What is the difference between V(peak voltage) and Vrms (root-mean-square) of AC voltage source?

Answers

Answer 1

Answer:

Answer:

V(peak voltage) is the highest voltage that the waveform will ever attain and the Vrms(root-mean-square) is the effective voltage of the total waveform representing the AC source.

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Points A, B, and C are at the corners of an equilateral triangle of side 8 m. Equal positive charges of 4 mu or micro CC are at A and B. (a) What is the potential at point C? 8.990 kV * [2.5 points] 2 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] 8.990 OK (b) How much work is required to bring a positive charge of 5 mu or micro CC from infinity to point C if the other charges are held fixed? .04495 J * [2.5 points] 1 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] .04495 OK (c) Answer parts (a) and (b) if the charge at B is replaced by a charge of -4 mu or micro CC. Vc= kV [2.5 points] 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] W =

Answers

Answer:

a) 8.99*10³ V b) 4.5*10⁻² J c) 0 d) 0

Explanation:

The electrostatic potential V, is the work done per unit charge, by the electrostatic force, producing a displacement d from infinity (assumed to be the reference zero level).
For a point charge, it can be expressed as follows:

$V =(k*q)/(d)$

As the electrostatic force is linear with the charge (it is raised to first power), we can apply superposition principle.
This means that the total potential at a given point, is just the sum of the individual potentials due to the different charges, as if the others were not there.
In our case, due to symmetry, the potential, at any corner of the triangle, is just the double of the potential due to the charge located at any other corner, as follows:

$V = (2*q*k)/(d) = (2*8.99e9N*m2/C2*4e-6C)/(8m) =\n \n V= 8.99e3 V$

The potential at point C is 8.99*10³ V

The work required to bring a positive charge of 5μC from infinity to the point C, is just the product of the potential at this point times the charge, as follows:

$W = V * q = 8.99e3 V* 5e-6C = 4.5e-2 J$

The work needed is 0.045 J.

If we replace one of the charges creating the potential at the point C, by one of the same magnitude, but opposite sign, we will have the following equation:

$V = (8.99e9N*m2/C2*(4e-6C))/(8m) + ((8.99e9N*m2/C2*(-4e-6C))/(8m)) = 0$

This means that the potential due to both charges is 0, at point C.

If the potential at point C is 0, assuming that at infinity V=0 also, we conclude that there is no work required to bring the charge of 5μC from infinity to the point C, as no potential difference exists between both points.

A thin Nichrome wire connected to an ammeter surrounds a region of time-varying magnetic flux, and the ammeter reads 13 amperes. If instead of a single wire we use a coil of thin Nichrome wire containing 23 turns, what does the ammeter read?

Answers

Answer:

The current would be same in both situation.

Explanation:

Given that,

Current I = 13 A

Number of turns = 23

We need to calculate the induced emf

Using formula of induced emf is

$\epsilon=NA(dB)/(dt)$

For N = 1

$\epsilon=A(dB)/(dt)$

We need to calculate the current

Using formula of current

$i=(\epsilon)/(R)$

Put the value of emf

$i=(A(dB)/(dt))/(R)$

Now, if the number of turn is 22 , then induced emf would be

$\epsilon'=NA(dB)/(dt)$

Then the current would be

$i'=(\epsilon')/(NR)$

$i'=(NA(dB)/(dt))/(NR)$

$i'=(A(dB)/(dt))/(R)$

$i'=i$

Hence, The current would be same in both situation.

On December 26, 2004, a great earthquake occurred off the coast of Sumatra and triggered immense waves (tsunami) that killed some 200000 people. Satellites observing these waves from space measured 800 from one wave crest to the next and a period between waves of 1.0 hour.Part AWhat was the speed of these waves in m/s?Express your answer using two significant figures.=Part BWhat was the speed of these waves in km/h ?Express your answer using two significant figures.=

Answers

Answers:

a) 222.22 m/s

b) 800.00 km/h

Explanation:

The speed of a wave is given by the following equation:

$v=f \lambda$

Where:

$v$ is the speed

$f=(1)/(T)$ is the frequency, which has an inverse relation with the period $T=1 h$

$\lambda=800 km$ is the wavelength

Solving with the given units:

$v=(1)/(T)\lambda$

$v=(1)/(1 h)800 km$

$v=800.00 km/h$ This is the speed of the wave in km/h

Transforming this speed to m/s:

$v=800.00 (km)/(h) (1 h)/(3600 s) (1000 m)/(1 km)$

$v=222.22 m/s$ This is the speed of the wave in m/s

Which of the following describes the net force acting on an object?The sum of all forces acting on an object
The gravitational force minus any contact forces acting on an object
The difference between the normal force and the gravitational force acting on an object
The sum of all the forces acting on an object in the same direction

Answers

The sum of all forces acting on an object in the same direction is described for the net force acting on an object.

What is a Net force?

When the forces are acting in the same direction of movement of the object it can be said as sum of the two individual forces will be equal to the "Net Force" .

The net force is the combined force of all individual forces acting on an object.
If the object with the forces in the opposite direction, then the net force will not be equal to the sum of the forces.

