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3 years ago

An engineer is designing a runway for an airport. Of the planes that will use the airport, the

Physics

1 answer:

Arada [10]3 years ago

6 0

Answer:

816m

Explanation:

70^2m/s=2(3m/ss)(x)

x=816

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max2010maxim [7]

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3 years ago

While passing through a human cell, an x-ray photon interacts with and inactivates the cell’s master molecule. What is the conse

Yuliya22 [10]

Answer:

Cell Death

Explanation:

Cell death is defined as the biological process which ceases the function of the cell to carry out. This can be caused due to the formation of new cells in place of old cells.

Or it can be cause due to some serious disease or may be caused due to the injury or due to the death of that organism to which these cells belong.

And another case is that when X-ray photon interact with the human cell while it passes through the cell, it will damage the cell and cease it to function well and a more drastic condition occurs and that cell become dead.

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3 years ago

If the distance to a distant galaxy is recorded as 10,000 light years, then light must have been traveling for 5000 years to rea

satela [25.4K]

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3 years ago

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FREE BRAINLEST JUST COMMENT MRBEAST

Pepsi [2]

Answer:

MRBEAST-

Explanation:

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3 years ago

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A stone is dropped into a river from a bridge at a height h above the water. Another stone is thrown vertically down at a time t

Mumz [18]

Answer:

$v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

Explanation:

We will apply the equations of kinematics to both stones separately.

First stone:

Let us denote the time spent after the second stone is thrown as 'T'.

$y - y_0 = v_{y_0}(t+T) + \frac{1}{2}a(t+T)^2\\0 - h = 0 + \frac{1}{2}(-g)(t+T)^2\\(t+T)^2 = \frac{2h}{g}\\T = \sqrt{\frac{2h}{g}}-t$

Second stone:

$y - y_0 = v_{y_0}T + \frac{1}{2}aT^2\\0 - h = v_{y_0}T -\frac{1}{2}gT^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\sqrt{\frac{2h}{g}} - t)^2\\-h = v_{y_0}(\sqrt{\frac{2h}{g}} - t) - \frac{g}{2}(\frac{2h}{g} + t^2 - 2t\sqrt{\frac{2h}{g}})\\-h = v_{y_0}\sqrt{\frac{2h}{g}} - v_{y_0}t - h -\frac{g}{2}t^2 + gt\sqrt{\frac{2h}{g}}\\v_{y_0}(\sqrt{\frac{2h}{g}} - t) = \frac{g}{2}t^2 - gt\sqrt{\frac{2h}{g}}\\v_{y_0} = \frac{\frac{g}{2}t(t - 2\sqrt{\frac{2h}{g}})}{\sqrt{\frac{2h}{g}} - t}$

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3 years ago

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