ENGINEERING COLLEGE

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer 1

Answer:

Answer:

$v = 250[1 - {e^(-6000t)}]$ mV

Explanation:

The voltage across a capacitor at a time t, is given by:

$v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)$ ----------------(i)

Where;

v(t) = voltage at time t

$t_(0)$ = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3 $e^(-6000t)$ mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt$

$v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt$ [Solve the integral]

$v = (3)/(2*10^(-6)*(-6000)) {e^(-6000t)}|_0^t$

$v = (-3000)/(12) {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]$

$v = -250 {e^(-6000t)} - [-250]$

$v = -250 {e^(-6000t)} + 250$

$v = 250 -250 {e^(-6000t)}$

$v = 250[1 - {e^(-6000t)}]$

Therefore, the voltage across the capacitor is $v = 250[1 - {e^(-6000t)}]$ mV

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What are two advantages of forging when compared to machining a part from a billet?

Answers

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

At steady state, air at 200 kPa, 325 K, and mass flow rateof 0.5 kg/s enters an insulated duct having differing inlet
and exit cross-sectional areas. The inlet cross-sectional area is
6 cm26cm
2. At the duct exit, the pressure of the air is 100 kPa and the velocity is 250 m/s. Neglecting potential energy
effects and modeling air as an ideal gas with constant cp=1.008 kJ/kg⋅Kc
p =1.008kJ/kg⋅K, determine
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm2
(a) the velocity of the air at the inlet, in m/s.
(b) the temperature of the air at the exit, in K.
(c) the exit cross-sectional area, in cm

Answers

Letra A

A letra

A.
Thank

Cody’s car accelerates from 0m/s to 45 m/s northward in 15 seconds. What is the acceleration of the car

Answers

Answer:

3 m/s²

Explanation:

Acceleration is calculated as :

a= Δv/ t

where ;

Δv = change in velocity

Δv = 45 - 0 = 45 m/s

t= 15 s

a= 45 /15

a= 3 m/s²

Describe the difference between design guidelines or criteria and design performance. Explain the relationship between the use of guideline/criteria tools and performance tools during the design process

Answers

Answer:

PART A

Design guidelines are sets of procedures to be followed in order to enhance the designing of an object or other things.

Design Performance is the actual process of carrying out the design process of an object using the design guidelines or criteria.

PART B

(1) Design guidelines tools helps to enhance design Performance.

(2) Design guidelines tools helps the designing performance tools to be effective.

Explanation:Design guidelines are the various steps which has special tools used to guide the designer in order to enhance the designing performance tools and ensure that the design process is done devoid of errors.

Design Performance tools are tools which helps to enhance the actual design Activities.

1.19. A gas is confined in a 0.47 m diameter cylinder by a piston, on which rests a weight. The mass of the piston and weight together is 150 kg. The local acceleration of gravity is 9.813 m·s−2, and atmospheric pressure is 101.57 kPa. (a) What is the force in newtons exerted on the gas by the atmosphere, the piston, and the weight, assuming no friction between the piston and cylinder? (b) What is the pressure of the gas in kPa? (c) If the gas in the cylinder is heated, it expands, pushing the piston and weight upward. If the piston and weight are raised 0.83 m, what is the work done by the gas in kJ? What is the change in potential energy of the piston and weight?\

Answers

In this exercise we have to use the knowledge of force to calculate the required energies, so we have to:

a) 19094 N

b) 110.055 kPa

c) 1222 J

What is the concept of force?

In the field of physics, force is a physical action that causes deformation or that changes the state of rest or movement of a given object.

a) Knowing that the force formula is defined by:

$F = P + p * A\nF = m * g + p *\pi /4 * d^2\nF = 150 * 9.813 + 101570 * \pi /4 * 0.47^2 = 19094 N$

b) Knowing that the force exerted by an area is equal to the pressure in that area, we have:

$p_1 = F / A\np_1 = F / (\pi /4 * d^2)\np_1 = 19094 / (\pi /4 * 0.47^2) = 110055 Pa = 110.055 kPa$

c)So calculating the potential energy we have:

$\Delta E_p = m * g * \Delta h\n\Delta E_p = 150 * 9.813 * 0.83 = 1222 J$

See more about force at brainly.com/question/26115859

Answer:

a) 19094 N

b) 110.055 kPa

c) 1222 J

Explanation:

The force on the gas is the weight plus the atmospheric pressure multiplied by the piston area

F = P + p * A

F = m * g + p * π/4 * d^2

F = 150 * 9.813 + 101570 * π/4 * 0.47^2 = 19094 N

The pressure is the force divided by the area of the piston

p1 = F / A

p1 = F / (π/4 * d^2)

p1 = 19094 / (π/4 * 0.47^2) = 110055 Pa = 110.055 kPa

variation of gravitational potential energy is defined as

ΔEp = m * g * Δh

ΔEp = 150 * 9.813 * 0.83 = 1222 J

WHEN A CAR WITH BRIGHT HEADLIGHTS COMES TOWARD YOU AT NIGHT, YOU SHOULD:A. Move toward the right edge of your lane

B. Look above the oncoming headlights

C. Look below the oncoming headlights

D. Look toward the right edge of your lane
Help

Answers

Look below the oncoming headlights when a car with bright headlights comes toward you at night.

When a car with bright headlights comes toward you at night, you should look below the oncoming headlights.

This is because looking directly into the bright lights can temporarily impair your vision and make it difficult to see the road ahead. By looking slightly below the headlights, you can still maintain visibility of the road while minimizing the direct glare.

Moving toward the right edge of your lane or focusing solely on the right edge of your lane can limit your overall visibility and may not be safe or necessary in this situation.

To learn more about driving at night, refer:

brainly.com/question/30160858

#SPJ4

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Answer: D

I hope this helped!

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