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2 years ago

1. What is an op-amp? List the characteristics of an ideal op-amp

Engineering

1 answer:

Alexus [3.1K]2 years ago

8 0

Answer:

An opamp is an operation amplifier. It takes an input signal and amplifies it on the output side.

An ideal opamp should have infinite impedance at its input, infinite gain on the output, and zero impedance on the output

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Could I please get help with this

alex41 [277]

Answer:

1. $I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. $I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = 37. $\overline 3$ in.⁴

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = 149. $\overline 3$ in.⁴

3. The maximum deflection of the beam is 2.55552 inches

Explanation:

1. The height of the beam having a rectangular cross section is h = 2.5 in.

The breadth of the beam, is = 5.5 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 5.5 × 2.5³/12 = 1375/192 = 7.161458 $\overline 3$

$I_{xc}$ = 7.161458 $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 2.5 × 5.5³/12 = 6655/192 = 36.661458 $\overline 3$

$I_{yc}$ = 36.661458 $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 5.5 × 2.5³/3 = 625/24 = 28.6458 $\overline 3$

Iₓ = 28.6458 $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 6655/48= 138.6548 $\overline 3$

$I_y$ = 138.6548 $\overline 3$ in.⁴

2. The height of the beam having a rectangular cross section is h = 7 in.

The breadth of the beam, b = 4 in.

The moment of inertia of a rectangular beam through its centroid is given as follows;

$I_{xc}$ = b·h³/12 = 4 × 7³/12 = 114. $\overline 3$

$I_{xc}$ = 114. $\overline 3$ in.⁴

$I_{yc}$ = h·b³/12 = 7 × 4³/12 = 37. $\overline 3$

$I_{yc}$ = 37. $\overline 3$ in.⁴

The moment of inertia about the base is given as follows;

Iₓ = b·h³/3 = 4 × 7³/3 = 457. $\overline 3$

Iₓ = 457. $\overline 3$ in.⁴

$I_y$ = h·b³/3 = 2.5 × 5.5³/3 = 149. $\overline 3$

$I_y$ = 149. $\overline 3$ in.⁴

3. The deflection, $\delta _{max}$ , of a simply supported beam having a point load at the center is given as follows;

$\delta_{max} = \dfrac{W \times L^3}{48 \times E \times I}$

The given parameters of the beam are;

The length of the beam, L = 22 ft. = 264 in.

The applied load at the center, W = 750 lbs

The modulus of elasticity for Cedar = 10,000,000 psi

The height of the wood, h = 3 in.

The breadth of the wood, b = 5 in.

The moment of inertia of the wood, $I_{xc}$ = b·h³/12 = 5 × 3³/12 = 11.25 in.⁴

By plugging in the given values, we have;

$\delta_{max} = \dfrac{750 \times 264^3}{48 \times 10,000,000 \times 11.25} = 2.55552$

The maximum deflection of the beam, $\delta _{max}$ = 2.55552 inches

5 0

3 years ago

Cho biết tác dụng chung của các hệ giằng khung ngang nhà công nghiệp nhẹ 1 tầng 1 nhịp.

Dmitry_Shevchenko [17]

I don’t know how to speak the laungue or know this language

4 0

2 years ago

Consider a single crystal of nickel oriented such that a tensile stress is applied along a [001] direction. If slip occurs on a

Elena L [17]

Answer:

$\mathbf{\tau_c =5.675 \ MPa}$

Explanation:

Given that:

The direction of the applied tensile stress =[001]

direction of the slip plane = [ $\bar 1$ 01]

normal to the slip plane = [111]

Now, the first thing to do is to calculate the angle between the tensile stress and the slip by using the formula:

$cos \lambda = \Big [\dfrac{d_1d_2+e_1e_2+f_1f_2}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_2^2+e_2^2+f_2^2) }} \Big]$

where;

$[d_1\ e_1 \ f_1]$ = directional indices for tensile stress

$[d_2 \ e_2 \ f_2]$ = slip direction

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_2$ = -1 , $e_2$ = 0 , $f_2$ = 1

$cos \lambda = \Big [\dfrac{(0\times -1)+(0\times 0) + (1\times 1) }{\sqrt{(0^2+0^2+1^2)+((-1)^2+0^2+1^2) }} \Big]$

$cos \ \lambda = \dfrac{1}{\sqrt{2}}$

Also, to find the angle $\phi$ between the stress [001] & normal slip plane [111]

