Two a.c V1=100sin(wt) and V2 = 150cos(wt) are fed into one circuit, determine the combined output of the two as a single a.c

Answers

Answer 1

Answer:

Explanation:

V = 100sin(ωt) + 150cos(ωt)

let x = ωt

V = 100sin(x) + 150cos(x)

a maximum or minimum will occur when the derivative is zero

V' = 100cos(x) - 150sin(x)

0 = 100cos(x) - 150sin(x)

100cos(x) = 150sin(x)

100/150 = sin(x)/cos(x)

0.6667 = tan(x)

x = 0.588 rad

V = 100sin(0.588) + 150cos(0.588)

V = 180.27756

as the maximum will not occur until ωt = 0.588 radians, for a cosine function we subtract that amount as a phase angle φ

V = 180.3 cos(ωt - 0.588)

or as a sine function, the phase angle lags the cosine by a difference of π/2

V = 180.3sin(ωt - (0.588 - π/2)

V = 180.3sin(ωt + 0.983)


Related Questions

What does the horizontal line through the center of the wave on a graph represent?

Answers

Answer:

This is the midline or the medium which is the exact middle of the graphs minimum and maximum points(which are the amplitude)

A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the ground). Neglect drag and the initial height of the football.
How long does the football need to rise?

What height will the football reach?

With what speed does the punter need to kick the football?

At what angle (θ), with the horizontal, does the punter need to kick the football?

Answers

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

if the momentum of a 1,400 kg car is the same as the truck in question 17, what is the velocity of the car?

Answers

Answer:

Explanation:

momentum is mass times velocity

p = mv

so take the momentum of the truck in question 17 and divide by the mass of this car

v = p/m = p / 1400

Understanding what motivates anyone is not easy because each individual has different

Answers

Has different what????

A disgruntled physics student, frustrated with
finals, releases his tensions by bombarding the
adjacent building, 13.5 m away, with water
balloons. He fires one at 38◦
from the horizontal with an initial speed of 23.6 m/s.
The acceleration of gravity is 9.8 m/s
2
.
For how long is the balloon in the air?

Answers

Answer:

Explanation:

The balloon would require a time of

t = d/v = 13.5/ (23.6cos38) = 0.7259...s

to travel the horizontal distance.

the vertical position relative to the throw point at that time is

h = 0 + (23.6sin38)(0.7259) + ½(-9.8)(0.7259²)

h = 7.9652...

so as long as the adjacent building is at least 8.0 m higher than the student position, the balloon is in the air for 0.726 s.

If the building is shorter than 8.0 m above the student, the balloon will land on the building roof and will be in the air for a longer period of time

What is the net force here?


11 N left
6 N right
1 N right
4 N right

Answers

answer = 6n to the right

Explanation:

2n plus 4n equals 6n

since 6n is more than 5n it goes 6n to the right

what type of data do you need to collect in a ADI​

Answers

full name.
address.
driving licence number.
email address.
telephone number.
ethnicity (optional)
website address.
convictions (motoring and non-motoring)
……………….

Dagmar says that diffusion happens really quickly. Is he right or wrong? Explain.

Answers

Answer:

Diffusion in gases is quick because the particles in a gas move quickly. It happens even faster in hot gases because the particles of gas move faster.

Find the net torque .

Answers

Answer:

Explanation:

I will ASSUME this means torque about the dot.

3) 20(3) + 10(6) - 30(4) = 0 N•m

4) 10(0.5) - 6sin45(1) = -0.7573593... or about 0.76 N•m CCW

5) 25(3) - 40sin30(4) = -5 N•m or 5 N•m CCW

6) 15(3) - 12(2) - 10sin45(6) = -21.4264068... or about 21 N•m CCW

The mass of fifteen washers is _____ kg, which exerts a force of _____ N

Answers

Answer:

It could be related with the lesson from which this question belongs as far we did not read the lesson

Sorry

59. (II) The crate shown in Fig. 4-60 lies on a plane tilted at an angle A = 25.0° to the horizontal, with Mk 0.19. (a) Determine the acceleration of the crate as it slides down the plane. (b) If the crate starts from rest 8.15 m up along the plane from its base, what will be the crate's speed when it reaches the bottom of the incline?​

Answers

Explanation:

a) We need to write down first Newton's 2nd law as applied to the given system. The equations of motion for the x- and y-axes can be written as follows:

[tex]x:\;\;\;\;\;mg\sin 25° - \mu_kN = ma\;\;\;\;\;\;(1)[/tex]

[tex]y:\;\;\;\;\;N - mg\cos 25° = 0\;\;\;\;\;\;\;\;\;(2)[/tex]

From Eqn(2), we see that

[tex]N = mg\cos 25°\;\;\;\;\;\;\;(3)[/tex]

so using Eqn(3) on Eqn(1), we get

[tex]mg\sin 25° - \mu_kmg\cos 25° = ma[/tex]

