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Ronch [10]
3 years ago
10

The displacement (in meters) of an object moving in a straight line is given by s=1+2t+1/4t^2^2, where t is measured in seconds.

Find the avg velocity over each time period, a) [1,3]
Physics
1 answer:
erica [24]3 years ago
4 0
 <span>(A) 
  s{ t } = (¼)t² + 2t + 1  ...  differentiate to get velocity versus time 

  v{ t } = (½)t + 2  ...  linearly increases with time 

(1) 

 Vavɢ{  a→b  } = [ v{ b } + v{ a } ] ⁄ 2 
 Vavɢ{ 1→3  } = [ v{ 3 } + v{1 } ] ⁄ 2 
 Vavɢ{ 1→3  } = [ (½)(3) + 2 + (½)(1) + 2 ]  ⁄  2 
 Vavɢ{ 1→3  } = 3 m/sec 

(II) and (III) are done the same way. 

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~... 

(B) 
    v{ t } = (½)t + 2 

   v{1 } = (½)(1) + 2 
   v{1 } = 2.5 m/sec</span>
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To find the direction of the force, which will tell us the direction in which the wire deflects or moves, first, we need to point with our thumb in the direction of the current, in this case, to the right.

Now, with the hand open, using the tip of our other fingers we point in the direction of the magnetic field.

For example, if the magnetic field is in the positive z-axis, we will point upwards.

Now the palm of our hand tells us in which direction the force is applied.

This is the right-hand rule.

For example, in the case that the current goes to the right and the magnetic field is upwards, we could see that the force is to the front.

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3 years ago
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What is the shortest-wavelength x-ray radiation in m that can be generated in an x-ray tube with an applied voltage of 93.3 kV?
VikaD [51]

(a) 1.33\cdot 10^{-11} m

The x-rays in the tube are emitted as a result of the collisions of electrons (accelerated through the potential difference applied) on the metal target. Therefore, all the energy of the accelerated electron is converted into energy of the emitted photon:

e \Delta V = \frac{hc}{\lambda}

where the term on the left is the electric potential energy given by the electron, and the term on the right is the energy of the emitted photon, and where:

e=1.6\cdot 10^{-19}C is the electron's charge

\Delta V = 93.3 kV = 93300 V is the potential difference

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

c=3.00\cdot 10^8 m/s is the speed of light

\lambda is the wavelength of the emitted photon

Solving the formula for \lambda, we find:

\lambda=\frac{hc}{e\Delta V}=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{(1.6\cdot 10^{-19})(93300)}=1.33\cdot 10^{-11} m

(b) 93300 eV (93.3 keV)

The energy of the emitted photon is given by:

E=\frac{hc}{\lambda}

where

h is Planck constant

c is the speed of light

\lambda=1.33\cdot 10^{-11} m is the wavelength of the photon, calculated previously

Substituting,

E=\frac{(6.63\cdot 10^{-34})(3\cdot 10^8)}{1.33\cdot 10^{-11}}=1.50\cdot 10^{-14} J

Now if we want to convert into electronvolts, we have to divide by the charge of the electron:

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(c) The following statements are correct:

The maximum photon energy is just the applied voltage times the electron charge. (1)

The value of the voltage in volts equals the value of the maximum photon energy in electron volts.

In fact, we see that statement (1) corresponds to the equation that we wrote in part (a):

e \Delta V = \frac{hc}{\lambda}

While statement (2) is also true, since in part (b) we found that the photon energy is 93.3 keV, while the voltage was 93.3 kV.

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3 years ago
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