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;

solving this two equations together;

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

Therefore, the time it will take the particle to pass through point (P) again is 1.639 ns.
Answer:
36.408cm3
Explanation:
Since we acknowledge that density is d= m/v, once we switch it up to maintain v as the number to be found it will change to v=m/d. Therefore, 275.32/7.562 is 36.408 and the unit is cm cube!
Hope that helped!!
Answer:
The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking.
The amount of calories contained in the food we eat every day must represent the amount of calories eliminated by the body in that time to have a steady weight.
Explanation:
The average number of calories needed daily represents the average quantity of calories eliminated by human body due to metabolism and must be compensated by eating and drinking. If total quantity of calories in the food we consume every day is higher that the average number of calories needed daily, then weight increases by fat accumulation.
Answer:
Explanatioyour answers look right, but if there has , has to be another answer its a , but your answers are right
Answer:1200
Explanation:
Given data
Upper Temprature
Lower Temprature 
Engine power ouput
Efficiency of carnot cycle is given by





rounding off to two significant figures
