Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
Answer:
Explanation:
Acceleration is equal to the change in velocity over the change in time, or
where the change in velocity is final velocity minus initial velocity. Filling in:
Note that I made the backward velocity negative so the forward velocity in our answer will be positive.
Simplifying that gives us:
and then isolating the final velocity, our unknown:
3.0(6.0) = v + 3.0 and
3.0(6.0) - 3.0 = v and
18 - 3.0 = v so
15 m/s = v and because this answer is positive, that means that the car is no longer rolling backwards (which was negative) but is now moving forward.
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: final Velocity v = 10.2m/s
Explanation:
Final speed v(t) is given as
v(t) = u + at .......1
Where; u = the initial speed
a = acceleration
t = time taken
The total distance travelled d is given as
d = ut + 1/2(at^2)
Given
d = 5.0m
u = 2.0m
a = g = 10m/s2 (acceleration due to gravity)
Substituting into the equation above we have
5 = 2t + 5t^2
5t^2 +2t -5 = 0
Applying the quadratic formula. We have;
t = 0.82s & t = -1.22s
t cannot be negative
t = 0.82s
From equation 1 above
v = 2.0m/s + 10(0.82)m/s
v = 10.2m/s
Answer:
I think D it could maybe B