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Alina [70]
1 year 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]1 year 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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A catapult launches a test rocket vertically upward from a well, giving the rocket an initial speed of 79.6 m/s at ground level.
Dimas [21]

Answer:

The rocket is motion above the ground for 44.7 s.

Explanation:

The equations for the height of the rocket are as follows:

y = y0 + v0 · t + 1/2 · a · t²

and, after the engine fails:

y = y0 + v0 · t + 1/2 · g · t²

Where:

y = height of the rocket at time t

y0 =  initial height

v0 = initial speed

t=  time

a = acceleration due to the engines

g = acceleration due to gravity

First, let´s calculate the time the rocket is being accelerated by the engines:

(The center of the framer of reference is located at the ground, y0 = 0).

y = y0 + v0 · t + 1/2 · a · t²

1190 m = 0 m + 79.6 m/s · t + 1/2 · 4.10 m/s² · t²

0 = 2.05  m/s² · t² + 79.6 m/s · t - 1190 m

Solving the quadratic equation:

t = 11.5 s  (the other value of t is discarded because it is negative).

At that time, the engines fail and the rocket starts to fall. The equation of the height will be:

y = y0 + v0 · t + 1/2 · g · t²

The initial velocity (v0) will be the velocity acquired during 11.5 s of acceleration plus the initial velocity of launch:

v = v0 + a · t

v = 79.6 m/s + 4.10 m/s² · 11.5 s

v = 126.8 m/s

Now, we can calculate the time it takes the rocket to reach the ground (y = 0) from a height of 1190 m:

y = y0 + v0 · t + 1/2 · g · t²

0 m = 1190 m + 126.8 m/s · t - 1/2 · 9.80 m/s² · t²

Solving the quadratic equation:

t = 33.2 s

Then, the total time the rocket is in motion is (33.2 s + 11.5 s) 44.7 s

 

5 0
3 years ago
The process of how rocks can tell us how old the earth is
Ludmilka [50]
The Earth is a constantly changing planet. Its crust is continually being created, modified, and destroyed. As a result, rocks that record its earliest history have not been found and probably no longer exist. Nevertheless, there is substantial evidence that the Earth and the other bodies of the Solar System are 4.5-4.6 billion years old, and that the Milky Way Galaxy and the Universe are older still. The principal evidence for the antiquity of Earth and its cosmic surroundings is:The oldest rocks on Earth, found in western Greenland, have been dated by four independent radiometric dating methods at 3.7-3.8 billion years. Rocks 3.4-3.6 billion years in age have been found in southern Africa, western Australia, and the Great Lakes region of North America. These oldest rocks are metamorphic rocks but they originated as lava flows and sedimentary rocks. The debris from which the sedimentary rocks formed must have come from even older crustal rocks. The oldest dated minerals (4.0-4.2 billion years) are tiny zircon crystals found in sedimentary rocks in western Australia.
The oldest Moon rocks are from the lunar highlands and were formed when the early lunar crust was partially or entirely molten. These rocks, of which only a few were returned by the Apollo missions, have been dated by two methods at between 4.4-4.5 billion years in age. 
The majority of the 70 well-dated meteorites have ages of 4.4-4.6 billion years. These meteorites, which are fragments of asteroids and represent some of the most primitive material in the solar system, have been dated by 5 independent radiometric dating methods. 
The "best" age for the Earth is based on the time required for the lead isotopes in four very old lead ores (galena) to have evolved from the composition of lead at the time the Solar System formed, as recorded in the Canyon Diablo iron meteorite. This "model lead age" is 4.54 billion years. 
The evidence for the antiquity of the Earth and Solar System is consistent with evidence for an even greater age for the Universe and Milky Way Galaxy. a) The age of the Universe can be estimated from the velocity and distance of galaxies as the universe expands. The estimates range from 7 to 20 billion years, depending on whether the expansion is constant or is slowing due to gravitational attraction. b) The age of the Galaxy is estimated to be 14-18 billion years from the rate of evolution of stars in globular clusters, which are thought to be the oldest stars in the Galaxy. The age of the elements in the Galaxy, based on the production ratios of osmium isotopes in supernovae and the change in that ratio over time due to radioactive decay, is 8.6-15.7 billion years. Theoretical considerations indicate that the Galaxy formed within a billion years of the beginning of the Universe. c) Combining the data from a) and b), the "best, i.e., most consistent, age of the universe is estimated to be around 14 billion years. For more current information on the age of the universe.
7 0
2 years ago
One car has two and a half times the mass of a second car, but only half as much kinetic energy. When both cars increase their s
Leokris [45]

Answer:

v_1 = 7.96 m/s

v_2 = 17.8 m/s

Explanation:

Let the mass of the other car is "m" and its kinetic energy is

K = \frac{1}{2}mv^2

now the mass of the first car is two and half times and its kinetic energy is half that of other car

so we will have

\frac{1}{2}(2.5m)v_1^2 = \frac{1}{2}(\frac{1}{2}mv^2)

2.5 v_1^2 = 0.5 v^2

v_1 = 0.447 v

now speed of both cars is increased by value of 9 m/s

so now we will have same kinetic energy for both cars

\frac{1}{2}(2.5 m)(0.447v + 9)^2 = \frac{1}{2}m(v + 9)^2

2.5(0.447 v + 9)^2 = (v + 9)^2

1.58(0.447v + 9) = v + 9

0.293v = 5.22

v = 17.8 m/s

so speed of first car is

v_1 = 0.447 v = 7.96 m/s

v_2 = 17.8 m/s

3 0
3 years ago
a load of 800 newton is lifted by an effort of 200 Newton. if the load is placed at a distance of 10 cm from the fulcrum. what w
nataly862011 [7]

Answer:

40 cm

Explanation:

We are given that

Load=800 N

Effort=200 N

Load  distance=10 cm

We have to find the effort distance.

We know that

load\times load\;distance=Effort\times effort\;distance

Using the formula

800\times 10=200\times effort\;distance

Effort distance=\frac{800\times 10}{200}

Effort distance=\frac{8000}{200}

Effort distance=40 cm

Hence,  the effort distance will be 40 cm.

7 0
2 years ago
I think I know the answer but I am not sure
Romashka [77]
The answer is a

Hope this helps!
5 0
3 years ago
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