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anzhelika [568]
3 years ago
10

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

pendulum is pulled to one side so that the rod is at an angle θ from the vertical, and released from rest.
a. Show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere?
b. Repeat part (a) for the instant when the pendulum rod is at an angle 9/2 from the vertical.
c. Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

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Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?
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The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

<h3>What is an ideal diode?</h3>
  • A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
  • A semiconductor diode is the kind of diode that is used the most commonly.
  • It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.
<h3>How to find the current in ideal diode?</h3>
  • Here we have given with the values,

                       V_2=65V\\V_1=0V\\R_1=490Ohm.

  • We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

                I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

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It is a measure of the strength of the bonds between ions.

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The first two choices are wrong because it is actually the opposite.

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As for the other third option, it is wrong because lattice energy is the energy RELEASED not absorbed.

3 0
3 years ago
Read 2 more answers
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