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Amiraneli [1.4K]
2 years ago
7

Someone help me answer this.

Physics
1 answer:
horrorfan [7]2 years ago
8 0

The angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

<h3>What is angular velocity?</h3>

The velocity of a particle when moving in the circular path.

Let speed of the bug with respect to ground is u.

Speed of bag with respect to ring will be

v = u - (- Rω) =

Then, u = v- Rω...............(1)

Angular momentum of ring and bug will remain conserved.

Initial  momentum: L ring + Lbug =0

Final  momentum:  -2m₁ R²ω + m₂uR =0...............(2)

Using equation (1) and (2), the angular velocity expression will be

ω =[m₂v / {(2m₁ +m₂)R}] in positive z direction

Thus, the angular velocity of the ring when the bug is halfway around and the angular velocity of the ring when the bug is back at the pivot is [m₂v / {(2m₁ +m₂)R}].

Learn more about angular velocity.

brainly.com/question/17592191

#SPJ1

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What was the greatest contribution of the monasteries? A. Illuminated manuscripts. B. Cloisters. C. Sculpture.d. Transepts
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Please help!!!
yuradex [85]
Gravitational potential energy changes into kinetic energy. The equation for gravitational potential energy is GPE = mgh, where m<span> is the mass in kilograms, </span>g<span> is the </span>acceleration<span> due to </span>gravity<span> (9.8 on Earth), and </span>h<span>is the height above the ground in meters.
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Therefore for the problem:

m = 70 kg
g = 10m/s^2
h = 10m

<span>GPE = mgh = (70kg)(10m/s^2)(10m) = 7000 J or 7 kJ
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I hope my answer has come to your help. God bless you and have a nice day ahead!


5 0
4 years ago
Compare the time periodof two pendulums of length 4m and 9m​
svlad2 [7]

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T

1

=2π

g

1

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4 0
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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