A2.31 kg rope is stretched between supports 10.4 m apart. if one end of the rope is tweaked, how long will it take for the resulting disturbance to reach the other end? assume that the tension in the rope is 55.7 n.
a. 0.615 s
b. 0.664 s
c. 0.720 s
d. 0.699 s

t = 0.657 s

Explanation:

First, let's use the appropiate equations to solve this:

V = √T/u

This expression gives us a relation between speed of a disturbance and the properties of the material, in this case, the rope.

Where:

V: Speed of the disturbance

T: Tension of the rope

u: linear density of the rope.

The density of the rope can be calculated using the following expression:

u = M/L

Where:

M: mass of the rope

L: Length of the rope.

We already have the mass and length, which is the distance of the rope with the supports. Replacing the data we have:

u = 2.31 / 10.4 = 0.222 kg/m

Now, replacing in the first equation:

V = √55.7/0.222 = √250.9

V = 15.84 m/s

Finally the time can be calculated with the following expression:

V = L/t > t = L/V

Replacing:

t = 10.4 / 15.84

t = 0.657 s

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

the closer the planet is to the sun, the larger the gravity