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Tcecarenko [31]
2 years ago
5

A disk, with a radius of 0.25 m, is to be rotated like a merry-go-round through 1000 rad, starting from rest, gaining angular sp

eed at the constant rate through the first 500 rad and then losing angular speed at the constant rate until it is again at rest. The magnitude of the centripetal acceleration of any portion of the disk is not to exceed 400 m/s².
(a) What is the least time required for the rotation?
(b) What is the corresponding value of α₁?

Physics
1 answer:
aniked [119]2 years ago
6 0

Answer:

(a) The least time required for the rotation is 50 seconds

(b) The corresponding value of α₁ is 1.6 rad/s²

Explanation: Please see the attachments below

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A transformer is designed to provide 6 V from a 150 V supply. If the primary coil has 1000 turns, how many turns does the second
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The applicable relationship is N1/N2 = V1/V2, meaning the ratio of primary voltage to secondary voltage is equal to the ratio of primary turns to secondary turns.

Here N1 = 1000, V1 = 250, V2 = 400V and N2 = TBD.

Rewriting the above relationship, N2 = N1 V2/V1 = 1000 x 400/250 = 1600 turns.
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What charge accumulates on the plates of a 2.0-μF air-filled capacitor when it is charged until the potential difference across
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Answer:

0.0002 C.

Explanation:

Charge: This can be defined as the ratio of current to time flowing in a circuit. The S.I unit of charge is Coulombs (C)

Mathematically, charge can be expressed as

Q = CV ................................. Equation 1

Where Q = amount of charge, C = capacitance of the capacitor, V = potential difference across the plates.

Given: C = 2.0-μF = 2×10⁻⁶ F, V = 100 V.

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Q = 2×10⁻⁶× 100

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The amount of charge accumulated = 0.0002 C

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What is unique about the movement of uranus compared to the other outer planets
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Uranus is tilted so far that it essentially orbits the sun on its side, with the axis of its spin nearly pointing at the star
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If the mass of an object i if the mass of an object is 44 kg in its velocity is 10 m/s East how much kinetic energy
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3 years ago
A proton is released from rest at the origin in a uniform electric field that is directed in the positive xx direction with magn
elena-s [515]

Answer:

The change in potential energy is  \Delta  PE =  -  3.8*10^{-16} \ J

Explanation:

From the question we are told that

     The  magnitude of the uniform electric field  is  E =  950 \ N/C

      The  distance traveled by the electron is  x =  2.50 \ m

Generally the force on this electron is  mathematically represented as

     F =  qE

Where F is the force and  q is the charge on the electron which is  a constant value of  q =  1.60*10^{-19} \ C

    Thus  

      F  =  950  * 1.60 **10^{-19}

      F  = 1.52 *10^{-16} \ N

Generally the work energy theorem can be mathematically represented as

          W =  \Delta  KE

Where W is the workdone on the electron by the  Electric field and  \Delta  KE  is the change in kinetic energy

Also  workdone on the electron can also  be represented as

        W =  F* x  *cos(  \theta )

Where  \theta  =  0 ^o considering that the movement of the electron is along the x-axis  

        So

             \Delta  KE  =  F  * x  cos  (0)

substituting values

         \Delta  KE  =  1.52 *10^{-16}  * 2.50   cos  (0)

          \Delta  KE   =  3.8*10^{-16} J

Now From the law of energy conservation

       \Delta PE  =  -  \Delta  KE

Where \Delta  PE is the change  in  potential energy  

Thus  

        \Delta  PE =  -  3.8*10^{-16} \ J

               

7 0
3 years ago
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