Example : If two forces (2 kids pushing in the same direction to move the object big box) act on an object (big box) in the same direction, then the net force is equal to the sum of the two forces. If the kids pushed in the opposite direction, the net force will not occur.

Hence, Option D is the correct answer.

Learn more about Net force,

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Answer:

The sum of all the forces acting on an object in the same direction.

G Water enters a house through a pipe 2.40 cm in diameter, at an absolute pressure of 4.10 atm. The pipe leading to the second-floor bathroom, 5.20 m above, is 1.20 cm in diameter. The flow speed at the inlet pipe is 4.75 m/s a) What is the algebraic expression for flow speed in the bathroom?
b) Calculate the flow speed in the bathroom.
c) What is algebraic expression for the pressure in the bathroom?
d) Calculate the water pressure in the bathroom. Report your answer in the (atm) unit.

Answers

Answer:

A) A₁ V₁ = A₂V₂

B) V₂ = 19 m /s

C) P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

D) P₂ = 1.88 atm

Explanation:

A) From the piaget's theory of conservation of volume, we can calculate the rate of flow of water from;

A₁ V₁ = A₂V₂

Where;

A₁ and A₂ are area of cross section V₁ and V₂ are velocity of flow at two places along pipe.

B) Using the formula given in A above, we obtain;

π x 1.2² x 4.75 = π x 0.6² x V₂

V₂ x 0.36 = 6.84

V₂ = 6.84/0.36

V₂ = 19 m /s

c ) To find pressure we shall apply Bernoulli's theorem in fluid dynamics;

P₁ + (1/2)ρv₁² = P₂ + (1/2)ρv₂² + (h₂ - h₁ )ρg

Where;

P₁ and P₂ are pressure at ground and second floor respectively

v₁ and v₂ are velocity at ground and second floor respectively

h₁ and h₂ are height at ground and second floor respectively ρ is density of water.

Thus, plugging in the relevant values to obtain;

4.1 x 10⁵ + (1/2 x 1000 x 4.75²) = P₂ + (1/2 x 1000 x 19²) + (5.2 x 1000 x 9.8)

(4.1 x 10⁵) + (0.11 x 10⁵) = P₂ + (1.8 X 10⁵) + (0.51 X 10

P₂ = 1.9 X 10⁵ N/m² = 1.88 atm

A traffic light weighing 200N hangs from a vertical cable tied to two other cables that are fastened to to a support ,as shown . The upper cables make angles 41° and 63° with the horizontal . Calculate the tension in of the three cables

Answers

Answer:

93.6 N in the 41° cable
155.6 N in the 63° cable
200 N in the vertical cable

Explanation:

Let T and U represent the tensions in the 41° and 63° cables, respectively. In order for the system to be stationary, the horizontal components of these tensions must balance, and the vertical components of these tensions must total 200 N.

Tcos(41°) =Ucos(63°) . . . . . balance of horizontal components

U = Tcos(41°)/cos(63°) . . . . write an expression for U

The vertical components must total 200 N, so we have ....

Tsin(41°) +Usin(63°) = 200

Tsin(41°) +Tcos(41°)sin(63°)/cos(63°) = 200

T(sin(41°)cos(63°) +cos(41°)sin(63°))/cos(63°) = 200

T = 200cos(63°)/sin(41° +63°) ≈ 93.6 . . . newtons

U = 200cos(41°)/sin(41° +63°) ≈ 155.6 . . . newtons

The vertical cable must have sufficient tension to balance the weight of the traffic light, so its tension is 200 N.

Then the tensions in the 3 cables are ...

41°: 93.6 N

63°: 155.6 N

90°: 200 N

The tension in each of the three cables are 94.29, 155.56 and 200 Newton respectively.

Given the following data:

Force = 200 Newton.

Angle 1 = 41°

Angle 2 = 63°

How to calculate the tension.

First of all, we would determine the third tension force based on the vertical component as follows:

$\sum F_y = 0\n\nT_3 - F_g =0\n\nT_3 - F_g=200\;N$

Next, we would apply Lami's theorem to resolve the forces acting on the traffic light at equilibrium:

For the horizontal component:

$\sum F_x = -T_1cos41+T_2cos 63=0\n\n0.7547T_1=0.4540T_2\n\nT_1=(0.4540T_2)/(0.7547)\n\nT_1 = 0.6016T_2$ ....equation 1.

For the vertical component:

$\sum F_y = T_1sin41+T_2sin 63-T_3=0\n\n\sum F_y = T_1sin41+T_2sin 63-200=0\n\n0.6561T_1+0.8910T_2 =200$ ...equation 2.

Substituting eqn. 1 into eqn. 2, we have:

$0.6561 * (0.6016T_2)+0.8910T_2 =200\n\n0.3947T_2+0.8910T_2 =200\n\n1.2857T_2 =200\n\nT_2 = (200)/(1.2857) \n\nT_2 = 155.56\;Newton$

For the first tension:

$T_1 = 0.6061T_2\n\nT_1 = 0.6061 * 155.56\n\nT_1 = 94.29\;Newton$

Read more on tension here: brainly.com/question/4080400

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