The number of charge drifts are 3.35 X 10⁻⁷C
<u>Explanation:</u>
Given:
Potential difference, V = 3 nV = 3 X 10⁻⁹m
Length of wire, L = 2 cm = 0.02 m
Radius of the wire, r = 2 mm = 2 X 10⁻³m
Cross section, 3 ms
charge drifts, q = ?
We know,
the charge drifts through the copper wire is given by
q = iΔt
where Δt = 3 X 10⁻³s
and i = 
where R is the resistance
R = 
ρ is the resistivity of the copper wire = 1.69 X 10⁻⁸Ωm
So, i = 
q = 
Substituting the values,
q = 3.14 X (0.02)² X 3 X 10⁻⁹ X 3 X 10⁻³ / 1.69 X 10⁻⁸ X 0.02
q = 3.35 X 10⁻⁷C
Therefore, the number of charge drifts are 3.35 X 10⁻⁷C
Answer:
3.2 m/s²
Explanation:
Acceleration can be calculated as:
v = u + at (where v is final velocity, u is initial velocity, a is acceleration and t is time)
25 m/s = 9 m/s + a(5 s) (a is unknown)
16 m/s = a(5 s)
a = 3.2 m/s²
We assume that this is a uniform acceleration (meaning that the velocity increases at an equal rate for those 5 seconds).
For any electromagnetic wave, the relationship between frequency and wavelength is given by
where
is the wavelength
c is the speed of light (the speed of the electromagnetic wave)
f is the frequency
For the X-rays in our problem, the frequency is
, while the speed of light is
, so the wavelength of this radiation is
C. The slope of the graph
C. a lot of inertia.............................. :-)
