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kupik [55]
3 years ago
6

A satellite of mass M moves in a circular orbit of radius R at a constant speed v around the Earth which has mass ME.

Physics
1 answer:
Gwar [14]3 years ago
3 0

-- "Work" is the product of (force exerted on the object) x (distance the object moves in the direction of the force).

Since the orbit is a circle, the gravitational force toward the center is always perpendicular to the orbit.  The object never moves in the direction of the force.  If it did, it wouldn't be 'R' away from the center of the circle.

So the product of (gravitational force) x (distance in the direction of the force) is always zero.

-- Even if the orbit ISN't a circle . . . there are some parts of the orbit that aren't quite perpendicular to the gravitational force.  If the satellite is traveling through one of those parts AND getting closer to the central body, then gravity is doing positive work on the satellite.  If the satellite is traveling through one of those parts and getting FARTHER from the central body, then the satellite is the one doing positive work, and gravity is doing 'negative work'.  The work done by gravity ... and the work done by the satellite ... is zero over a complete revolution, although not zero at every point.

This is exactly the definition of a "Conservative Force" ... a force that does zero work through one trip around any CLOSED path.  Gravity is a conservative force, and so is the Electrostatic force.

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In an LC circuit containing a 40-mH ideal inductor and a 1.2-mF capacitor, the maximum charge on the capacitor is 45mC during os
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E) 6.5 A

Explanation:

Given that

L = 40 m H

C= 1.2 m F

Maximum charge on capacitor ,Q= 45 m C

The maximum current I given as

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ω =angular frequency

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A uniform beam with mass M and length L is attached to the wall by a hinge, and supported by a cable. A mass of value 3M is susp
Jobisdone [24]

Answer:

The tension is  T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by hinge   Fx= \frac{11}{4\sqrt{3} } Mg

Explanation:

   From the question we are told that

          The mass of the beam  is   m_b =M

          The length of the beam is  l = L

           The hanging mass is  m_h = 3M

            The length of the hannging mass is l_h = \frac{3}{4} l

            The angle the cable makes with the wall is \theta = 60^o

The free body diagram of this setup is shown on the first uploaded image

The force F_x \ \ and \ \ F_y are the forces experienced by the beam due to the hinges

      Looking at the diagram we ca see that the moment of the force about the fixed end of the beam along both the x-axis and the y- axis is zero

     So

           \sum F =0

Now about the x-axis the moment is

              F_x -T cos \theta  = 0

     =>     F_x = Tcos \theta

Substituting values

            F_x =T cos (60)

                 F_x= \frac{T}{2} ---(1)

Now about the y-axis the moment is  

           F_y  + Tsin \theta  = M *g + 3M *g ----(2)

Now the torque on the system is zero because their is no rotation  

   So  the torque above point 0 is

          M* g * \frac{L}{2}  + 3M * g \frac{3L}{2} - T sin(60) * L = 0

            \frac{Mg}{2} + \frac{9 Mg}{4} -  T * \frac{\sqrt{3} }{2}    = 0

               \frac{2Mg + 9Mg}{4} = T * \frac{\sqrt{3} }{2}

               T = \frac{11Mg}{4} * \frac{2}{\sqrt{3} }

                   T= \frac{11}{2\sqrt{3} } Mg

The horizontal force provided by the hinge is

             F_x= \frac{T}{2} ---(1)

Now substituting for T

              F_{x} = \frac{11}{2\sqrt{3} } * \frac{1}{2}

                  Fx= \frac{11}{4\sqrt{3} } Mg

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