If it's Kepler's law of equal areas you're talking about,
then the first of the four statements is true.
D.Radiant energy does not require a medium through which to travel.
Answer:
Explanation:
The relation between the de Broglie wavelength and the momentum of the particle is given by

where, m is the mas of the particle and v be the velocity of the particle and h be the Plank's constant.
So, the de broglie wavelength of proton is given by
.... (1)
The de broglie wavelength of electron is given by
.... (2)
Divide equation (2) by equation (1), we get

As the mass of proton is much more than the mass of electron, so the de broglie wavelength of electron is more than the de Broglie wavelength of proton.
First we need to convert the mm to inches to make our computation
easier.
1mm = 0.0393701
32mm * 0.0393701 = 1.25 in
Solution:
C = 1/2d = ½ (1.25) = 0.625 in^4
Tension: tension = Te/J = 2T/ piC^3
= (2)(2500)/pi (0.0625)^3 = 6.519 x 10^3 psi = 6.519 ksi
Bending:
I = pi/4 * c^4 = 119.842 x 10^-3 in^4
M = (5)(600) = 3600 lb in
G = My/I = (3600)(0.625)/119.842 x 10^-3 = -18.775 x 10^2
psi = -18.775ksi
Gx = -18.775 ksi
Gy = 0
Txy = 6.519 ksi
G ave – ½ (Gx + Gy) = -9.387 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2 + (6.519)^2 = 11.429 ksi
1.
G1 = Gave + R = -9/387 + 11.429 = 2.04 ksi
G2 = Gave - R = -9/387 - 11.429 = -20.8
Tan 2ϴp = 2txy/Gx – Gy = 2(6.519)/-9.387 =
-1.3889
ϴp = -27.1 degrees and 62.9 degrees
2.
Tmax = R = 11.43 ksi
R = sqrt (Gx – Gy/2)^2 + Txy^2 = sqrt(-9.387)^2
+ (6.519)^2 = 11.429 ksi
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