Answer:
Magnetic field, ![B=2\times 10^{-4}\ T](https://tex.z-dn.net/?f=B%3D2%5Ctimes%2010%5E%7B-4%7D%5C%20T)
Explanation:
Given that,
Velocity of electron, ![v=5\times 10^7\ m/s](https://tex.z-dn.net/?f=v%3D5%5Ctimes%2010%5E7%5C%20m%2Fs)
It enters a region of space where perpendicular electric and a magnetic fields are present.
Magnitude of electric field, ![E=10^4\ V/m](https://tex.z-dn.net/?f=E%3D10%5E4%5C%20V%2Fm)
We need to find the magnetic field will allow the electron to go through the region without being deflected.
Magnetic force on the electron,
.......(1)
Electric force on the electron, F = q E........(2)
From equation (1) and (2) we get:
![qvB\ sin\theta=qE](https://tex.z-dn.net/?f=qvB%5C%20sin%5Ctheta%3DqE)
![B=\dfrac{E}{v}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7BE%7D%7Bv%7D)
![B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}](https://tex.z-dn.net/?f=B%3D%5Cdfrac%7B10%5E4%5C%20V%2Fm%7D%7B5%5Ctimes%2010%5E7%5C%20m%2Fs%7D)
B = 0.0002 T
or
![B=2\times 10^{-4}\ T](https://tex.z-dn.net/?f=B%3D2%5Ctimes%2010%5E%7B-4%7D%5C%20T)
Hence, this is the required solution.
(1.00 atm) (0.1156 L) = (n) (0.08206 L atm / mol K) (273 K) I hoped that helped
25km/h = 6.94 m/s
suvat
s=16
u=6.94
v=0
a=a
v^2=u^2+2as
(v^2-u^2)/2s = a =1.5ms^-2
Answer: The hottest star is Archenar( blue) and the coolest star is Betelgeuse
Explanation:
Objects emit radiation that depends exclusively on their temperature. At an ambient temperature, the radiation emitted by an object is in the infrared spectrum (we could only see it with a special camera). If we heat it we will see that it first turns red (whose state we call “red hot”) because it is the lowest and least energetic wavelength of all.
If we continue to heat it, the wavelength that it emits to one with more energy will continue to increase and we will see that it turns yellow and then white. This is a signal that is emitting at all frequencies (but mainly in blue).
If we continue to warm a body that is "white hot", it would emit in the ultraviolet spectrum, with what would become ... black! then we would not see it emits light in the visible spectrum (well, we would see a very faint bluish light corresponding to the tail of the distribution of the spectrum it emits, but the peak of that spectrum would be in the ultraviolet).