Complete question:
At a particular instant, an electron is located at point (P) in a region of space with a uniform magnetic field that is directed vertically and has a magnitude of 3.47 mT. The electron's velocity at that instant is purely horizontal with a magnitude of 2×10⁵ m/s then how long will it take for the particle to pass through point (P) again? Give your answer in nanoseconds.
[<em>Assume that this experiment takes place in deep space so that the effect of gravity is negligible.</em>]
Answer:
The time it will take the particle to pass through point (P) again is 1.639 ns.
Explanation:
F = qvB
Also;
![F = \frac{MV}{t}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7BMV%7D%7Bt%7D)
solving this two equations together;
![\frac{MV}{t} = qVB\\\\t = \frac{MV}{qVB} = \frac{M}{qB}](https://tex.z-dn.net/?f=%5Cfrac%7BMV%7D%7Bt%7D%20%3D%20qVB%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7BMV%7D%7BqVB%7D%20%3D%20%5Cfrac%7BM%7D%7BqB%7D)
where;
m is the mass of electron = 9.11 x 10⁻³¹ kg
q is the charge of electron = 1.602 x 10⁻¹⁹ C
B is the strength of the magnetic field = 3.47 x 10⁻³ T
substitute these values and solve for t
![t = \frac{M}{qB} = \frac{9.11 *10^{-31}}{1.602*10^{-19}*3.47*10^{-3}} = 1.639 *10^{-9} \ s \ = 1.639 \ ns](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7BM%7D%7BqB%7D%20%3D%20%5Cfrac%7B9.11%20%2A10%5E%7B-31%7D%7D%7B1.602%2A10%5E%7B-19%7D%2A3.47%2A10%5E%7B-3%7D%7D%20%3D%201.639%20%2A10%5E%7B-9%7D%20%20%5C%20s%20%5C%20%3D%201.639%20%5C%20ns)
Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer:
83.3kg
Explanation:
GPE = m × g × h
GPE = mass of leopard × 10 × 36m
29988J = 360 × mass
mass = 83.3kg
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K = C + 273, so 27°C = 27+273 = 300 K
1 dg = 100 mg, so 20 dg = 20×100 = 2,000 mg
The atoms which make up the ion are covalently bonded to one another. 19) It is possible for a compound to possess both ionic and covalent bonding. a. If one of the ions is polyatomic then there will be covalent bonding within it.