Answer:
x = A sin ω t describes the displacement of the particle
v = A ω cos ω t
a = -A ω^2 sin ω t
a (max) = -A ω^2 is the max acceleration (- can be ignored here)
ω = (K/ m)^1/2 for SHM
F = - K x^2 restoring force of spring
K = 4.34 / .0745^2 = 782 N / m
ω = (782 / .297)^1/2 = 51.3 / sec
a (max) = .0745 * 782 / .297 = 196 m / s^2
Answer:
Woke done, W = 4156.92 Joules
Explanation:
The work done by the force can be calculated as :
is the angle between force and the displacement
It is assumed to find the work done for the given parameters i.e.
Force, F = 30 N
Distance travelled, s = 160 m
Angle between force and displacement,
Work done is given by :
W = 4156.92 Joules
So, the work done by the object is 4156.92 Joules. Hence, this is the required solution.
Answer:
b) q large and m small
Explanation:
q is large and m is small
We'll express it as :
q > m
As we know the formula:
F = Eq
And we also know that :
F = Bqv
F =
Bqv =
or Eq =
Assume that you want a velocity selector that will allow particles of velocity v⃗ to pass straight through without deflection while also providing the best possible velocity resolution. You set the electric and magnetic fields to select the velocity v⃗ . To obtain the best possible velocity resolution (the narrowest distribution of velocities of the transmitted particles) you would want to use particles with q large and m small.
Answer
a) For the rock
b) for maximum range
c) The value of θ is the same on every planet as g divides out.