When one body(sun) exerts a force on a second body(planet), the second body simultaneously exerts a force equal in magnitude and opposite in direction of the first body. Which makes the planet orbit in path C.
Hope this helps!!
Answer:
same 0.81m
Explanation:
in this problem if we assume there no resistance of any sort. and we apply the energy conservation
change in Potential energy = change in kinetic energy
mgh = 0.5mv^2
gh = 0.5v^2
the above relation suggests that the speed at the bottom is only depending on the height it is released from not on the shape, mass or radius.
so at the bottom
put h = 0.81m
9.81 * 0.81 * 2 = v^2
v=3.99 m/s
both CYLINDER and SPHERE will have same velocity at the bottom if released from the same height irrespective of shape and size
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.
Explanation:
The time taken by a wave crest to travel a distance equal to the length of wave is known as wave period.
The relation between wave period and frequency is as follows.
T = \frac{1}{f}T=
f
1
where, T = time period
f = frequency
It is given that wave period is 18 seconds. Therefore, calculate the wave period as follows.
T = \frac{1}{f}T=
f
1
or, f = \frac{1}{T}f=
T
1
= \frac{1}{18 sec}
18sec
1
= 0.055 per second (1cycle per second = 1 Hertz)
or, f = 5.5 \times 10^{-2} hertz5.5×10 −2 hertz
<h3>Thus, we can conclude that the frequency of the wave is 5.5 \times 10^{-2} hertz5.5×10 −2 hertz .</h3>
Potential energy U = mgh
Given h = 123 m,
mg = F = 780 N
Then
U = (123)(780)
= 95940
= 9.59 x 10^4