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:
The answer is: To accelerate an object <u>the force applied to the object</u> has to increase.
Explanation:
the acceleration of an object <u>increases with increased force</u> and <u>decreases with increased mass.</u>
Answer:
6 atm.
Explanation:
Let the mass of both be m
Then moles of He = m/ 4
Moles of Ne = m/ 20
mole fraction of He = Moles of He/ Total moles = m/4/ (m/4 + m/20) = 0.25 m/0.3m = 0.83
Pressure of He = Mole fraction×total pressure = 0.83 × 6 atm = 5 atm
Explanation:
Mass of the astronaut, m₁ = 170 kg
Speed of astronaut, v₁ = 2.25 m/s
mass of space capsule, m₂ = 2600 kg
Let v₂ is the speed of the space capsule. It can be calculated using the conservation of momentum as :
initial momentum = final momentum
Since, initial momentum is zero. So,



So, the change in speed of the space capsule is 0.17 m/s. Hence, this is the required solution.