Solving for the acceleration, we see that

[tex]a = g(\sin 25° - \mu_k\cos 25°)[/tex]

[tex]\;\;\;\;= 2.45\:\text{m/s}^2[/tex]

b) Now that we have the acceleration, we can now solve for the velocity of the crate at the bottom of the plane. Using the equation

[tex]v^2 = v_0^2 + 2ax[/tex]

Since the crate started from rest, [tex]v_0 = 0.[/tex] Thus our equation reduces to

[tex]v^2 = 2ax \Rightarrow v = \sqrt{2ax}[/tex]

[tex]v = \sqrt{2(2.45\:\text{m/s}^2)(8.15\:\text{m})}[/tex]

[tex]\;\;\;\;= 6.32\:\text{m/s}[/tex]

An object of mass 6.36 kg is released from rest and drops 2.05 m to the floor. The collision is completely inelastic. How much kinetic energy is lost during the collision

Answers

Answer:

Essentially all of it

Explanation:

The potential energy was

PE = mgh = 6.36(9.81)(2.05) = 127.90278 = 128 J

ignoring air resistance, this PE converts to KE

With no rebound final velocity is zero, so Kinetic energy lost = 128 J

AnswAnswer This!!!!!!
I'll give brainliest to whoever gets it right.

Answers

answer: 1.0 mol

8/2 = 4/2 = 2/2 = 1

3. A 1500 kg car moving at 30 m/s strikes a 6000 kg van initially at rest. If the car
comes to a complete stop after the collision, what is the final velocity of the van?

Answers

Answer:

7.5m/s

Explanation:

Force= mass × velocity

Energy is conserved, the car and van should have the same overall force.

1500kg × 30m/s= 6000kg × final velocity

Final velocity = 7.5m/s

In a police ballistics test, 2.00-g bullet traveling at 700 m/s suddenly hits and becomes embedded in a stationary 5.00-kg wood block. What is the speed of the block immediately after the bullet has stopped moving relative to the block

Answers

Answer:

Here we use the conservation of momentum theorem.

m stands for mass, and v stands for velocity. The numbers refer to the respective objects.

m1v1 + m2v2 = m1vf1 + m2vf2

Since the equation is perfectly inelastic, the final velocity of both masses is the same. Let’s account for this in our formula.

m1v1 + m2v2 = vf(m1 + m2)

Let’s substitute in our givens.

(0.002 kg)(700 m/s) + (5 kg)(0 m/s) = vf(0.002 kg + 5 kg)

I assume you are proficient in algebra I, so I will not include the steps to simplify this equation.

Note that I have considered the bullet’s velocity to be in the positive direction,

The answer is vf = 0.280 m/s

A 5.0 m length of rope, with a mass of 0.52 kg, is pulled taut with a tension of 46 N. Find the speed of waves on the rope

Answers

Answer:

Speed of waves on the rope is 21 m/s

Explanation:

Length of the rope (l) = 5.0 m

Mass of the rope (m) = 0.52 kg

Tension in the rope (T) = 46 N

Formula of speed of waves on the rope:

[tex] \bold{v = \sqrt{\dfrac{T}{\mu}}} [/tex]

[tex] \mu [/tex] = Mass per unit length of the rope (m/l)

By substituting the values in the formula we get:

[tex] \implies \rm v = \sqrt{\dfrac{T}{ \dfrac{m}{l} }} \\ \\ \implies \rm v = \sqrt{\dfrac{Tl}{m}} \\ \\ \implies \rm v = \sqrt{ \dfrac{46 \times 5}{0.52} } \\ \\ \implies \rm v = \sqrt{ \dfrac{230}{0.52} } \\ \\ \implies \rm v = \sqrt{442.3} \\ \\ \implies \rm v = 21 \: m {s}^{ - 1} [/tex]

Speed of waves on the rope (v) = 21 m/s

Can you solve this question?

Answers

Hi there!

In this instance, the object's centripetal force is provided by the horizontal component of the tension, so:

Tsinθ = mv²/r

**We use sine because in this situation, the angle is with the vertical**

We can plug in the known values for tension and theta:

60sin(60) = mv²/r

51.96 = mv²/r

The radius is equivalent to the sine of the string in respect to theta:

sin(60) = O/H = r/L

2sin(60) = 1.732 m

Now, solve for the velocity:

51.96 = mv²/r

51.96r / m = v²

51.96(1.732)/.400 = v²

v² = 225

v = 15 m/s

define parking orbit?​

Answers

A parking orbit is a temporary orbit used during the launch of a spacecraft. A launch vehicle boosts into the parking orbit, then coasts for a while, then fires again to enter the final desired trajectory.

Answer:

An orbit of a spacecraft from which the spacecraft or another vehicle may be launched on a new trajectory.

