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Lelu [443]
3 years ago
11

The photon energies used in different types of medical x-ray imaging vary widely, depending upon the application. Single dental

x rays use photons with energies of about 25 {\rm keV}. The energies used for x-ray microtomography, a process that allows repeated imaging in single planes at varying depths within the sample, is 2.5 times greater.
What are the wavelengths of the x rays used for these two purposes? answer in m
\lambda_{\rm dental}
\lambda_{\rm microtomography}
Physics
1 answer:
pav-90 [236]3 years ago
4 0

A) 5.0\cdot 10^{-11} m

The energy of an x-ray photon used for single dental x-rays is

E=25 keV = 25,000 eV \cdot (1.6\cdot 10^{-19} J/eV)=4\cdot 10^{-15} J

The energy of a photon is related to its wavelength by the equation

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

Re-arranging the equation for the wavelength, we find

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{4\cdot 10^{-15}J}=5.0\cdot 10^{-11} m

B) 2.0\cdot 10^{-11} m

The energy of an x-ray photon used in microtomography is 2.5 times greater than the energy of the photon used in part A), so its energy is

E=2.5 \cdot (4\cdot 10^{-15}J)=1\cdot 10^{-14} J

And so, by using the same formula we used in part A), we can calculate the corresponding wavelength:

\lambda=\frac{hc}{E}=\frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{1\cdot 10^{-14}J}=2.0\cdot 10^{-11} m

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Spaceship 1 and Spaceship 2 have equal masses of 300kg. They collide. Spaceship 1's final speed is 3 m/s, and Spaceship 2's fina
fiasKO [112]

Answer:

B. 1500 kg*m/s

Explanation:

Momentum p = m* v

In any type of collision, the total momentum is preserved!

The total momentum before and the total momentum after the collision is the same. We know the mass and speed after the collision so we can calculate the total momentum.

p1 + p2 =

m1*v1 + m2*v2

m1 = me = 300 kg

v1 = 3 m/s

v2 = 2 m/s

Substitute the given numbers:

300*3 + 300+2

900 + 600

1500 kg*m/s, which is answer B.

3 0
3 years ago
Two charges are located in the x – y plane. If ????1=−4.10 nC and is located at (x=0.00 m,y=0.600 m) , and the second charge has
faust18 [17]

Answer:

The x-component of the electric field at the origin = -11.74 N/C.

The y-component of the electric field at the origin = 97.41 N/C.

Explanation:

<u>Given:</u>

  • Charge on first charged particle, q_1=-4.10\ nC=-4.10\times 10^{-9}\ C.
  • Charge on the second charged particle, q_2=3.80\ nC=3.80\times 10^{-9}\ C.
  • Position of the first charge = (x_1=0.00\ m,\ y_1=0.600\ m).
  • Position of the second charge = (x_2=1.50\ m,\ y_2=0.650\ m).

The electric field at a point due to a charge q at a point r distance away is given by

\vec E = \dfrac{kq}{|\vec r|^2}\ \hat r.

where,

  • k = Coulomb's constant, having value \rm 8.99\times 10^9\ Nm^2/C^2.
  • \vec r = position vector of the point where the electric field is to be found with respect to the position of the charge q.
  • \hat r = unit vector along \vec r.

The electric field at the origin due to first charge is given by

\vec E_1 = \dfrac{kq_1}{|\vec r_1|^2}\ \hat r_1.

\vec r_1 is the position vector of the origin with respect to the position of the first charge.

Assuming, \hat i,\ \hat j are the units vectors along x and y axes respectively.

\vec r_1=(0-x_1)\hat i+(0-y_1)\hat j\\=(0-0)\hat i+(0-0.6)\hat j\\=-0.6\hat j.\\\\|\vec r_1| = 0.6\ m.\\\hat r_1=\dfrac{\vec r_1}{|\vec r_1|}=\dfrac{0.6\ \hat j}{0.6}=-\hat j.

Using these values,

\vec E_1 = \dfrac{(8.99\times 10^9)\times (-4.10\times 10^{-9})}{(0.6)^2}\ (-\hat j)=1.025\times 10^2\ N/C\ \hat j.

The electric field at the origin due to the second charge is given by

\vec E_2 = \dfrac{kq_2}{|\vec r_2|^2}\ \hat r_2.

\vec r_2 is the position vector of the origin with respect to the position of the second charge.

\vec r_2=(0-x_2)\hat i+(0-y_2)\hat j\\=(0-1.50)\hat i+(0-0.650)\hat j\\=-1.5\hat i-0.65\hat j.\\\\|\vec r_2| = \sqrt{(-1.5)^2+(-0.65)^2}=1.635\ m.\\\hat r_2=\dfrac{\vec r_2}{|\vec r_2|}=\dfrac{-1.5\hat i-0.65\hat j}{1.634}=-0.918\ \hat i-0.398\hat j.

Using these values,

\vec E_2= \dfrac{(8.99\times 10^9)\times (3.80\times 10^{-9})}{(1.635)^2}(-0.918\ \hat i-0.398\hat j) =-11.74\ \hat i-5.09\ \hat j\  N/C.

The net electric field at the origin due to both the charges is given by

\vec E = \vec E_1+\vec E_2\\=(102.5\ \hat j)+(-11.74\ \hat i-5.09\ \hat j)\\=-11.74\ \hat i+(102.5-5.09)\hat j\\=(-11.74\ \hat i+97.41\ \hat j)\ N/C.

Thus,

x-component of the electric field at the origin = -11.74 N/C.

y-component of the electric field at the origin = 97.41 N/C.

4 0
3 years ago
Main function of gametes?
vichka [17]
Sexual reproduction. thats the answer i think
3 0
3 years ago
At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coeffi
Flura [38]

The given data is incomplete. The complete question is as follows.

At an accident scene on a level road, investigators measure a car's skid mark to be 84 m long. It was a rainy day and the coefficient of friction was estimated to be 0.36.  Use these data to determine the speed of the car when the driver slammed on (and locked) the brakes. (why does the car's mass not matter?)

Explanation:

Let us assume that v is the final velocity and u is the initial velocity of the car. Let s be the skid marks and \mu be the friction coefficient and m be the mass of car.

Hence, the given data is as follows.

                v = 0,     s = 84 m,     \mu = 0.36

According to Newton's law of second motion the expression for acceleration is as follows.

                      F = ma

                 -\mu N = ma

                 -\mu mg = ma

                      a = -\mu g

Also,    

               v^{2} = u^{2} + 2as

              (0)^{2} = u^{2} + 2(-\mu g)s

                  u^{2} = 2(\mu g)s

                            = \sqrt{2(0.36)(9.81 m/s^{2})(84 m)}

                            = 24.36 m/s

Thus, we can conclude that the speed of the car when the driver slammed on (and locked) the brakes is 24.36 m/s.

4 0
3 years ago
Explain why the rust of an iron bar decreases the strength of the bar.
garri49 [273]
Rusting is a chemical change, which forms a new substance which is brittle. A solid that is formed by a chemical reaction.
7 0
4 years ago
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