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3 years ago

9

What do we mean when we say that two light rays striking a screen are in phase with each other?

1 answer:

andrey2020 [161]3 years ago

4 0

Answer:

When the electric field due to one is a maximum, the electric field due to the other is also a maximum, and this relation is maintained as time passes.

Explanation:

Phase of a wave or light ray is the instantaneous situation of the cycle in which the wave is at a given time.

When two waves are in phase means that the maximum and minimum of both coincide in time. They are in the same point of their cycle at the same time. And this relationship is maintained as time passes.

The waves can also be visualized as the oscillation of an electric field. (usually plotted like a sine function).

So the fact that two waves are in phase means that the maximums of their electric field coincide in time.

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The highest tides occur when

andrey2020 [161]

Answer:

Happens when lunar perigee occurs.

Explanation:

If the lunar perigee occurs when the Moon is between the Sun and the Earth, it produces unusually high Spring high tides. When it occurs on the opposite side from the Earth that where the Sun is located ( during full moon) it produces unusually low, Neap Tides.

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3 years ago

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Football is kicked at 23.4 m/s at a 38.5 angle. How far away does it land?

natulia [17]

Answer: go0gle will know the answer

Explanation: hey I’m not giving fault answers so yea

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3 years ago

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Using the midpoint and the distance formulas, calculate he coordinate of the midpoint and the length of the segment.

arsen [322]

Answer:

Explanation:

Formula to calculate the distance between two points $(x_1,y_1)$ and $(x_2,y_2)$ is,

d = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Therefore, length of CD having coordinates C(3,1) and D(3, 3),

CD = $\sqrt{(3-3)^2+(1-3)^2}$

= 2 units

Formula to find the midpoint of CD is,

M = $(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$

= $(\frac{3+3}{2},\frac{1+3}{2})$

= (3, 2)

Therefore, length of CD = 2 units and (3, 2) are the coordinates of the midpoint of CD.

7 0

4 years ago

A truck heading east has an initial velocity of 6 m/s. It accelerates at 2 m/s2 for 12 seconds. What distance does the truck tra

omeli [17]

Vf = vi + at
vf = 6 + (2)(12)
vf = 6 + 24
vf = 30

2as = vf² - vi²
2(2)s = (30)² - (6)²
4s = 900 - 36
4s = 864
s = 864/4
s = 216
<span>d = 216</span>

6 0

3 years ago

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A uniform line charge of density λ lies on the x axis between x = 0 and x = L. Its total charge is 7 nC. The electric field at x

DedPeter [7]

Answer:

The electric field at x = 3L is 166.67 N/C

Solution:

As per the question:

The uniform line charge density on the x-axis for x, 0< x< L is $\lambda$

Total charge, Q = 7 nC = $7\times 10^{- 9} C$

At x = 2L,

Electric field, $\vec{E_{2L}} = 500N/C$

Coulomb constant, K = $8.99\times 10^{9} N.m^{2}/C^{2}$

Now, we know that:

$\vec{E} = K\frac{Q}{x^{2}}$

Also the line charge density:

$\lambda = \frac{Q}{L}$

Thus

Q = $\lambda L$

Now, for small element:

$d\vec{E} = K\frac{dq}{x^{2}}$

$d\vec{E} = K\frac{\lambda }{x^{2}}dx$

Integrating both the sides from x = L to x = 2L

$\int_{0}^{E}d\vec{E_{2L}} = K\lambda \int_{L}^{2L}\frac{1}{x^{2}}dx$

$\vec{E_{2L}} = K\lambda[\frac{- 1}{x}]_{L}^{2L}] = K\frac{Q}{L}[frac{1}{2L}]$

$\vec{E_{2L}} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{2L}] = \frac{63}{L^{2}}$

Similarly,

For the field in between the range 2L< x < 3L:

$\int_{0}^{E}d\vec{E} = K\lambda \int_{2L}^{3L}\frac{1}{x^{2}}dx$

$\vec{E} = K\lambda[\frac{- 1}{x}]_{2L}^{3L}] = K\frac{Q}{L}[frac{1}{6L}]$

$\vec{E} = (9\times 10^{9})\frac{7\times 10^{- 9}}{L}[frac{1}{6L}] = \frac{63}{6L^{2}}$

Now,

If at x = 2L,

$\vec{E_{2L}} = 500 N/C$

Then at x = 3L:

$\frac{\vec{E_{2L}}}{3} = \frac{500}{3} = 166.67 N/C$

4 0

4 years ago