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Serjik [45]
3 years ago
5

is spring constant the ratio of external force to displacement from mean position or the ratio of restoring force to displacemen

t from mean position?​
Physics
1 answer:
Talja [164]3 years ago
8 0
The magnitude of the force required to change the length of a spring-like object is directly proportional to the spring constant and the displacement of the spring. Elastic potential energy is directly proportional to the square of the change in length and the spring constant


I think….?
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Three balls, with masses of 3m,2m and m, are fastened to a massless rod of length L. The rotational inertias about the ledt
Oduvanchick [21]

I = MR^2

The Attempt at a Solution:::

I total = (3M)(0)^2 + (2M)(L/2)^2 + (M)(L)^2

I total = 3ML^2/2

It says the answer is 3ML^2/4 though.

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Which of the following statements about psychodynamic perspective is not true?
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There are NO true statements on that list of choices.
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8 0
3 years ago
You are holding a positive charge and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east. Wh
lara [203]

If I hold a positive charge in my hand and there are positive charges of equal magnitude 1 mm to your north and 1 mm to your east then the direction of the force on the charge I am holding is towards the north-east direction.

Reasoning:

It is given that there is a positive charge in my hand. There are two more positive charges with the same magnitude. One is 1 mm far towards the east, and the other one is 1 mm far towards the north. It is required to find the direction of the force acting on the charge in my hand.

Let the magnitude of the charge in my hand is Q, and the magnitude of the other charges is q.

Thus the electric force applied on the charge in my hand due to each other is,

F=\frac{kQq}{r^2}

Here k is the Coulomb constant, and r is the distance between the charges.

It is also known that the force on a positive charge due to another positive charge is acted outwards.

Thus, the force on the charge due to the charge on the east is,

\vec{F_1}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{i}

And the force on the charge due to the charge on the north is,

\vec{F_2}=\frac{kQq}{( 10^{-3}\text{ m})^2}\hat{j}

As the forces are equal in magnitude and one is perpendicular to the other, thus the net force will be acted at an angle of 45^\circ from the north or from the north direction.

Thus the net force is acting in the north-east direction.

Learn more about the direction of the force here,

brainly.com/question/2037071

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3 0
2 years ago
How is energy and work related.<br>​
Anna007 [38]

Answer:

Varies

Explanation:

They both relate to the process of doing something.

5 0
3 years ago
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