The time elapsed when the vehicles are closest to each other is 20 min.
The given parameters:
The time elapsed when the cars are close to each other is calculated by applying the principles of relative speed.
Thus, the time elapsed when the vehicles are closest to each other is 20 min.
Learn more about relative velocity here: brainly.com/question/24430414
Answer:
A half-life is the time required for one half of the nuclei in a radio- active isotope to decay.
Explanation:
A radio-active isotope is an isotope which undergoes radioactive decay.
Radioactive decay is a spontaneous process in which the nucleus of an atom changes its state (turning into a different nucleus, or de-exciting), emitting radiation, which can be of three different types: alpha, beta or gamma.
The half-life of a radio-active isotope is the time required for half of the nuclei of the initial sample to decay.
The law of radio-active decay can be expressed as follows:
where
N(t) is the number of undecayed nuclei left at time t
N0 is the initial number of nuclei
t is the time
is the half-life
We see that when (that means, when 1 half-life has passed), the number of undecayed nuclei left is
So, half of the initial nuclei.
The current flowing in silicon bar is 2.02 10^-12 A.
<u>Explanation:</u>
Length of silicon bar, l = 10 μm = 0.001 cm
Free electron density, Ne = 104 cm^3
Hole density, Nh = 1016 cm^3
μn = 1200 cm^2 / V s
μр = 500 cm^2 / V s
The total current flowing in the bar is the sum of the drift current due to the hole and the electrons.
J = Je + Jh
J = n qE μn + p qE μp
where, n and p are electron and hole densities.
J = Eq (n μn + p μp)
we know that E = V / l
So, J = (V / l) q (n μn + p μp)
J = (1.6 10^-19) / 0.001 (104
1200 + 1016
500)
J = 1012480 10^-16 A / m^2.
or
J = 1.01 10^-9 A / m^2
Current, I = JA
A is the area of bar, A = 20 μm = 0.002 cm
I = 1.01 10^-9
0.002 = 2.02
10^-12
So, the current flowing in silicon bar is 2.02 10^-12 A.
Answer:
The beta decay takes place.
Explanation:
The reaction of radioactivity of carbon 14 to nitrogen 14 is
There is a beta decay.
The reaction is
Here some energy is released in form of neutrino.
Answer:
a) What are the characteristics of the radiation emitted by a blackbody?
The total emitted energy per unit of time and per unit of area depends in its temperature (Stefan-Boltzmann law).
The peak of emission for the spectrum will be displaced to shorter wavelengths as the temperature increase (Wien’s displacement law).
The spectral density energy is related with the temperature and the wavelength (Planck’s law).
b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave length of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?
The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.
Explanation:
A blackbody is an ideal body that absorbs all the thermal radiation that hits its surface, thus becoming an excellent emitter, as these bodies express themselves without light radiation, and therefore they look black.
The radiation of a blackbody depends only on its temperature, thus being independent of its shape, material and internal constitution.
If it is study the behavior of the total energy emitted from a blackbody at different temperatures, it can be seen how as the temperature increases the energy will also increase, this energy emitted by the blackbody is known as spectral radiance and the result of the behavior described previously is Stefan's law:
(1)
Where is the Stefan-Boltzmann constant and T is the temperature.
The Wien’s displacement law establish how the peak of emission of the spectrum will be displace to shorter wavelengths as the temperature increase (inversely proportional):
(2)
Planck’s law relate the temperature with the spectral energy density (shape) of the spectrum:
(3)
b) According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wavelength of 450 nm, than a object whose spectrum peaks in the red, at 700 nm?
It is need it to known the temperature of both objects before doing the comparison. That can be done by means of the Wien’s displacement law.
Equation (2) can be rewrite in terms of T:
(4)
Case for the object with the blackbody emission spectrum peak in the blue:
Before replacing all the values in equation (4), (450 nm) will be express in meters:
⇒
Case for the object with the blackbody emission spectrum peak in the red:
Following the same approach above:
⇒
Comparison:
The object with the blackbody emission spectrum peak in the blue is 1.55 times hotter than the object with the blackbody emission spectrum peak in the red.