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ki77a [65]
2 years ago
6

If you could travel 900 meters in 55 seconds, what is your speed? (Answer in details=brainliest)

Physics
1 answer:
Alika [10]2 years ago
7 0

Answer:

<em>The </em><em>rate </em><em>of </em><em>change </em><em>of </em><em>distance</em><em> </em><em>is </em><em>called </em><em><u>speed </u></em><em>it </em><em>can </em><em>also </em><em>be </em><em>defined</em><em> </em><em>as </em><em>distance</em><em> </em><em>travelled</em><em> </em><em>per </em><em>unit </em><em>time,</em><em>from </em><em>these </em><em>definition</em><em> </em><em>you </em><em>can </em><em>tell </em><em>that </em><em>the </em><em>formula</em><em> </em><em>will </em><em>be </em>

<em>speed=</em><em>distance</em><em>/</em><em>time</em>

<em>in </em><em>this </em><em>question</em><em> </em><em>the </em><em>distance</em><em> </em><em>is </em><em>9</em><em>0</em><em>0</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em> </em><em>and </em><em>the </em><em>time </em><em>is </em><em>5</em><em>5</em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em>s</em><em> </em><em>therefore</em><em> </em><em>the </em><em>speed </em><em>will </em><em>be:</em>

<em>s=</em><em>9</em><em>0</em><em>0</em><em>m</em><em>e</em><em>t</em><em>e</em><em>r</em><em>s</em><em>/</em><em>5</em><em>5</em><em>s</em><em>e</em><em>c</em><em>o</em><em>n</em><em>d</em><em>s</em>

<em>=</em><em>1</em><em>6</em><em>.</em><em>3</em><em>6</em><em>m</em><em>/</em><em>s</em>

<em>I </em><em>hope</em><em> this</em><em> helps</em>

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a high schooler who lives on the "5th" floor of an apartment complex dropped his phone off the balcony. If the balcony is 18.75
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Answer:

1.38 s

Explanation:

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7 0
3 years ago
Sketch the electric field around these two objects if they have the same sign of charge. Make a separate drawing showing equipot
diamong [38]

Answer:

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

Explanation:

The electric field around two objects with charge of the same sign, for simplicity suppose that the objects have positive point spherical charges,

          E = k q / r2

bold letters indicate vectors, therefore the total electric field is

           E_total = E1 + E2

the module of this field is

           E_total = E1- E2

therefore we can outline this field

* far from one of the charges, the field of the other charge is small and can be neglected

* on the outside of the loads the fields are added territorially

* between the charges the two fields tend to vanish

An outline of these shows in Attachment A

The equipotential surfaces are defined as being perpendicular to the electric field lines since the electric field and the power difference are related

              E = \frac{dV}{dx} i^ + \frac{dV}{dy} j^ + \frac{dV}{dz} k^ = \Delta V

We can schematize some characteristics of these surfaces

* very close to each load are spherical surfaces

* very far from the load is an elliptical surface, which envelops the loads

* between them there is a point of zero potential point C

See attached part B

5 0
3 years ago
Which wave diffracts the most when encountering an obstacle?
viva [34]

Answer;

-A wave with the longest wavelength.

Explanation;

-Diffraction is the apparent of wave through,around small obstacles and the spreading out of wave past small openings. When thinking of diffraction of a wave think of shining a flashlight around a corner. The light bends around the corner but there is a place where it is dark and the light does not hit. Diffraction of a wave is basically the wave bending around an object then dispersing out.

-The amount of diffraction (the sharpness of the bending) increases with increasing wavelength and decreases with decreasing wavelength. When the wavelength of the waves is smaller than the obstacle, no noticeable diffraction occurs.

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ludmilkaskok [199]

Answer:

from Late Latin convectionem

Explanation:

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2 years ago
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Answer:

Suppose the micrometeoroid weighed 1 g = .001 kg

Suppose also the spacecraft were moving at 18,000 mph (1.5 hrs per rev)

Usually, the smaller particle would be moving but for simplicity suppose that it were stationary wrt the ground

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