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notka56 [123]
3 years ago
9

Two narrow slits are 0.12 mm apart. light of wavelength 550 nm illuminates the slits, causing an interference pattern on a scree

n 1.0 m away. Light from each slit travels to the m=1 maximum on the right side of the central maximum. how much farther did the light from the left slit travel that the light from the right slit?
Physics
1 answer:
mamaluj [8]3 years ago
3 0

Answer:

500 nm

Explanation:

d_l = Distance traveled by light from left slit

d_r = Distance traveled by light from right slit

\Delta x = Path difference = d_l-d_r

m = Order = 1

\lambda = Wavelength = 550 nm

Path difference is given by

\Delta x=m\lambda\\\Rightarrow d_l-d_r=1\times 550\\\Rightarrow d_l-d_r=500\ nm

The required distance is 500 nm

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Mass (kg) 4.0
densk [106]

Answer:

25 m/s

Explanation:

First of all, we can find the acceleration the object by using Newton's second law of motion:

F=ma

where

F = 20.0 N is the net force applied on the object

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a is its acceleration

Solving for a, we find

a=\frac{F}{m}=\frac{20}{4}=5.0 m/s^2

Now we know that the motion of the object is a uniformly accelerated motion, so we can find its final velocity by using the following suvat equation:

v=u+at

where

v is the final velocity

u = 0 is the initial velocity

a=5.0 m/s^2 is the acceleration

t = 5 s is the time

By substituting,

v=0+(5.0)(5)=25 m/s

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3 years ago
What is the another name for the lower fixed point?
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I believe another name for lower point is Ice point
3 0
3 years ago
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A hollow conducting spherical shell has radii of 0.80 m and 1.20 m, The radial component of the electric field at a point that i
mars1129 [50]

Complete Question

The complete question is  shown on the first uploaded image  

 

Answer:

The electric field at that point is  E = 7500 \ N/C

Explanation:

From the question we are told that  

       The  radius of the inner circle is r_i  =  0.80  \ m

        The  radius of the outer circle is  r_o  =  1.20 \ m

       The  charge on the spherical shell q_n  =  -500nC  = -500*10^{-9} \ C

      The magnitude of the point charge at the center is  q_c =  + 300 nC  =  + 300 * 10^{-9} \ C

        The  position we are considering is  x =  0.60 m  from the center

Generally  the  electric field  at the distance x =  0.60 m  from the center  is mathematically represented as

                 E =  \frac{k *  q_c   }{x^2}

substituting values  

                  E =  \frac{k *  q_c   }{x^2}

where  k is  the coulomb constant with value k = 9*10^{9}  \ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.

     substituting values

                  E =  \frac{9*10^9  *  300 *10^{-9}}{0.6^2}

                 E = 7500 \ N/C

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3 years ago
Arrange the balls in order from greatest amount of gravitational potential energy to least
Aloiza [94]
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3 years ago
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A battery charges a parallel-plate capacitor fully and then is removed. The plates are immediately pulled apart. (With the batte
Assoli18 [71]

Answer:

<em>There will be an increase in potential difference.</em>

Explanation:

As we know that the potential difference depends upon the capacitance.

ΔV = Q/C

When battery is disconnected the charge remains constant on the plates but the capacitance decreases. As the capacitance has an inverse relation with the potential difference, there will be an increase in it.

In addition to that the potential difference can also be defined as the product of field and distance between the plates. As the charge is constant so the field is constant. Upon increasing the separation between the plates the potential difference will also increased.

4 0
3 years ago
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