Answer:
Explanation:
Given that
Superelation= 0.08ft/ft
Given curve= u•
Curve junction factor= 0.13
DR= 5729.57795
R = 5729.57795/D
R = 5729.57795/4
R = 1432.4ft
c + f = V^2/gG
0.08 + 0.13 = V^2 / (32*1432.4)
V^2 = 9625.728 or V = 98 ft/sec
The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed
Answer:
E. The period of oscillation increases.
Explanation:
The period of oscillation is:
T = 2π√(m/k)
Frequency is the inverse of period (f = 1/T), so as period increases, frequency decreases.
Increasing the mass will increase the period and decrease the frequency.
Answer: a) 274.34 nm; b) 1.74 eV c) 1.74 V
Explanation: In order to solve this problem we have to consider the energy balance for the photoelectric effect on tungsten:
h*ν = Ek+W ; where h is the Planck constant, ek the kinetic energy of electrons and W the work funcion of the metal catode.
In order to calculate the cutoff wavelength we have to consider that Ek=0
in this case h*ν=W
(h*c)/λ=4.52 eV
λ= (h*c)/4.52 eV
λ= (1240 eV*nm)/(4.52 eV)=274.34 nm
From this h*ν = Ek+W; we can calculate the kinetic energy for a radiation wavelength of 198 nm
then we have
(h*c)/(λ)-W= Ek
Ek=(1240 eV*nm)/(198 nm)-4.52 eV=1.74 eV
Finally, if we want to stop these electrons we have to applied a stop potental equal to 1.74 V . At this potential the photo-current drop to zero. This potential is lower to the catode, so this acts to slow down the ejected electrons from the catode.
