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2 years ago

What are some advantages of making electronic components like transistors increasingly smaller?

Engineering

1 answer:

Stells [14]2 years ago

4 0

Answer:

Dr. Engelbart, who would later help develop the computer mouse and other personal computing technologies, theorized that as electronic circuits were made smaller, their components would get faster, require less power and become cheaper to produce — all at an accelerating pace

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Select three types of lines that engineers use to help represent the shape of a design in a sketch.

Vikki [24]

Hidden lines

Used to describe the in shown lines (like diagonals inside cubes)

Extension lines:-

Used to explain the expansion of structures like building

Object lines

Used to describe the structure of objects and the lining to show borders

7 0

2 years ago

For a body moving with simple harmonic motion state the equations to represent: i) Velocity ii) Acceleration iii) Periodic Time

max2010maxim [7]

Answer with Explanation:

The general equation of simple harmonic motion is

$x(t)=Asin(\omega t+\phi)$

where,

A is the amplitude of motion

$\omega$ is the angular frequency of the motion

$\phi$ is known as initial phase

part 1)

Now by definition of velocity we have

$v=\frac{dx}{dt}\\\\\therefore v(t)=\frac{d}{dt}(Asin(\omega t+\phi )\\\\v(t)=A\omega cos(\omega t+\phi )$

part 2)

Now by definition of acceleration we have

$a=\frac{dv}{dt}\\\\\therefore a(t)=\frac{d}{dt}(A\omega cos(\omega t+\phi )\\\\a(t)=-A\omega ^{2}sin(\omega t+\phi )$

part 3)

The angular frequency is related to Time period 'T' as $T =\frac{2\pi }{\omega }$

where

$\omega$ is the angular frequency of the motion of the particle.

Part 4) The acceleration and velocities are plotted below

since the maximum value that the sin(x) and cos(x) can achieve in their respective domains equals 1 thus the maximum value of acceleration and velocity is $A\omega ^{2}$ and $A\omega$ respectively.

4 0

3 years ago

What is the resistance of a resistor if the current flowing through it is 3mA and the voltage across it is 5.3V?

Flura [38]

Answer: 1766.667 Ω = 1.767kΩ

Explanation:

V=iR

where V is voltage in Volts (V), i is current in Amps (A), and R is resistance in Ohms(Ω).

3mA = 0.003 A

Rearranging the equation, we get

R=V/i

Now we are solving for resistance. Plug in 0.003 A and 5.3 V.

R = 5.3 / 0.003

= 1766.6667 Ω

= 1.7666667 kΩ

The 6s are repeating so round off to whichever value you need for exactness.

6 0

1 year ago

What is the biggest expectation when engineers test out designs?

forsale [732]

The answer is B because it could be feasible but it’s not a need it and you got a time frame but it’s not a requirement and it doesn’t have to be unique.

6 0

4 years ago

Read 2 more answers

Using a forked rod, a 0.5-kg smooth peg P is forced to move along the vertical slotted path r = (0.5 θ) m, whereθ is in radians.

-BARSIC- [3]

Answer:

N_c = 3.03 N

F = 1.81 N

Explanation:

Given:

- The attachment missing from the question is given:

- The given expressions for the radial and θ direction of motion:

r = 0.5*θ

θ = 0.5*t^2 ...... (correction for the question)

- Mass of peg m = 0.5 kg

Find:

a) Determine the magnitude of the force of the rod on the peg at the instant t = 2 s.

b) Determine the magnitude of the normal force of the slot on the peg.

Solution:

- Determine the expressions for radial kinematics:

dr/dt = 0.5*dθ/dt

d^2r/dt^2 = 0.5*d^2θ/dt^2

- Similarly the expressions for θ direction kinematics:

dθ/dt = t

d^2θ/dt^2 = 1

- Evaluate each at time t = 2 s.

θ = 0.5*t^2 = 0.5*2^2 = 2 rad -----> 114.59°

r = 1 m , dr / dt = 1 m/s , d^2 r / dt^2 = 0.5 m/s^2

- Evaluate the angle ψ between radial and horizontal direction:

tan Ψ = r / (dr/dθ) = 1 / 0.5

Ψ = 63.43°

- Develop a free body diagram (attached) and the compute the radial and θ acceleration:

a_r = d^2r / dt^2 - r * dθ/dt

a_r = 0.5 - 1*(2)^2 = -3.5 m/s^2

a_θ = r * (d^2θ/dt^2) + 2 * (dr/dt) * (dθ/dt)

a_θ = 1(1) + 2*(1)*(2) = 5 m/s^2

- Using Newton's Second Law of motion to construct equations in both radial and θ directions as follows:

Radial direction: N_c * cos(26.57) - W*cos(24.59) = m*a_r

θ direction: F - N_c * sin(26.57) + W*sin(24.59) = m*a_θ

Where, F is the force on the peg by rod and N_c is the normal force on peg by the slot. W is the weight of the peg. Using radial equation:

N_c * cos(26.57) - 4.905*cos(24.59) = 0.5*-3.5

N_c = 3.03 N

F - 3.03 * sin(26.57) + 4.905*sin(24.59) = 0.5*5

F = 1.81 N

4 0

3 years ago