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scoray [572]
3 years ago
10

An electron enters a region of uniform perpendicular en and bn fields. it is observed that thevelocity nv of the electron is una

ffected. a possible explanation is:
Physics
1 answer:
konstantin123 [22]3 years ago
5 0
<span>v is perpendicular to both E and B and has a magnitude E/B</span>
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I have a combination of myopia and presbyopia—overall, the power of my visual system is too large, but I also have a very limite
e-lub [12.9K]

Answer:

The range of powers is    - 5 \ D \le P \le - 2.667\  D

Explanation:

From the question we are told that

       The far point of the left eye is n_f = 20 cm

       The near point of the left eye is  n =  15cm

       The near point with the glasses on is n_g =25 \ cm

     

From these parameter we can see that with the glass on that for near point the

         Object distance would be u = -25 \ cm

          Image distance would be  v =  -15 \ cm

To obtain the focal length we would apply the lens formula which is mathematically represented as

              \frac{1}{f} =  \frac{1}{v}  -  \frac{1}{u}

substituting values

              \frac{1}{f} =  \frac{1}{-15}  -  \frac{1}{-25}

               f =  - \frac{75}{2} cm

           converting to  meters

               f =  - \frac{75}{2} * \frac{1}{100}

               f =  - \frac{75}{200} \ m

   Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f}

Substituting values

                 P = -  \frac{200}{75}  m

                 P = - 2.667 \ D

   

From these parameter we can see that with the glass on that for far  point the

         Object distance would be u_f = - \infty \ cm

          Image distance would be  v_f =  -20  \ cm

To obtain the focal length of the lens we would apply the lens formula which is mathematically represented as

                    \frac{1}{f_f} =  \frac{1}{v_f}  -  \frac{1}{u_f}

substituting values

                  \frac{1}{f} =  \frac{1}{-20}  -  \frac{1}{- \infty}

                 \frac{1}{f} =  \frac{1}{-20}  -  0      

                  f_f =  \frac{20}{1}  \ cm

converting to  meters

                f_f =  - \frac{20}{1}  * \frac{1}{100}

               

Generally the power of the lens is mathematically represented as

                P  = \frac{1}{f_f}

Substituting values

                 P = -  \frac{100}{20}  m

                 P = - 5 \ D

This implies that the range of powers of the lens in his glass is

                  - 5 \ D \le P \le - 2.667\  D

   

               

               

           

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A mass of 1000 kg drops from a height of 10 m on a platform of negligible mass. It is desired to design a spring and dashpot on
Juliette [100K]

Answer:

k = 5\times 10^{4}\ N/m

b = 0.707\times 10^{3}

t = 7.1\times 10^{- 5}\ s

Solution:

As per the question:

Mass of the block, m = 1000 kg

Height, h = 10 m

Equilibrium position, x = 0.2 m

Now,

The velocity when the mass falls from a height of 10 m is given by the third eqn of motion:

v^{2} = u^{2} + 2gh

where

u = initial velocity = 0

g = 10m/s^{2}

Thus

v = \sqrt{2\times 10\times 10} = 10\sqrt{2}\ m/s

Force on the mass is given by:

F = mg = 1000\times 10 = 10000 N = 10\ kN

Also, we know that the spring force is given by:

F = - kx

Thus

k = \frac{F}{x} = \frac{10000}{0.2} = 5\times 10^{4}\ N/m

Now, to find the damping constant b, we know that:

F = - bv

b = \frac{F}{v} = \frac{10000}{10\sqrt{2}} = 0.707\times 10^{3}

Now,

Time required for the platform to get settled to 1 mm or 0.001 m is given by:

t = \frac{0.001}{v} = \frac{0.001}{10\sqrt{2}} = 7.1\times 10^{- 5}\ s

4 0
3 years ago
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