Can someone help me a bit on this? Will mark brainliest. ( no physical science option soooo)
1 answer:
Answer:
When a light wave goes through a slit, it is diffracted, which means the slit opening acts as a new source of waves. How much a light wave diffracts<em> (how much it fans out)</em> depends on the wavelength of the incident light. The wavelength must be larger than the width of the slit for the maximum diffraction. Thus, for a given slit, red light, because it has a longer wavelength, diffracts more than the blue light.
The corresponding relation for diffraction is
,
where is the wavelength of light,
is the slit width, and
is the diffraction angle.
From this relation we clearly see that the diffraction angle is directly proportional to the wavelength
of light—longer the wavelength larger the diffraction angle.
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Answer:
A) F(r) = [12a/(r^(13))] - [6b/(r^(7))]
ii) Graphs are attached
B) Equilibrium Distance = (2a/b)^(1/6)
C) Minimum Energy = b²/4a
D) a = 6.67 x 10^(-138) Jm^(12)
b = 6.41 x 10^(-78) Jm^(6)
Explanation:
I've attached the explanation of A-C alongside the graphs
D) i) From the question, we are to make r equal to the derivative of "r" we got when F(r) = 0 which was r = (2a/b)^(1/6)
Thus, since we are given equilibrium distance as: 1.13 x 10^(-10), hence;
(2a/b)^(1/6) = 1.13 x 10^(-10)m
So, (2a/b)= [1.13 x 10^(-10)]^(6)
a = (b/2)[1.13 x 10^(-10)]^(6)
From earlier, we saw that b²/4a = U(r)
Thus since U(r) = 1.54 x 10^(-18) from the question, b²/4a = 1.54 x 10^(-18)
Putting a = (b/2)[1.13 x 10^(-10)]^(6);
We have;
(b²) / [(4b/2)[1.13 x 10^(-10)]^(6)]] = 1.54 x 10^(-18)
b/2 = [1.13 x 10^(-10)]^(6)] x 1.54 x 10^(-18)
So, b = 6.41 x 10^(-78) Jm^(6)
ii) Putting (6.41 x 10^(-78))² for b in;
a = (b/2)[1.13 x 10^(-10)]^(6)
We have, a = (6.41 x 10^(-78))²/ ( 4 x 1.54 x 10^(-18)
So a = 6.67 x 10^(-138) Jm^(12)
Answer:
The cart moved away from the starting point between 8 s and 10 s.
Explanation:
Given that :
Which of these does NOT describe the cart’s motion on this graph?
The cart was at rest between 5 s and 7 s : From the distance of golf cart Vs time graph ; the car was at rest between 5s and 7s as the graph was flat with that time interval, meaning there was no change in distance.
The cart moved toward the starting point between 7 s and 12 s. : The graph depicts a negative slope at this time interval as the distance from starting point fell from about 26 m to 10 m
The cart moved away from the starting point between 8 s and 10 s. : At this time interval, the cart moved towards the starting not and not away. This could be seen in the decrease in Distance from starting point between the tune interval.
The cart moved away from the starting point between 2 s and 5 s. - - > The cart moved away from the starting point, with the positive slope signifying an increase in distance.
Therefore, The cart moved away from the starting point between 8 s and 10 s does not describe the motion of the cart on the graph.
