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4 years ago

12

A light ray is striking the surface of Earth. Which factor would make the light ray more likely to be absorbed than reflected? l

ight colors smooth surfaces the winter season the high angle of the Sun

2 answers:

vichka [17]4 years ago

7 0

Answer:

The answer is D

Explanation:

Because the other options do not make since.

Vladimir [108]4 years ago

5 0

Answer:

B

Explanation:

Trust

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A series of pulses, each of amplitude 0.1m , is sent down a string that is attached to a post at one end. The pulses are reflect

sp2606 [1]

The net displacement at a point on the string where the pulses cross is 0.2 m.

The term "displacement" refers to a shift in an object's position. It has a magnitude and a direction, making it a vector quantity. An arrow pointing from the starting point to the finishing point serves as its symbol.

A string that is connected to a post at one end is used to transmit a sequence of pulses, each measuring 0.1 meters in amplitude.

At the post, the pulses are reflected and return along the string without losing any of their amplitude.

Now, let's say the ends are free.

There is no inversion on reflection if the end is free. The amplitude at their intersection is 2A.

Now, since A = 0.1 m

Then, 2A = 2(0.1) = 0.2 m

As a result, the net displacement at the string's intersection of two pulses is 0.2 m.

The correct option is (c).

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4 0

2 years ago

Convert the following in to SI units without changing its values<br> a. 4000 g <br>b. 2.4 km

lana66690 [7]

hi <3

i believe i explained this answer properly in my last answer but it would be 4kg and 2400m as these are the SI units for these values.

hope this helps :)

3 0

3 years ago

Read 2 more answers

Material A is less optically dense than material B. How can you tell from the diagram above?

zimovet [89]

You can tell because the line bends and the closer it is to horizontal or past horizontal it is more dense

7 0

3 years ago

7. A double-slit experiment uses coherent light of wavelength 633 nm with a slit separation of 0.100 mm and a screen placed 2.0

Illusion [34]

Answer:

The distance between first-order and second-order bright fringes is 12.66mm.

Explanation:

The physicist Thomas Young establishes through its double slit experiment a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

$\Lambda x = L\frac{\lambda}{d}$ (1)

Where $\Lambda x$ is the distance between two adjacent maxima, L is the distance of the screen from the slits, $\lambda$ is the wavelength and d is the separation between the slits.

The values for this particular case are:

$L = 2.0m$

$\lambda = 633nm$

$d = 0.100mm$

Notice that is necessary to express L and $\lambda$ in units of milimeters.

$L = 2.0m \cdot \frac{1000mm}{1m}$ ⇒ $2000mm$

$\lambda = 633nm \cdot \frac{1mm}{1x10^{6}nm}$ ⇒ $6.33x10^{-4}mm$

Finally, equation 1 can be used:

$\Lambda x = (2000mm)\frac{(6.33x10^{-4}mm)}{(0.100mm)}$

$\Lambda x = 12.66mm$

Hence, the distance between first-order and second-order bright fringes is 12.66mm.

8 0

3 years ago

An object begins with a speed of 20 meters per second and slows down to a speed of 10 meters per second over a time of 5 seconds

Andreyy89

Answer:

V = 6 m/s

Explanation:

Given that,

Initial speed of an object is 20 m/s

Final speed of an object is 10 m/s

Time, t = 5 s

We need to find the average speed of the object during these 5 seconds. Let it is equal to V. Here, time is same. The average speed is given by :

$V=\dfrac{x+y}{2}\\\\\text{Putting values}\\\\V=\dfrac{20+10}{5}\\\\V=6\ m/s$

So, the average speed of the object is 6 m/s.

3 0

3 years ago