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TiliK225 [7]
3 years ago
7

When the trough of one light wave meets the crest of another light wave, __________ occurs.

Physics
2 answers:
alina1380 [7]3 years ago
7 0

Answer:

B Destructive Interference

padilas [110]3 years ago
5 0

I took the test, its constructive interference

Edit: this is kind of a trick question. took the test again it was the same question yet this time the answer was destructive. Good luck with that

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A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
The diagram shows monochromatic light passing through two openings.
rusak2 [61]

Answer:

constructive interference in which waves strengthen each other

Explanation:

Some definitions:

- Costructive interference occurs when two (or more) waves meet each other in phase, so with same displacement at the same point. In such situation, the two waves strengthen each other, and the amplitude of the resultant wave is the sum of the amplitudes of the individual waves

- Destructive interference occurs when two waves meet each other in anti-phase, so with opposite displacement at the same point. In such situation, the two waves cancel each other out, and the amplitude of the resultant wave is the difference of the amplitudes of the individual waves (which means zero if the two waves are identical)

For light waves interfering with each other, 'white' means costructive interference, while 'black' means destructive interference (because black is absence of colors, so this means that the waves cancel each other out). In this problem, we see that point X, Y and X are white, therefore they are point of constructive interference, where the waves strengthen each other.

5 0
3 years ago
Read 2 more answers
The model below shows a calcium atom. An image has a mix of red and blue balls in its center and 4 concentric black rings around
zubka84 [21]

Answer:

8 electrons in the third energy level

Explanation:

From the description,the third energy level has 8 electron (represented by the small green balls you describe)

4 0
3 years ago
1. Which wave refracts?
Digiron [165]

Answer:

I want to say your answer is A) an ocean wave approaching the shore at an angle. But not 100% sure. I'm so sorry if it's wrong I really tried to figure it out.

Explanation:

Refraction of waves involves a change in the direction of waves as they pass from one medium to another. Refaction , or the bending of the path of the waves, is accompanied by a change in speed and wavelength of the waves. Thus, if water waves are passing from deep water into shallow water, they will slow down. ( This is what I read on google maybe it'll help you out).

8 0
3 years ago
A 4 cm diameter ball is located 40 cm from a point source and 80 cm from a wall. What is the size of the shadow on the wall?
Slav-nsk [51]
<span>11.823 cm There is a slight ambiguity with this question in that I don't know if the measurements are from the surface of the ball, or the center of the ball. I will take this question literally and as such the point light source will be 124 cm from the wall. The key thing to remember is that ball won't be showing an effective diameter of 4 cm to the light source. Instead the shadow line is a tangent to the ball's surface. There is a right triangle where the hypotenuse is the distance from the center of the ball to the light source (42 cm), one leg of the triangle is the radius (2cm). That right triangle will define a chord that will be the effective diameter of the disk casting the shadow. The cosine of the half angle of the chord will be 2/42 = 1/21. The sine of the half angle then becomes sqrt(1-(1/21)^2) = sqrt(440/441) = 2sqrt(110) = 0.99886557. Now multiply that sine by 4 (radius of ball multiplied by 2 since it's the half angle and we want the full side of the chord) and we get an effective diameter of 3.995462279 cm. Now we need to calculate the effective distance that circle is from the wall. It will be slightly larger than 82 cm. The exact value will be 82 + cos(half angle) * radius. So 82 + 1/21 * 2 = 82 + 2/21 = 82.0952381 Now we have the following dimensions with a circle replacing the ball in the original problem. Distance from wall to effective circle = 82.0952381 cm Distance from effective circle to point source = 124 - 82.0952381 = 41.9047619 cm Effective diameter of circle = 3.995462279 cm And because the geometry makes similar triangles, the following ratio applies. 3.995462279/41.9047619 = X/124 Now solve for X 3.995462279/41.9047619 = X/124 124*3.995462279/41.9047619 = X 495.4373226/41.9047619 = X 11.82293611 = X The shadow cast on the wall will be a circle with a diameter of 11.823 cm</span>
7 0
3 years ago
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