The wavelength of the third line in the Lyman series, and identify the type of EM radiation
In this series, the spectral lines are obtained when an electron makes a transition from any high energy level (n=2,3,4,5... ). The wavelength of light emitted in this series lies in the ultraviolet region of the electromagnetic spectrum.
1 / lambda = R(h)* (
-
)
= 109678 (
-
)
= 109678 (8/9)
Lambda = 9 / (109678 * 8 )
= 102.6 *
m = 102.6 nm
To learn more about Lyman series here
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Answer:
320 N/m
Explanation:
From Hooke's law, we deduce that
F=kx where F is applied force, k is spring constant and x is extension or compression of spring
Making k the subject of formula then

Conversion
1m equals to 100cm
Xm equals 25 cm
25/100=0.25 m
Substituting 80 N for F and 0.25m for x then

Therefore, the spring constant is equal to 320 N/m
This problem uses the relationships among current
I, current density
J, and drift speed
vd. We are given the total of electrons that pass through the wire in
t = 3s and the area
A, so we use the following equation to to find
vd, from
J and the known electron density
n,
so:

<span>The current
I is any motion of charge from one region to another, so this is given by:
</span>

The magnitude of the current density is:

Being:

<span>
Finally, for the drift velocity magnitude vd, we find:
</span>
Notice: The current I is very high for this wire. The given values of the variables are a little bit odd
One, three and five are correct.
Although if the second statement is saying that stars use gravitational force to support nuclear fusion which in turn produces energy then that would be correct, but I don’t think so :)