A cylindrical Nickel rod (9 mm diameter, 50 m long) is pulled in tension with a load of 6,283 N. What would the elongation of th
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Answer:
- k = 167.33 N/m
Explanation:
- The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
- v = c = λ*f
where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =
4.92*10⁷ m.
Solving for f, we get the frequency of the radio waves:
f = 6.1 Hz
- Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by a fixed relationship between the spring constant k and the mass m, as follows:
- Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:
- We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
Answer:
<em>a) 318.2 W/m^2</em>
<em>b) 2.5 x 10^-4 J</em>
<em>c) 1.55 x 10^-8 v/m</em>
<em></em>
Explanation:
Power of laser P = 1 mW = 1 x 10^-3 W
exposure time t = 250 ms = 250 x 10^-3 s
If beam diameter = 2 mm = 2 x 10^-3 m
then
cross-sectional area of beam A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) Intensity I = P/A
where P is the power of the laser
A is the cros-sectional area of the beam
I = ( 1 x 10^-3)/(3.142 x 10^-6) = <em>318.2 W/m^2</em>
<em></em>
b) Total energy delivered E = Pt
where P is the power of the beam
t is the exposure time
E = 1 x 10^-3 x 250 x 10^-3 = <em>2.5 x 10^-4 J</em>
<em></em>
c) The peak electric field is given as
E =
where I is the intensity of the beam
E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = <em>1.55 x 10^-8 v/m</em>