How does Earth's surface and the structures on the surface change as a result of an earthquake?
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The magnetic force experienced by the proton is given by
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
the angle between the direction of v and B. Since the proton moves perpendicularly to the magnetic field, this angle is 90 degrees, so 
and we can ignore it in the formula.
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
where q is the proton charge, v its velocity, B the magnitude of the magnetic field and
For Netwon's second law, the force is also equal to the proton mass times its acceleration:
So we have
from which we can find the magnitude of the field:
Answer:
x = 240 m
Explanation:
This is a kinematics exercise
Let's fix our frame of reference on car A
x = x₀ₐ+ v₀ₐ t + ½ aₐ t²
the initial position of car a is zero
x = 0 + v₀ₐ t + ½ 0.8 t²
for car B
x = x_{ob} + v_{ob} t - ½ a_b t²
car B's starting position is 30 m
x = 30 + v_{ob} t - ½ 0.4 t²
at the point where they meet, the position of the two vehicles is the same
0 + v₀ₐ t + ½ 0.8 t² = 30 + v_{ob} t - ½ 0.4 t²
let's reduce the speeds to the SI system
v₀ₐ = 14.4 km / h (1000 m / 1 km) (1h / 3600s) = 4 m / s
v_{ob} = 23.4 km / h = 6.5 m / s
4 t + 0.4 t² = 30 + 6.5 t - 0.2 t²
0.2 t² - 2.5 t - 30 = 0
t² - 12.5 t - 150 = 0
we solve the quadratic equation
t =
t =
t₁ = 20 s
t₂ = -7.5 s
time must be a positive quantity so the correct result is t = 20 s
let's look for the distance
x = 4 t + ½ 0.8 t²
x = 4 20 + ½ 0.8 20²
x = 240 m