Solution :
Given :
Mass attached to the spring = 4 kg
Mass dropped = 6 kg
Force constant = 100 N/m
Initial amplitude = 2 m
Therefore,
a).
= 10 m/s
Final velocity, v at equilibrium position, v = 5 m/s
Now,
A' = amplitude = 1.4142 m
b).
m' = 2m
Hence,
c).
Therefore, factor
Thus, the energy will change half times as the result of the collision.
Answer:
Explanation:
When a charged particle moves perpendicularly to a magnetic field, the force it experiences is:
where
q is the charge
v is its velocity
B is the strength of the magnetic field
Moreover, the force acts in a direction perpendicular to the motion of the charge, so it acts as a centripetal force; therefore we can write:
where
m is the mass of the particle
r is the radius of the orbit of the particle
The equation can be re-arranges as
where in this problem we have:
is the magnitude of the charge of the electron
is the strength of the magnetic field
The beam penetrates 3.45 mm into the field region: therefore, this is the radius of the orbit,
is the mass of the electron
So, the electron's speed is
Answer:
T = 676 N
Explanation:
Given that: f = 65 Hz, L = 2.0 m, and ρ = 5.0 g = 0.005 kg
A stationary wave that is set up in the string has a frequency of;
f =
⇒ T = 4M
Where: t is the tension in the wire, L is the length of the wire, f is the frequency of the waves produced by the wire and M is the mass per unit length of the wire.
But M = L × ρ = (2 × 0.005) = 0.01 kg/m
T = 4 × ×
× 0.01
= 4 × 4 ×4225 × 0.01
= 676 N
Tension of the wire is 676 N.