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Alina [70]
2 years ago
15

The paths of the light waves that interfere to cause first-order lines (2 points) Group of answer choices differ in length by th

e wavelength of the light are parallel lines are the same length differ in length by one-half of the wavelength of the light
Physics
1 answer:
Zarrin [17]2 years ago
6 0

The paths of the light waves that interfere cause first-order lines to differ in length by the wavelength of the light.

The phenomenon of wave interference occurs when two waves meet while traveling in the same medium.

As the two light waves interfere in the first order they interfere by differing the consecutive lengths by the wavelength of the light. The wavelength of the light can be defined as the distance between identical points (adjacent crests) in the adjacent cycles of a wave signal propagated in space or along a wire.

Hence, it can be concluded that the paths of the light waves that interfere cause first-order lines to differ in length by the wavelength of the light.

Learn more about waves here:

brainly.com/question/15663649

#SPJ10

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Answer:

Explanation:

the Earth is 6 xx 10^(24) kg and its radius is 6400 km. Find the acceleration due to gravity on the surface of the Earth. Updated On: 7-4-2021

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4 years ago
uring the investigation of a traffic accident, police find skid marks 89.9 m long. They determine the coefficient of friction be
aivan3 [116]

Answer:

V_{0}=29.68m/s

Explanation:

In order to solve this problem, we must first do a drawing of the situation (see attached picture).

When the brakes of the car were applied, we can see that there was only one horizontal force affecting the vehicle's movement, which was the force of friction. When analyzing the free body diagram we can apply Newton's laws to determine the equations we will use to solve this.

\sum F_{x}=ma

so:

-f=ma

we also know that:

f=N\mu_{k}

so

-N\mu_{k}=ma

we can now solve for the acceleration, so we get:

a=\frac{-N\mu_{k}}{m}

we don't know what the normal force is, so we can find it out by analyzing the vertical forces applied to the car:

\sum F_{y}=0

so we get:

N-W=0

which means that:

N=W

we also know that:

W=mg

so

N=mg

we can substitute this into the first equation so we get:

a=\frac{-mg\mu_{k}}{m}

which simplifies to:

a=-g\mu_{k}

we can now substitute the provided data:

a=(-9.8m/s^{2})(0.5)

which yields:

a=-4.9m/s^{2}

once we got the acceleration, we can use kinematics formulas to solve this, we got the following formula:

a=\frac{V_{f}^{2}-V_{0}^{2}}{2x}

we know the final velocity must be zero, since that's where the car got to a stop, so the formula then becomes:

a=-\frac{V_{0}^{2}}{2x}

we can now solve for the initial velocity, which yields:

V_{0}=\sqrt{-2xa}

so we can now substitute the daa we know, so we get:

V_{0}=\sqrt{-2(89.9m)(-4.9m/s^{2})}

so we get:

V_{0}=29.68m/s

So from this we know that the velocity of the car must have been of at least 29.68m/s when the brakes were applied.

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If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon-population sy
Genrish500 [490]

The question is incomplete. The complete question is :

If the entire population of Earth were transferred to the Moon, how far would the center of mass of the Earth-Moon population system move? Assume the population is 7 billion, the average human has a mass of 65 kg, and that the population is evenly distributed over both the Earth and the Moon. The mass of the Earth is 5.97×1024 kg and that of the Moon is 7.34×1022 kg. The radius of the Moon’s orbit is about 3.84×105 m.

Solution :

Given :

Mass of earth, $M_e = 5.97 \times 10^{24} \ kg$

Mass of moon, $M_m = 7.34 \times 10^{22} \ kg$

Mass of each human, $m_p =65 \ kg$

Therefore mass of total population, $M_p = 65  \times 7 \times 10^{9} \ kg$

                                                           $M_p = 4.55 \times 10^{11} \ kg$

Let the earth is at the origin of the coordinate system. Then,

Since $M_e>> M_p$

         $M_m>> M_p$

Hence if we shift all the population on the moon there will be negligible change in the mass of the moon and earth. Hence there will not be any significant shift on the centre of mass. i.e.

      $X_{cm} = \frac{5.97 \times 10^{24}+ 7.34 \times 10^{22} \times 3.84 \times 10^5}{5.97 \times 10^{24}+ 7.34 \times 10^{22}}$

              $= 4.68 \times 10^6 \ m$

$ 4.68 \times 10^3 \ km$ from the earth.

         

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3 years ago
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