Then;

$cos \ \phi = \Big [\dfrac{d_1d_3+e_1e_3+f_1f_3}{\sqrt{(d_1^2+e_1^2+f_1^2)+(d_3^2+e_3^2+f_3^2) }} \Big]$

replacing their values;

i.e $d_1$ = 0 , $e_1$ = 0 $f_1$ = 1 & $d_3$ = 1 , $e_3$ = 1 , $f_3$ = 1

$cos \ \phi= \Big [ \dfrac{ (0 \times 1)+(0 \times 1)+(1 \times 1)} {\sqrt {(0^2+0^2+1^2)+(1^2+1^2 +1^2)} } \Big]$

$cos \phi= \dfrac{1} {\sqrt{3} }$

However, the critical resolved SS(shear stress) $\mathbf{\tau_c}$ can be computed using the formula:

$\tau_c = (\sigma )(cos \phi )(cos \lambda)$

where;

applied tensile stress $\sigma =$ 13.9 MPa

∴

$\tau_c =13.9\times ( \dfrac{1}{\sqrt{2}} )( \dfrac{1}{\sqrt{3}})$

$\mathbf{\tau_c =5.675 \ MPa}$

3 0

3 years ago

Define the terms (a) thermal conductivity, (b) heat capacity and (c) thermal diffusivity

IceJOKER [234]

Explanation:

(a)

The measure of material's ability to conduct thermal energy (heat) is known as thermal conductivity. For examples, metals have high thermal conductivity, it means that they are very efficient at conducting heat. The SI unit of heat capacity is W/m.K.

The expression for thermal conductivity is:

$q=-\kappa \bigtriangledown T$

Where,

q is the heat flux

$\kappa$ is the thermal conductivity

$\bigtriangledown T$ is the temperature gradient.

(b)

Heat capacity for a substance is defined as the ratio of the amount of energy required to change the temperature of the substance and the magnitude of temperature change. The SI unit of heat capacity is J/K.

The expression for Heat capacity is:

$C=\frac{E}{\Delta T}$

Where,

C is the Heat capacity

E is the energy absorbed/released

$\Delta T$ is the change in temperature

(c)

Thermal diffusivity is defined as the thermal conductivity divided by specific heat capacity at constant pressure and its density. The Si unit of thermal diffusivity is m²/s.

The expression for thermal diffusivity is:

$\alpha=\frac{\kappa}{C_p \times \rho}$

Where,

$\alpha$ is thermal diffusivity

$\kappa$ is the thermal conductivity

$C_p$ is specific heat capacity at constant pressure

$\rho$ is density

6 0

3 years ago

A ball thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s. Determine (a) how hig

Masteriza [31]

Answer:

A.) 62.5 ft

B.) 3.58 seconds

C.) 8.58 seconds

Explanation:

A.) Given that a ball is thrown vertically upward from the top of a building of 60ft with an initial velocity of vA=35 ft/s

To determine how high above the top of the building the ball will go before it stops at B, let us use the third equation of motion.

V^2 = U^2 - 2gH

Since the ball is going up, g will be negative. And at maximum height, V = 0

Substitute all the parameters into the formula

0 = 35^2 - 2 × 9.8 × H

19.6H = 1225

H = 1225/19.6

H = 62.5 ft

(B) The time tAB it takes to reach its maximum height will be achieved by using second equation of motion

H = Ut - 1/2gt^2

Substitutes all the parameters into the formula

62.5 = 35t - 1/2 × 9.8 × t^2

62.5 = 35t - 4.9t^2

4.9t^2 - 35t + 62.5 = 0

Let's use quadratic equations to find t

Divide all by 4.9

t^2 - 7.143t + 12.755 = 0

t^2 - 7.143t + 3.57^2 = - 12.755 + 3.57^2

( t - 3.57)^2 = 0.000102

( t - 3.57 ) = +/-( 0.01 )

t = 3.57 + 0.01

t = 3.58 seconds

Ignore the negative one.

When the object is falling back from B, the initial velocity = 0. And the height h will be 60 + 62.5 = 122.5 ft

Using equation 2 of equations of motion again.

h = 1/2gt^2

122.5 = 1/2 × 9.8 × t^2

122.5 = 4.9t^2

t^2 = 122.5/4.9

t^2 = 25

t = 5

Total time = 5 + 3.58 = 8.58 seconds

3 0

2 years ago