A block of mass m = 3.0 kg is pushed a distance d = 2.0 m along a frictionless horizontal table by
a constant applied force of magnitude F= 20.0 N directed at an angle 0= 30.0° below the horizontal
as shown in Figure. Determine the work done by (a) the applied force, (b) the normal force exerted
by the table, and (d) the net force on the block.

Answers

Explanation:

We apply the definition of work by a constant force in the first three parts, but then in the fourth part we add up the answers. The total (net) work is the sum of the amounts of work done by the individual forces, and is the work done by the total (net) force. This identification is not represented by an equation in the chapter text, but is something you know by thinking about it, without relying on an equation in a list.

The definition of work by a constant force is W=FΔrcosθ.

(a) The applied force does work given by

W=FΔrcosθ=(16.0N)(2.20m)cos25.00=31.9J

(b), (c) The normal force and the weight are both at 900 to the displacement in any time interval. Both do 0 work.

(d) ∑W=31.9J+0+0=31.9J

what is the pressure exerted by a force of 25 N on an area of 5m square

Answers

Answer:

pressure = force / area

then pressure = 25 / 5 = "5" N/m^2

Describe a vibration that is not periodic. NO LINKS PLEASE

Answers

Answer:

1)The position change of almost any manually operated room light switch.

2) Sunlight striking a point on the ground on a partly cloudy and windy day

Explanation:

I need ideas of what kind of simple motor i can build and how i can build it. The simple motor MUST spin without using your own force. What materials would i use and how would i create it. what would i create

Answers

Answer:

i don't know but my father i think he can't answer this

Name the energy possessed by hot air

Answers

Answer:

geothermal energy

Explanation:

the energy is obtained from the heat within the surface of earth

Answer:

heat energy

Explanation:

CAN SOMEONE PLZ HELP

Answers

Answer:

magnetic force.

Explanation:bc it makes sense, and can i please get brainliest answer i never asked. its ok if you say no. have a great day <3.

In a Little League baseball game, the 145 g ball enters the strike zone with a speed of 17.0 m/s . The batter hits the ball, and it leaves his bat with a speed of 20.0 m/s in exactly the opposite direction. Part A What is the magnitude of the impulse delivered by the bat to the ball

Answers

Hi there!

Impulse = Change in momentum

I = Δp = mΔv = m(vf - vi)

Where:

m = mass of object (kg)

vf = final velocity (m/s)

vi = initial velocity (m/s)

Begin by converting grams to kilograms:

1 kg = 1000g ⇒ 145g = .145kg

Now, plug in the given values. Remember to assign directions since velocity is a vector. Let the initial direction be positive and the opposite be negative.

I = (.145)(-20 - 17) = -5.365 Ns

The magnitude is the absolute value, so:

|-5.365| = 5.365 Ns

If an object accelerates from rest, what will its velocity be after 1.3 s if it has a constant acceleration of 9.1 m/s^2?

Answers

[tex]\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time, t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0 +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}[/tex]

what torque required stopping awheel of moment of inertia 6 × 10^-3kgm2 from speed of 40rad/s in 20 sec.​

Answers

solution:

the formula is T = F * r * sin(theta) so just input the numbers and solve it.

Explanation:

Torque is the twisting force that tends to cause rotation. The point where the object rotates is known as the axis of rotation. Mathematically, torque can be written as T = F * r * sin(theta), and it has units of Newton-meters

An automobile moving along a straight track changes its velocity from 40 m/s to 80 m/s in a distance of 200 m. What is the (constant) acceleration of the vehicle during this time

Answers

Answer:

[tex]\huge\boxed{\sf a = 1200\ m/s\²}[/tex]

Explanation:

Given Data:

Initial Velocity = Vi = 40 m/s

Final Velocity = Vf = 80 m/s

Distance = S = 200 m

Required:

Acceleration = a = ?

Formula:

2aS = Vf² - Vi² (THIRD EQUATION OF MOTION)

Solution:

2a (200) = (80)² - (40)²

400a = 6400 - 1600

400a = 4800

Divide 400 to both sides

a = 4800 / 400

a = 1200 m/s²

[tex]\rule[225]{225}{2}[/tex]

Hope this helped!

~AH1807

A spring in a dart gun is compresscht a distance of 0.05 m. The spring has a spring constant
of 1,115 N/m. If the dart has a mass of 0.025 kg, determine the velocity of the dart as it
leaves the dart gun.

Answers

Answer:

Explanation:

ASSUMING that the dart is fired horizontally so that gravity potential energy considerations are not needed. Also ignoring friction work.

The spring potential will convert to kinetic.

KE = PS

½mv² = ½kx²

     v = [tex]\sqrt{kx^2/m}[/tex]

     v = [tex]\sqrt{1115(0.05^2)/0.025}[/tex]

     v = 10.55935...

     v = 11 m/s

Can someone help label these?

Answers

A. reactants
B. subscript
C. coefficient
D. products
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