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Rashid [163]
3 years ago
11

Hello, my younger sister is stuck on this and I just got on here to ask this question. I have my original account but I just mad

e a new one for no particular reason. She uses a program called Edgenuity and this is a lab report about some kind of motion lab that includes graphs and toy cars, racetracks, etc. But anyways, it would help my sister if you could answer, explain, and do anything you know about this, and she wants to know what "trends" mean. If you use Edgenuity and you were/are in the 6th grade or over, and if you have went through this before, it would be great if you could take the time to answer this: If you constructed graphs, what trends do they indicate in your data?
Yes this is funny and I wrote like a whole paragraph just to say this question, but please answer it anyways.
Physics
1 answer:
nignag [31]3 years ago
4 0

Answer:

Hi there!

I'm sorry your sister is struggling!

I am an edgenuity student in grade 11, I could probably help!

A trend is a line on a graph that all points seem to follow. This is best explained when thinking about line of best fit. If all your points go upward each time and are closed together, we can find the line that gets closest to each point!

Think of a trend as a pattern. The line of best fit, created by analyzing the trends, helps us guess at what the data points beyond what we have will equal

Is there any clarification she needs?

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lions [1.4K]

Hi there!

Recall that:
\Delta V = -\int\limits^a_b {E \cdot } \, dx

Given an electric field, the potential difference can be solved by using integration. Similarly:
E = -\frac{dV}{dx}

We can differentiate the electric potential equation to solve for the electric field.

Use the power rule:
\frac{dy}{dx} x^n = nx^{n - 1}

Differentiate the given equation.

-\frac{dV}{dx}\frac{7}{x^2} =- \frac{dV}{dx}7x^{-2} = -(-14x^{-3}) = \frac{14}{x^3}

Or:
\boxed{E(x) = \frac{14}{x^3}}

6 0
2 years ago
Worth 25 points on my exam
I am Lyosha [343]
  • Initial velocity=20m/s
  • Final velocity=0m/s(As the car stops)
  • Acceleration=-8m/s^2
  • Distance=s=26m

We need to verify the thrid equation of kinematics here

\\ \tt\longmapsto v^2-u^2=2as

\\ \tt\longmapsto 20^2=2(-8)s

\\ \tt\longmapsto 400=-16s

\\ \tt\longmapsto s=|400/-16|

\\ \tt\longmapsto s=25m

The squirrel has a good luck ,Car gets stopped just 1m away from the squirrel .

7 0
2 years ago
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A vehicle of mass 100kg has a kinetic energy of 5000 J at an instant. The velocity at that instant is​
snow_lady [41]

Answer:

  • \bf\pink{10\:m/s}

Explanation:

<h2><u>Given</u> :-</h2>

  • \sf\red{Mass \ of \ vehicle \ (m) = 100 \ kg}
  • \sf\orange{Kinetic \ energy \ (K.E.) = 5000 \ J}

<h2><u>To Find</u> :-</h2>

  • \sf\green{Velocity\: of\: the \:vehicle \:at \:the \:instant}

<h2><u>Formula to be used</u> :-</h2>

  • \sf\blue{K.E. = \dfrac{1}{2} mv^{2}}

Where,

  • K.E. = Kinetic energy possessed by the body
  • M = Mass of the body
  • V = Velocity of the body

<h2><u>Solution</u> :-</h2>

\to\:\:\sf\red{K.E. = \dfrac{1}{2}mv^{2}}

\to\:\:\sf\orange{5000 = \dfrac{1}{2} \times 100 \times v^{2}}

\to\:\:\sf\green{5000 = 50 \times v^{2}}

\to\:\:\sf\blue{\dfrac{5000}{50} = v^{2}}

\to\:\:\sf\purple{100 = v^{2}}

\to\:\:\sf\red{\sqrt{100} = v}

\to\:\:\sf\orange{ 10 = v}

\to\:\:\bf\pink{v = 10\:m/s}

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7 0
2 years ago
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White raven [17]
So, there should be two forces acting on the refrigerator: the applied force and the friction force.

The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.

We have an applied force of 400 N to the right, this means that:
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3 years ago
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The Sun has a mass of 1.99x10^30 kg and a radius of 6.96x10^8 m. Calculate the acceleration due to gravity, in meters per second
just olya [345]

Answer:

g=274\ m/s^2

Explanation:

Mass of the Sun, M=1.99\times 10^{30}\ kg

The radius of the Sun, r=6.96\times 10^8\ m

We need to find the acceleration due to gravity on the surface of the Sun. It is given by the formula as follows :

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So, the value of acceleration due to gravity on the Sun is 274\ m/s^2.

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