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ANEK [815]
3 years ago
6

HELPPPPP MEEEEE PLEASE I NEED TO SUBMIT IN LESS THAN 10 MINSS

Physics
1 answer:
DerKrebs [107]3 years ago
6 0

Answer:

X = 2146.05 m

Explanation:

We need to understand first what is the value we need to calculate here. In this case, we want to know how far from the starting point the package should be released. This is the distance.

We also know that the plane is flying a certain height with an specific speed. And the distance we need to calculate is the distance in X with the following expression:

X = Vt   (1)

However we do not know the time that this distance is covered. This time can be determined because we know the height of the plain. This time is referred to the time of flight. And the time of flight can be calculated with the following expression:

t = √2h/g   (2)

Where g is gravity acceleration which is 9.8 m/s². Replacing the data into the expression we have:

t = √(2*2500)/9.8

t = 22.59 s

Now replacing into (1) we have:

X = 95 * 22.59

<h2>X = 2146.05 m</h2>

This is the distance where the package should be released.

Hope this helps

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<h3>What is magnification?</h3>

Magnification is the process in which the size of an object is enlarging but not physical size of something. Those instruments having high magnification power can shows us the image larger and due to which we can see that image clearly.

So we can conclude that we can see the image in most detail that having 400x magnification.

Learn more about telescope here: brainly.com/question/18300677

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<h3>Work done</h3>

The work required to assemble this charge arrangement starting with each of the charges as very large distance from any of the other charges is calculated as follows;

W = qV

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W = q (\frac{kq}{r} )

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W = q (\frac{kq}{r} + \frac{kq}{r\sqrt{2} } + \frac{kq}{r} )\\\\W = \frac{kq^2}{r} ( 2 + \frac{1}{\sqrt{2} } )\\\\W = \frac{(9\times 10^9 \times (4\times 10^{-6})^2}{0.2} ( 2.707)\\\\W = 1.9 \ J

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