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Masja [62]
3 years ago
12

Please help I'll mark brainliest!!!!

Physics
2 answers:
madam [21]3 years ago
7 0

This question is based on the fundamental assumption of  vector direction.

A vector is  a physical quantity which has  magnitude as well direction  for its complete specification.

The magnitude of a physical quantity is simply a  numerical number .Hence it can not be negative.

A negative vector is a vector which comes into existence when it is opposite to our assumed direction with respect to any other vector.  For instance, the vector is taken positive if it is along + X axis and negative if it is along - X axis.

As per the first option it is given that a vector is negative if its magnitude is greater than 1. It is not correct as magnitude play no role in it.

The second option tells that the magnitude of the vector is less than 1. Magnitude can not be negative. So this is also wrong.

Third one tells that a vector is negative if its displacement is along north. It does not give any detail information about the negativity of a vector.

In a general sense we assume that vertically downward motion  is negative and vertically upward is positive. In case of a falling object the motion is  vertically downward. So the velocity of that object is negative .

So last   option is  partially  correct  as  the vector can be negative depending on our choice of co-ordinate system.





exis [7]3 years ago
4 0

The velocity of a falling object is a vector that should be negative

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3 years ago
Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light.
pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

=>      x arcminute =>   0.3 arcsecond

So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

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