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3 years ago

4. A spring is stretched 0.5 m from equilibrium. The force constant (k) of the

Physics

2 answers:

makkiz [27]3 years ago

8 0

Answer:

31.25

Explanation:

The formula for the potential energy of a spring is $\frac{1}{2}kx^2$ , where x is the distance in meters the spring is stretched. Plugging in the numbers that you are given, you get $\frac{1}{2}\cdot 250\cdot 0.25=31.25$ . Hope this helps!

hodyreva [135]3 years ago

8 0

31.25 hope this helps:)

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Answer:

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a body dropped from a height reaches a velocity of 13m/s just before touching the ground. What is the initial height of the ball

SashulF [63]

Hi there!

We can use the following (derived) equation to solve for the final velocity given height:

vf = √2gh

We can rearrange to solve for height:

vf² = 2gh

vf²/2g = h

Plug in the given values (g = 9.81 m/s²)

(13)²/2(9.81) = 8.614 m

We can calculate time using the equation:

vf = vi + at, where:

vi = initial velocity (since dropped from rest, = 0 m/s)

a = acceleration (in this instance, due to gravity)

Plug in values:

13 = at

13/a = t

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2 years ago

An object in circular motion has velocity that is constantly changing. The direction of the acceleration is

Keith_Richards [23]

<h2>Answer: Toward the center of the circle.</h2>

This situation is characteristic of the uniform circular motion , in which the movement of a body describes a circumference of a given radius with constant speed.

However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

Let's say $\vec{V}$ is the velocity vector, whose direction is perpendicular to the radius $r$ of the trajectory, therefore the acceleration $\vec{a}$ is directed toward the center of the circumference.