A small Keplerian telescope has an objective with a 1.33 m focal length. Its eyepiece is a 2.82 cm focal length lens. It is used to look at a 25000 km diameter sunspot on the sun, a distance 1.5*108 km from Earth. Required:
What angle is subtended by the sunspot's telescopic image in degree?

The angle is  

Explanation:

From the question we are told that  

     The objective  focal length  

     The  eyepiece focal length is  

      The diameter of the sunlight is  

       The distance of the sun from from the earth is  

  Generally the magnification of the object is mathematically evaluated as

The negative  sign is because the lens of the telescope is  diverging light

 substituting values  

Now we can obtain the angle made by the object (sunlight ) with respect to the telescope  as follows  

substituting values

The  magnification can  also be mathematically represented as

Where is the angle the image made with telescope

Since the negative sign indicate direction of light movement we will remove it from the calculation below

      =>  

     =>    

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hope this !

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//