Answer:
The young tree, originally bent, has been brought into the vertical position by adjusting the three guy-wire tensions to AB = 7 lb, AC = 8 lb, and AD = 10 lb. Determine the force and moment reactions at the trunk base point O. Neglect the weight of the tree.
C and D are 3.1' from the y axis B and C are 5.4' away from the x axis and A has a height of 5.2'
Explanation:
See attached picture.
Answer:
t = 103.45 n m
Explanation:
given,
refractive index of cornea = 1.38
refractive index of eye drop = 1.45
wavelength of refractive index = 600 nm
refractive index of eye drop is greater than refractive index of cornea and the air.
Formula used in this case
for constructive interference

At m = 0 for the minimum thickness, so
t = 103.45 n m
the minimum thickness of the film of eyedrops t = 103.45 n m
Answer:
The frequency of the photon decreases upon scattering
Explanation:
Here we note that when a photon is scattered by a charged particle, it is referred to as Compton scattering.
Compton scattering results in a reduction of the energy of the photon and hence an increase in the wavelength (from λ to λ') of the photon known as Compton effect.
Therefore, since the wavelength increases, we have from
λf = λ'f' = c
f = c/λ
Where:
f and f' = The frequency of the motion of the photon before and after the scattering
c = Speed of light (constant)
We have that the frequency, f, is inversely proportional to the wavelength, λ as follows;
f = c/λ
As λ = increases, and c is constant, f decreases, therefore, the frequency of the photon decreases upon scattering.
Answer:
I = 215.76 A
Explanation:
The direction of magnetic field produced by conductor 1 on the location of conductor 2 is towards left. Based on Right Hand Rule -1 and taking figure 21.3 as reference, the direction of force Fm due to magnetic field produced at C_2 is shown above. The force Fm balances the weight of conductor 2.
Fm = u_o*I^2*L/2*π*d
where I is the current in each rod, d = 0.0082 m is the distance 27rId
between each, L = 0.85 m is the length of each rod.
Fm = 4π*10^-7*I^2*1.1/2*π*0.0083
The mass of each rod is m = 0.0276 kg
F_m = mg
4π*10^-7*I^2*1.1/2*π*0.0083=0.0276*9.8
I = 215.76 A
note:
mathematical calculation maybe wrong or having little bit error but the method is perfectly fine
Answer:
v_average = 500 m / min
Explanation:
Average speed is defined
v = (x_{f} -x₀) / Δt
let's look in each section
section 1
the variation of the distance is 800 in a time of 1.4 min
v₁ = 800 / 1.4
v₁ = 571.4 m / min
section 2
distance interval 500 in a 1.6 min time interval
v₂ = 500 / 1.6
v₂ = 312.5 m / min
section 3
distance interval 1200 m in a time 2 min
v₃ = 1200/2
v₃ = 600 m / min
taking the speed of each section we can calculate the average speed
the distance traveled
Δx = 800 + 500 + 1200
Δx = 2500 m
the time spent
Δt = 1.4 + 1.6+ 2
Δt = 5 min
v_average = Δx / Δt
v_average = 2500/5
v_average = 500 m / min