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mrs_skeptik [129]
3 years ago
9

An image seen in a smooth dinner plate is an example of ____ reflection.

Physics
1 answer:
spayn [35]3 years ago
8 0

The correct answer is letter C, regular reflection.

<span>An image seen in a smooth dinner plate is an example of a regular reflection. This phenomenon only occurs in smooth and polished surfaces. During this stage light occurs at a certain angle and is reflected back at the same angle producing reflections in a way like a mirror does this commonly.</span>

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Explanation:

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When the computer you are working on uses energy it also produces waste energy. What type of energy is this waste energy?
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Find the magnitude of the vector v given its initial and terminal points. Round your answer to four decimal places.
Sliva [168]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value is |v| = 6.93

Explanation:

From the question we are told that

    The initial point is (x_1 , y_1 , z_1 ) = (-1 , 7 , 4 )

    The  terminal point is  (x_2 , y_2 , z_2) = (-5 , 11, 8 )

Generally the magnitude of the vector is mathematically represented as

     |v| = \sqrt{(x_2 -x_1 )^2 + (y_2 - y_1 )^2 + (z_2 -z_1)^2}

=>   |v| = \sqrt{(-5 -(-1) )^2 + (11 - 7 )^2 + (8 -4)^2}

=>   |v| = 6.93

3 0
3 years ago
Juan is on a morning jog. His speed is represented in the graph. At what rate of speed is Juan running between 4min and 6min? Ac
Nesterboy [21]

Answer:

<em>a) The speed of Juan is 250 m/min </em>

<em>b) In that interval, he's at rest </em>

<em>c) Juan would have needed 12 minutes to run 3000 m</em>

Explanation:

<u>Distance vs Time Graph Analysis </u>

If time is on the horizontal axis, and the distance is in the vertical axis, then the slope of the graph represents the instantaneous speed of the moving object. The graph shows three different zones which reflect Juan's jogging at that morning.

a) The period between 4 and 6 minutes clearly shows Juan is moving in such a way the distance increases when time passes. The speed of the motion can be obtained by computing the slope of the line. We can locate the points (4,1000) and (6,1500). We compute the speed  

\displaystyle v=\frac{1500-1000}{6-4}=\frac{500}{2}=250 m/min

The speed of Juan is 250 m/min

b) We can see between 7 and 11 minutes, Juan's distance is not changing because he stopped running. In that interval, he's at rest

c) We have already determined the speed on the first 7 minutes (the same as between 4 and 6 minutes). We know that

\displaystyle v=\frac{x}{t}

Where v is the speed, x is the distance, and t is the time

We need to know the time needed to travel x=3000 m at the initial speed v=250 m/min, so we solve the equation for t

\displaystyle t=\frac{x}{v}

\displaystyle t=\frac{3000}{250}=12\ minutes

Juan would have needed 12 minutes to run 3000 m

3 0
4 years ago
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