Answer:
Part A:
The proton has a smaller wavelength than the electron.
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Part B:
The proton has a smaller wavelength than the electron.
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Explanation:
The wavelength of each particle can be determined by means of the De Broglie equation.
(1)
Where h is the Planck's constant and p is the momentum.
(2)
Part A
Case for the electron:

But 


Case for the proton:


Hence, the proton has a smaller wavelength than the electron.
<em>Part B </em>
For part b, the wavelength of the electron and proton for that energy will be determined.
First, it is necessary to find the velocity associated to that kinetic energy:


(3)
Case for the electron:

but


Then, equation 2 can be used:

Case for the proton :

But 


Then, equation 2 can be used:

Hence, the proton has a smaller wavelength than the electron.
Answer:
D = 9.9 10⁶ mi
Explanation:
In the exercise they give the expression for maximum viewing distance
D = 2 r h + h²
Ask for this distance for a height of 1100 feet
Let's calculate
D = 2 3960 1100 + 1100²
D = 8.712 10⁶ + 1.21 10⁶
D = 9.92 10⁶ mi
D = 9.9 10⁶ mi
The unit of measurement of work is the Joules. B.
11: 75% = 0.75 12 / 0.75 = 16 The Knicks attempted 16 free throws.
12: 320 - (0.05 x 320) = 304 304 Claims were accepted