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3 years ago

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Hospital X-ray generators emit X-rays with wavelength of about 15.0 nanometers (nm), where 1nm=10−9m. What is the energy of a ph

oton of the X-rays? Express your answer to three significant figures and include the appropriate units.

1 answer:

Sergio [31]3 years ago

4 0

Answer:

$E=1.32\times 10^{-17}\ J$

Explanation:

Given that,

The wavelength of x ray, $\lambda=15\ nm=15\times 10^{-9}\ m$

We need to find the energy if a photon on the x-rays. The formula for the energy of the photon is given by :

$E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.63\times 10^{-34}\times 3\times 10^8}{15\times 10^{-9}}$

$E=1.32\times 10^{-17}\ J$

So, the energy of a photon of the X-rays is $1.32\times 10^{-17}\ J$ . Hence, this is the required solution.

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In order to find each of those, we need to start by finding the velocities of the ball in points A and B. We will analyze the trajectory of the ball when bouncing back to the top of the building. Let's start by finding the velocity of the ball in A:

We can use the following formula to determine the velocity of the ball in part A:

$\Delta y=V_{A}t+\frac{1}{2}at^{2}$

which can be solved for $V_{A}$

so we get:

$V_{A}=\frac{\Delta y-\frac{1}{2}at^{2}}{t}$

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$V_{A}=\frac{(1.30m)-\frac{1}{2}(-9.8m/s^{2})(0.115s)^{2}}{0.115s}$

which yields:

$V_{A}=11.87m/s$

once we got the velocity at point A, we can now find the velocity at point B. We can do so by using the following formula:

$a=\frac{V_{A}-V_{B}}{t}$

which can be solved for $V_{B}$ which yields:

$V_{B}=V_{A}-at$

so we can substitute values now:

$V_{B}=11.87m/s-(-9.8)(0.115s)$

Which yields:

$V_{B}=13m/s$

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Let's start with $y_{1}$

We can use the following formula to find it:

$y_{1}=\frac{V_{f}^{2}-V_{A}^{2}}{2a}$

we know the final velocity of the rebound will be zero, so we can simplify our formula:

$y_{1}=\frac{-V_{A}^{2}}{2a}$

so we can substitute now:

$y_{1}=\frac{-(11.87m/s)^{2}}{2(-9.8m/s^{2})}$

which solves to:

$y_{1}=7.19m$

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the value of $y_{2}$ can be found by using the following formula:

$y_{2}=V_{B}t-\frac{1}{2}at^{2}$

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$t=\frac{2.04s}{2}=1.02s$

so now we can substitute all the values in the given formula:

$y_{2}=(13m/s)(1.02s)-\frac{1}{2}(-9.8m/s^{2})(1.02s)^{2}$

which yields:

$y_{2}=18.36m$

so now that we have all the values we need, we can go ahead and calculate the height of the building:

$h=y_{1}+window+y_{2}$

when substituting we get:

$h=7.19m+1.3m+18.36m$

So the answer is:

h=26.85m

The building is 26.85m tall.

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3 years ago