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liberstina [14]
3 years ago
6

Maximum voltage produced in an AC generator completing 60 cycles in 30 sec is 250V. (a) What is period of armature? (b) How many

cycles are completed in T/2 sec? (c) What is the maximum emf produced when the armature completes 180 0 rotation?
Physics
1 answer:
Vlada [557]3 years ago
8 0

Answer:

a. 2 Hz b. 0.5 cycles c . 0 V

Explanation:

a. What is period of armature?

Since it takes the armature 30 seconds to complete 60 cycles, and frequency f = number of cycles/ time = 60 cycles/ 30 s = 2 cycles/ s = 2 Hz

b. How many cycles are completed in T/2 sec?

The period, T = 1/f = 1/2 Hz = 0.5 s.

So, it takes 0.5 s to complete 1 cycles. At t = T/2 = 0.5/2 = 0.25 s,

Since it takes 0.5 s to complete 1 cycle, then the number of cycles it completes in 0.25 s is 0.25/0.5 = 0.5 cycles.

c. What is the maximum emf produced when the armature completes 180° rotation?

Since the emf E = E₀sinθ and when θ = 180°, sinθ = sin180° = 0

E = E₀ × 0 = 0

E = 0

So, at 180° rotation, the maximum emf produced is 0 V.

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The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other.

<h3>Why is graphite a solid lubricant?</h3>

We know that graphite is composed of layers. These hexagonal layers are held together by weak Van Der Walls forces and as such are able to slide over each other. The carbon atom in each layer are held together by strong covalent bonds.

The fact that the layers of graphite are held together by only weak Van der Walls forces implies that they can slide over each other and as such make the graphite fluid.

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3 0
2 years ago
A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N
kotegsom [21]

Answer:

(A) = 34,560 J

(B) Work done during the round trip is not zero, hence the resistive force is not a conservative force.

Explanation:

This question is incomplete, here is the complete question:

A girl rides her bike 5.4 km due east. While riding she experiences a resistive force from the air that has a magnitude of 3.1 N and points due west. She then turns around and rides due west, back to her starting point. The resistive force from the air on the return trip has a magnitude of 3.3 N and points due east.? (a) Find the work done by the resistive force during the round trip. (b) Based on your answer to part (a), is the resistive force a conservative force? Explain.

SOLUTION:

distance covered by the girl due east (De) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fw) = 3.1 N due west

distance covered by the girl due west (Dw) = 5.4 km = 5400 m

magnitude of resistive force by the air (Fe) = 3.3 N due east

(A) to find the work done by the resistive force during the round trip we have to get the work done by the resistive force due west and add it to the work done by the resistive force due east

  • work done by the resistive force due west = (Fw.cosθ) x De

        where θ is the angle between the displacement and the force. The    

        displacement is due east while the force is due west, hence θ = 180°

       work done (W1) = (3.1 x cos 180) x 5400 = -16,740 j

  • work done by the resistive force due east = (Fe.cosθ) x Dw

        where θ is the angle between the displacement and the force. The    

        displacement is due west while the force is due east, hence θ = 180°

       work done (W2) = (3.3 x cos 180) x 5400 = -17,820 j

Hence work done during the round trip = W1 + W2 = (-16,740 ) + (-17,820)

= 34,560 J

(B) A conservative force is a force in which the work done by the force in moving an object around a closed path is zero. From part (A) above the work done during the round trip is not zero, hence the resistive force is not a conservative force.

3 0
2 years ago
What is the velocity of the object?
dmitriy555 [2]
<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2>\huge\boxed{\text{V = 9.5 m/s}}</h2><h2>_____________________________________</h2>

<h2>DATA:</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

{\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

                           W\ =\ \frac{1}{2}\:6\:x\:30

                           W   =  90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2

\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}

\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

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