Answer:
The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false.
Explanation:
The equation for the magnetic force on a charge q moving at velocity v on a magnetic field B is given by the (vectorial) Lorentz Force Law 
From it we can clearly see that the <em>magnitude of the magnetic force </em>exerted on the particle is <em>proportional to the magnitude of the charge q and to the speed v of the particle</em>, and that it is also <em>perpendicular to the particle's velocity</em>. This means that at each instant it moves perpendicularly to the force, so <em>the work done by the magnetic force on the particle is zero</em>.
The statement "if the magnetic force is always perpendicular to the velocity, the path of the particle is a straight line" is false not only for this but for any force, a force always perpendicular to a velocity will curve the trajectory.
Answer with Explanation:
We are given that


Differentiate x and y w.r.t t





Substitute t=1


Magnitude of velocity=

Hence, the magnitude of the missile's velocity=16.49 m/s


Substitute t=1



Hence, the magnitude of acceleration when t=1 s=
Answer:
the distance in meters traveled by a point outside the rim is 157.1 m
Explanation:
Given;
radius of the disk, r = 50 cm = 0.5 m
angular speed of the disk, ω = 100 rpm
time of motion, t = 30 s
The distance in meters traveled by a point outside the rim is calculated as follows;

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m
They are trailing the same speed as it states in the question they are heeding toward each other a 70 mph <span />