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creativ13 [48]
3 years ago
13

Una rapidez de 7 mm/us es igual a: A) 7000m/s B) 70 m/s C) 7 m/s D) 0.07 m/s

Physics
1 answer:
Ivan3 years ago
3 0

For this conversion you should know that 1 meter has 1000 millimeters and that 1s equals 1000000 micro seconds, then:


7mm/us * 1m/1000mm * 1000000us/1s = 7000 m/s

The mm are canceled just like the micro seconds and the result remains in m / s .

Therefore option A is correct

A) 7000m / s




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A proton, initially traveling in the +x-direction with a speed of 5.05×10^5 m/s , enters a uniform electric field directed verti
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Answer:

The strength of the electric field is 1.35\times10^{4}\ N/C.

Explanation:

Given that,

Speed v= 5.05\times10^{5}\ m/s

Time t= 3.90\times10^{-7}\ s

Angle = 45°

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Using equation of motion

v = u+at

5.05\times10^{5}=0+a\times3.90\times10^{-7}

a =\dfrac{5.05\times10^{5}}{3.90\times10^{-7}}

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We need to calculate the strength of the electric field

Using relation of newton's second law and electric force

F= ma=qE

ma = qE

E=\dfrac{ma}{q}

Put the value into the formula

E=\dfrac{1.67\times10^{-27}\times1.29\times10^{12}}{1.6\times10^{-19}}

E=1.35\times10^{4}\ N/C

Hence, The strength of the electric field is 1.35\times10^{4}\ N/C.

3 0
3 years ago
Una ave vuela a una velocidad constante de 15m/s en una trayectoria rectilínea. Si dura una hora volando ¿cuanta distancia habrá
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3 years ago
Why do you not come to thermal equilibrium on a cold day
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3 0
3 years ago
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

   

               

               

           

3 0
3 years ago
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