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Ray Of Light [21]

2 years ago

13

Please write a haiku for this assignment also this is for science but I don't think brainly has a subject for that. Fake answers

will be reported.

2 answers:

grandymaker [24]2 years ago

7 0

Results filter in (5 syllables)
Where should I put the data? (7 syllables)
Inside of a graph! (5 syllables)

Verizon [17]2 years ago

5 0

Graphing is useful
Help analyzing data
Of Math and science

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Definition of fluoresence

denis23 [38]

Answer:

the property of absorbing light of short wavelength and emitting light of longer wavelength.

Explanation:

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Which statement best describes frustration?

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C. when you can't achieve your goal due to events beyond your control

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An ant crawls 50 cm to the right, stops, and then crawls 30 cm to the left. Calculate the distance that the ant travels, and cal

borishaifa [10]

Answer and Explanation:

If the ant was to crawl 50cm to the right, then come back 30 cm, then the total distance walked would be <u>80cm</u>.

- Combine 50cm and 30cm to get 80 cm.

For displacement, the answer is <u>20 cm.</u>

- When calculating displacement, you use the initial (starting) distance. and subtract that from the final distance, giving you the displacement, or the amount traveled from the starting point to the final point if you were to make a straight line from the starting point to end point. (0 to 50, then back 30 the same direction, so subtract 30 from 50 to get 20)

<em><u>#teamtrees #PAW (Plant And Water)</u></em>

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2 years ago

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A wave is incident on the surface of a mirror at an angle of 41° with the normal. what can you say about its angle of reflection

Umnica [9.8K]

It's angle of reflection must be 41 degrees

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Constant Acceleration Kinematics: Car A is traveling at 22.0 m/s and car B at 29.0 m/s. Car A is 300 m behind car B when the dri

Ainat [17]

Answer:

The taken is $t_A = 19.0 \ s$

Explanation:

Frm the question we are told that

The speed of car A is $v_A = 22 \ m/s$

The speed of car B is $v_B = 29.0 \ m/s$

The distance of car B from A is $d = 300 \ m$

The acceleration of car A is $a_A = 2.40 \ m/s^2$

For A to overtake B

The distance traveled by car B = The distance traveled by car A - 300m

Now the this distance traveled by car B before it is overtaken by A is

$d = v_B * t_A$

Where $t_B$ is the time taken by car B

Now this can also be represented as using equation of motion as

$d = v_A t_A + \frac{1}{2}a_A t_A^2 - 300$

Now substituting values

$d = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

Equating the both d

$v_B * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

substituting values

$29 * t_A = 22t_A + \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A = \frac{1}{2} (2.40)^2 t_A^2 - 300$

$7 t_A =1.2 t_A^2 - 300$

$1.2 t_A^2 - 7 t_A - 300 = 0$

Solving this using quadratic formula we have that

$t_A = 19.0 \ s$

7 0

3 years ago