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rusak2 [61]
2 years ago
15

Hi, I'm stuck on the problem: Consider a resistor (R=1000 kΩ) and a capacitor (C=1μF) connected in series. This configuration is

connected in series to a battery with an emf of 10 V. Find the maximum voltage in [V] across the capacitor. Find the maximum charge on the capacitor in [µC]. How long will it take until the capacitor has a potential difference of 5 V?
Physics
1 answer:
galina1969 [7]2 years ago
4 0

Answer:

Following are the responses to the given choices:

Explanation:

The maximum pressure is 10 V from across condenser.

The highest charge mostly on condenser.

vc = 10\times 1\ mf\\\\

    = 10^{-5}\ C

The time it takes for the condenser to be 5 V different.

time = ln_2 \times 1000  \times 1mf

        =0 .69 \times  10^{-3}\  sec

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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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