Ans: Time <span>taken by a pulse to travel from one support to the other
= 0.348s</span>
Explanation:First you need to find out the speed of the wave.
Since
Speed = v =

Where
T = Tension in the cord = 150N
μ = Mass per unit length = mass/Length = 0.65/28 = 0.0232 kg/m
So
v =

= 80.41 m/s
Now the time-taken by the wave = t = Length/speed = 28/80.41=
0.348s
Answer:
<em>The force is now 9 times the original force</em>
Explanation:
<u>Coulomb's Law
</u>
The electrostatic force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's formula is:

Where:

q1, q2 = the particles' charge
d= The distance between the particles
Suppose the distance is reduced to d'=d/3, the new force F' is:




The force is now 9 times the original force
Answer:
0.853 m/s
Explanation:
Total energy stored in the spring = Total kinetic energy of the masses.
1/2ke² = 1/2m'v².................... Equation 1
Where k = spring constant of the spring, e = extension, m' = total mass, v = speed of the masses.
make v the subject of the equation,
v = e[√(k/m')].................... Equation 2
Given: e = 39 cm = 0.39 m, m' = 0.4+0.4 = 0.8 kg, k = 1.75 N/cm = 175 N/m.
Substitute into equation 2
v = 0.39[√(1.75/0.8)
v = 0.39[2.1875]
v = 0.853 m/s
Hence the speed of each mass = 0.853 m/s
Answer:
The difference between frictionless ramp and a regular ramp is that on a frictionless ramp the ball cannot roll it can only slide, but on a regular ramp the ball can roll without slipping.
We will use conversation of energy.

Note that initial potential energy is zero because the ball is on the bottom, and the final kinetic energy is zero because the ball reaches its maximum vertical distance and stops.
For the ball B;


The initial velocities of the balls are equal. Their maximum climbing point will be proportional to their final potential energy. Since their initial kinetic energies are equal, their final potential energies must be equal as well.
Hence, both balls climb the same point.
Explanation: