDEG

Explanation:

got it right on the test on edge because i used my b r a i n

5 0

3 years ago

What is the activation energy (Q) for a vacancy formation if 10 moles of a metal have 2.3 X 10^13 vacancies at 425°C?

Yakvenalex [24]

Answer:

$Activation\ Energy=2.5\times 10^{-19}\ J$

Explanation:

Using the expression shown below as:

$N_v=N\times e^{-\frac {Q_v}{k\times T}$

Where,

$N_v$ is the number of vacancies

N is the number of defective sites

k is Boltzmann's constant = $1.38\times 10^{-23}\ J/K$

${Q_v}$ is the activation energy

T is the temperature

Given that:

$N_v=2.3\times 10^{13}$

N = 10 moles

1 mole = $6.023\times 10^{23}$

So,

N = $10\times 6.023\times 10^{23}=6.023\times 10^{24}$

Temperature = 425°C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T = (425 + 273.15) K = 698.15 K

T = 698.15 K

Applying the values as:

$2.3\times 10^{13}=6.023\times 10^{24}\times e^{-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$ln[\frac {2.3}{6.023}\times 10^{-11}]=-\frac {Q_v}{1.38\times 10^{-23}\times 698.15}$

$Q_v=2.5\times 10^{-19}\ J$

4 0

3 years ago

The 15-kg block A slides on the surface for which µk = 0.3. The block has a velocity v = 10 m/s when it is s = 4 m from the 10-k

sammy [17]

Answer:

s_max = 0.8394m

Explanation:

From equilibrium of block, N = W = mg

Frictional force = μ_k•N = μ_k•mg

Since μ_k = 0.3,then F = 0.3mg

To determine the velocity of Block A just before collision, let's apply the principle of work and energy;

T1 + ΣU_1-2 = T2

So, (1/2)m_a•(v_ao)² - F•s =(1/2)m_a•(v_a1)²

Plugging in the relevant values to get ;

(1/2)•(15)•(10)² - (0.3•15•9.81•4) =(1/2)(15)•(v_a1)²

750 - 176.58 = 7.5(v_a1)²

v_a1 = 8.744 m/s

Using law of conservation of momentum;

Σ(m1v1) = Σ(m2v2)

Thus,

m_a•v_a1 + m_b•v_b1 = m_a•v_a2 + m_b•v_b2

Thus;

15(8.744) + 10(0) = 15(v_a2) + 10(v_b2)

Divide through by 5;

3(8.744) + 2(0) = 3(v_a2) + 2(v_b2)

Thus,

3(v_a2) + 2(v_b2) = 26.232 - - - (eq1)

Coefficient of restitution has a formula;

e = (v_b2 - v_a2)/(v_a1 - v_b1)

From the question, e = 0.6.

Thus;

0.6 = (v_b2 - v_a2)/(8.744 - 0)

0.6 x 8.744 = (v_b2 - v_a2)

(v_b2 - v_a2) = 5.246 - - - (eq2)

Solving eq(1) and 2 simultaneously, we have;

v_b2 = 8.394 m/s

v_a2 = 3.148 m/s

Now, to find maximum compression, let's apply conservation of energy on block B;

T1 + V1 = T2 + V2

Thus,

(1/2)m_b•(v_b2)² + (1/2)k(s_1)² = (1/2)m_b•(v_b'2)² + (1/2)k(s_max)²

(1/2)10•(8.394)² + (1/2)1000(0)² = (1/2)10•(0)² + (1/2)(1000)(s_max)²

500(s_max)² = 352.29618

(s_max)² = 352.29618/500

(s_max)² = 0.7046

s_max = 0.8394m

8 0

3 years ago

Levart [38]

Answer:

This is a for loop.

Explanation:

6 0

3 years ago