What type of force pulls in two opposite directions?a. tension (associated with normal faults)
b. normal (associated with longitudinal faults)
c. transform (associated with transform faults)
d. compression (associated with reverse faults)
Which is not a type of fault?
a. normal
b. reverse
c. diagonal
d. transform

Answers

Answer 1

Answer:

Answer

a. tension (associated with normal faults)

c. diagonal

Explanation

A pull of spring or of string on both ends of an object is called tension. So for the question one, the answer is tension (associated with normal faults)

There are three types of faults. Faults are produced by stress or strain by moving plates. These faults are: normal faults, reverse faults and transcurrent or Strike-slip. Strike-slip faults can also be called transform fault. The answer to the second question is c. diagonal.

Answer 2

Answer:

Answer:

Tension Force (associated with normal faults)

Explanation:

As we know that tension force is an internal force between the molecules which opposes the tendency of molecules to separate out.

The tension force of string is always in opposite directions at two ends of string and it is always along the length

Answer:

Diagonal

Explanation:

As we know that there are four types of vaults

1) Normal vaults

2) Longitudinal vaults

3) Reverse vaults

4) Transform vaults

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If the tension in a rope is increased what will happen to the speed of a wave traveling through the rope

Answers

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A small 175-g ball on the end of a light string is revolving uniformly on a frictionless surface in a horizontal circle of diameter 1.0 m. The ball makes 2.0 revolutions every 1.0 s. What are the magnitude and direction of the acceleration of the ball? What's the tension in the string?

Answers

Answer :

Explanation :

It is given that:

mass of the ball, $m=175\ g=0.175\ Kg$

Radius of circle, $r=(diameter)/(2)=0.5\ m$

The ball makes 2.0 revolutions every 1.0 s. So, angular speed is $\omega=4\pi\ radian/sec$

Since, it is moving in circular path so centripetal acceleration will act here.

So, centripetal acceleration $\alpha$ = $m\omega^2r$

$\alpha=0.175\ Kg* (4\pi)^2* 0.5$

So, $\alpha=13.803\ m/s^2$

Hence, the acceleration is $13.803\ m/s^2$ and it is directed towards the center of rotation.

Tension is a force which is given by :

$F=ma$

$F=0.175\ Kg*13.803\ m/s^2$

$F=2.415\ N$

This is the required answer.

For circular motion.

Centripetal acceleration = mv²/r = mω²r

Where v = linear velocity, r = radius = diameter/2 = 1/2 = 0.5m

m = mass = 175g = 0.175kg.

Angular speed, ω = Angle covered / time

   = 2 revolutions / 1 second

   = 2 * 2π radians / 1 second

   = 4π radians / second

Centripetal Acceleration = mω²r = 0.175*(4π)² * 0.5 Use a calculator

   ≈13.817 m/s²

The magnitude of acceleration ≈13.817 m/s² and it is directed towards the center of rotation.

Tension in the string = m*a

   = 0.175*13.817

   = 2.418 N

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