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kramer
3 years ago
9

Help with 2 Physics questions, WILL CHOOSE BRAINLIEST

Physics
1 answer:
Tju [1.3M]3 years ago
3 0

1) D

2) D.) Greater than \theta_c

Explanation:

1)

The phenomenon of total internal reflection occurs when a ray of light hitting the interface between two mediums is totally reflected back into the original medium, therefore no refraction into the second medium occurs.

This phenomenon occurs only if two conditions are satisfied:

  • The index of refraction of the first medium is larger than the index of refraction of the 2nd medium
  • The angle of incidence is greater than a certain angle called critical angle

In picture 1, we have 4 different diagrams. In the diagrams:

  • The red arrow represents the incident ray
  • The green arrow represents the refracted ray
  • The blue arrow represents the reflected ray

Total internal reflection occurs when there is no refraction, therefore when there is no green arrow: this occurs only in figure D, so this is the correct option. (in figure C, there is a refracted ray but it is parallel to the interface: this condition occurs when the angle of incidence is exactly equal to the critical angle, however in this problem, the angle of incidence is greater than the critical angle, so the correct option is D)

2)

As we stated in problem 1), total internal reflection occurs when the angle of incidence is equal or greater than the critical angle. Therefore in this case, the angle of incidence must be

D.) Greater than \theta_c

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A blue ball is thrown upward with an initial speed of 24.1 m/s, from a height of 0.5 meters above the ground. 2.9 seconds after
Aleks04 [339]

Answer:

1. Speed=0

2. 2.46 s

3.30.1 m

4. 22.0 m

5.1.004 s

Explanation:

We are given that

Initial speed of blue ball, u=24.1 m/s

Height of blue ball from ground y_0=0.5 m

Initial speed of red ball , u'=7.2 m/s

Height of red from ground=y'0=32 m

Gravity, g=9.81ms^{-2}

1.When the ball reaches its maximum height then the speed of the blue ball is zero.

2.v=0

v=u+at

Using the formula and substitute the values

0=24.1-9.81t

Where g is negative because motion of ball is against gravity

24.1=9.81t

t=\frac{24.1}{9.81}=2.46s

3.y=y_0+ut+\frac{1}{2}at^2

Using the formula

y=0.5+24.1(2.46)-\frac{1}{2}(9.81)(2.46)^2

y=30.1 m

4.Time of flight for red ball=3.77-2.9=0.87s

y'=32-7.2(0.87)-\frac{1}{2}(9.81)(0.87)^2

y'=22.0m

Hence, the height of red ball 3.77 s after the blue ball is 22.0 m.

5.According to question

0.5+24.1(t+2.9)-\frac{1}{2}(9.81)(2.9+t)^2=32-7.2t-\frac{1}{2}(9.81)t^2

0.5+24.1t+69.89-4.905(t^2+5.8t+8.41)=32-7.2t-4.905t^2

0.5+24.1t+69.89-4.905t^2-28.449t-41.25105=32-7.2t-4.905t^2

0.5+69.89-41.25105-32=-24.1t+28.449t-7.2t

-2.86105=-2.851t

t=\frac{2.86105}{2.851}=1.004 s

Hence,  1.004 s  after the blue ball is thrown  are the two balls in the air at the same height.

8 0
3 years ago
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