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REY [17]
3 years ago
5

A beam of visible light is passed through a plexiglass container filled with water. If a beam of light comes in at a slight angl

e, describe the change in the path of the beam of light as it passes through the container, moving in from one side and then out the other.
A) The beam of light will bend twice: entering and leaving the container.
B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.
D) The beam of light will bend three times: large angle of refraction through container wall; smaller angle of refraction through water; greater angle passing out of a container.
Physics
2 answers:
pentagon [3]3 years ago
8 0

B) The beam of light will bend three times: small angle of refraction through container wall; larger angle of refraction through water; small angle passing out of a container.
Eliminate
patriot [66]3 years ago
8 0

Answer:

C) The beam of light will refract four times; largest entering container; smaller angles of refraction moving through water, through container wall again, and finally into the air on the outside.

Explanation:

Just got it right on the test

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An object has a mass of 50.0 g and a volume of 10.5 cm3. What is the object's density?
seraphim [82]
Volume=mass/density

volume=455.6/19.3

volume=23.6 mL

4 0
3 years ago
A baseball player friend of yours wants to determine his pitching speed. You have him stand on a ledge and throw the ball horizo
Alla [95]

Answer:

v_{ox}= 19.6\ m/s

Explanation:

Data provided in the question:

Height above the ground, H= 5.0m

Range of the ball, R= 20 m

Initial horizontal velocity = v_{ox}

Initial vertical velocity= v_{oy}  (Since ball was thrown horizontally only)

Acceleration acting horizontally, a_x = 0 m/s²  [ Since no acceleration acts horizontally) ]

Vertical Acceleration, a_y = 9.8 m/s² (Since only gravity acts on it)

Let 't' be the time taken to reach ground

Therefore, using equations of motion, we have

H= v_{oy}t+\frac{1}{2}a_yt^2

5= (0)t+\frac{1}{2}(9.8)t^2

t= \frac{10}{9.8}=1.02 s

Then using Equations of motion for horizontal motion,

R= v_{ox}t+\frac{1}{2}a_xt^2

20= v_{ox}(1.02)+\frac{1}{2}(0)(1.02)^2

v_{ox}= 19.6\ m/s

4 0
3 years ago
Can some one help me with this question its hard :( down below ill give brainliest
eduard

Answer:  B

Explanation:

8 0
3 years ago
What is the path of a projectile called? Friction Track Trajectory acceleration
sergiy2304 [10]
It is trajectory acceleration. A friction track is a device to study motion in low friction environments, I believe. Does this help?
3 0
3 years ago
Read 2 more answers
Your friend's 10.8 g graduation tassel hangs on a string from his rearview mirror. When he accelerates from a stoplight, the tas
Nitella [24]

The tension in the string holding the tassel and the vertical will the tension in the string

  • T = 0.1953 N
  • Ф = 34.4 °

<h3>What is the tension in the string holding the tassel. ?</h3>

Generally, the equation for Tension is  mathematically given as

TCos\theta = mg

Therefore

TCos6.58^{o} = 19.8*10^{-3}*9.8

T = 0.1953 N

b).

Where

T* sin \theta = ma

0.1953*Sin6.58 \textdegree  = 19.8*10^{-3}*a

a = 1.13 m/s^2

In conclusion

T* sinФ = ma

2msinФ = ma

2sinФ = a

sin\theta = \frac{a}{2}

\theta = sin^{-1}\frac{a}{2} \\\\\theta= sin^{-1}\frac{1.13}{2}

Ф = 34.4 °

In conclusion, The tension in the string holding the tassel and the vertical will the tension in the string

T = 0.1953 N

Ф = 34.4 °

Read more about tension

brainly.com/question/15880959

#SPJ1

4 0
2 years ago
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