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.
The shape of a liquid can change because the atoms in it are not close together to form a solid, they flow freely.
Answer:
7.6 s
Explanation:
Considering kinematics formula for final velocity as

Where v and u are final and initial velocities, a is acceleration and s is distance moved.
Making v the subject then

Substituting 8.8 m/s for u, 138 m for s and 2.45 m/s2 for a then

Also, v=u+at and making t the subject of the formula

Substituting 27.45 m/s for v, 8.8 m/s for u and 2.45 m/s for a then

Therefore, it needs 7.6 seconds to travel
Answer:
4.384 * 10^13
Explanation:
Given the expression :
[(6.67 * 10^-11) * (1.99 * 10^30)] ÷ [(1.74*10^3)*(1.74*10^3)]
Applying the laws of indices
[(6.67 * 1.99) *10^(-11 + 30)] ÷ [(1.74 * 1.74) * 10^3+3]
13.2733 * 10^19 ÷ 3.0276 * 10^6
(13.2733 / 3.0276) * 10^(19 - 6)
4.3840996 * 10^13
= 4.384 * 10^13
Answer:
v₂ = 5.7 m/s
Explanation:
We will apply the law of conservation of momentum here:

where,
Total Initial Momentum = 340 kg.m/s
m₁ = mass of bike
v₁ = final speed of bike = 0 m/s
m₂ = mass of Sheila = 60 kg
v₂ = final speed of Sheila = ?
Therefore,

<u>v₂ = 5.7 m/s </u>