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:
80%
Explanation:
Efficiency = Power output / Power input × 100 %
To calculate efficiency we need to find power output of electric pump.
We can use,
Work done = Energy change
Work done per second = Energy change per second
Work done per second = Power
Therefore, Power = Energy change per second
= Change in potential energy of water per second
=mgh / t
= 200× 10×6 / 10
= 1200 W = 1.2 kW
Now use the first equation to find efficiency,
Efficiency = × 100%
= 80 %
Answer:
I believe the answer is B.
Explanation:
Newton's First Law of Gravity states, "The greater the weight (or mass) of an object, the more inertia it has. Heavy objects are harder to move than light ones because they have more inertia.
"
B. Elastic potential to kinetic energy
The elastic potential energy in the slingshot will be transferred to the stone as kinetic energy as the stone is launched.
The given question is incomplete. The complete question is as follows.
A parallel-plate capacitor has capacitance = 8.50 pF when there is air between the plates. The separation between the plates is 1.00 mm.
What is the maximum magnitude of charge that can be placed on each plate if the electric field in the region between the plates is not to exceed V/m?
Explanation:
It is known that relation between electric field and the voltage is as follows.
V = Ed
Now,
Q = CV
or, Q =
Therefore, substitute the values into the above formula as follows.
Q =
=
=
Hence, we can conclude that the maximum magnitude of charge that can be placed on each given plate is .