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Alecsey [184]
2 years ago
6

a ray of light incident on a mirror, at an angle of 45°. Another mirror is placed at an angle of 45° to the first ones as shown.

Sketch the patch of the ray until it emerges from the two mirrors (2mks) 45° www.www 45° ​
Physics
1 answer:
Gwar [14]2 years ago
3 0

Answer:

If the ray of light is deflected by 45 degrees by the first mirror its total deflection by mirror (I) is 90 deg. (incident = 45 and exit ray equals 45 deg)

The second mirror will cause a net deflection of 90 degrees and the total deflection will be 180 deg or in opposite  direction to the  incident ray.

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How many stars are contained in the milking way galaxy A) 0 b)1 c) billions d)infinite
skelet666 [1.2K]

Answer:

c) billions

Explanation:

there are more than 0 stars, because our sun is a star

and  there are more than one because the milky way has a lot of suns

there aren't infinite because the milky way is finite it has some volume and even you made the smallest star possible you wouldn't have infinite stars

so the only option is billions

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The amplitude is the number of waves that pass a given point per unit time.<br> a. True<br> b. False
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In which situation does heating a substance generally change the temperature?
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3 years ago
At the same moment, one rock is dropped and one is theown downwand with an iniial velocily of 29 us frm op of a building that is
Helen [10]

Answer:

The thrown rock will strike the ground 2.42s earlier than the dropped rock.

Explanation:

<u>Known Data</u>

  • y_{i}=300m
  • y_{f}=0m
  • v_{iD}=0m/s
  • v_{iT}=-29m/s, it is negative as is directed downward

<u>Time of the dropped Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300m-\frac{(9.8m/s^{2})t_{D}^{2}}{2}, then clearing for t_{D}.

t_{D}=\sqrt[2]{\frac{300m}{4.9m/s^{2}}} =\sqrt[2]{61.22s^{2}} =7.82s

<u>Time of the Thrown Rock</u>

We can use y_{f}=y_{i}+v_{iy}t-\frac{gt^{2}}{2}, to find the total time of fall, so 0=300-29t_{T}-\frac{(9.8)t_{T}^{2}}{2}, then, 0=-4.9t_{T}^{2}-29t_{T}+300, as it is a second-grade polynomial, we find that its positive root is t_{T}=5.4s

Finally, we can find how much earlier does the thrown rock strike the ground, so t_{E}=t_{D}-t_{T}=7.82s-5.4s=2.42s

6 0
4 years ago
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