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PilotLPTM [1.2K]
3 years ago
14

What happens when you push a spring? How is this different than pulling it? (Hooke’s Law)

Physics
2 answers:
Yuki888 [10]3 years ago
7 0

A spring is an object that can be deformed by a force and then return to its original shape after the force is removed.Springs come in a huge variety of different forms, but the simple metal coil spring is probably the most familiar. Springs are an essential part of almost all moderately complex mechanical devices; from ball-point pens to racing car engines.

Furkat [3]3 years ago
4 0

When you push a spring you are putting force on it, or rather forcing the spring onto its self. This is different than pulling the spring because when you push it and let go it will spring but if you pull it its would not move as far as it would if you push it. Pulling the spring would most likely deform the spring will pushing it would cause it to spring back into place.

<u>Hooke's law states that the force that extends or compresses a spring by some distance scales linearly with respect to that distance.</u>

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If Ms. Ramirez drops a watermelon from the top of the second-floor fire escape outside her room, and it takes 1.5 seconds to hit
mamaluj [8]

The velocity of the watermelon as it hits the ground is 14.715 m/s.

Given the data in the question;

Since the Watermelon was dropped from rest,

  • Initial velocity of the Watermelon;u = 0m/s
  • Time taken for the Watermelon to hit the ground; t = 1.5s
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Now, we determine the velocity of the watermelon as it hits the ground which is the final velocity.

From the First Equation of Motion:

v = u + at

Now, under gravity, acceleration ''a" will Acceleration due to gravity;g = 9.81m/s^2, "v" is the final velocity, "u" is the initial velocity and "t" is the time taken.

So, we substitute our values into the equation

v = 0m/s\ +\ ( 9.81m/s^2\ *  1.5s )\\\\v = ( 9.81m/s^2\ *  1.5s )\\\\v = 14.715 m/s

Therefore, the velocity of the watermelon as it hits the ground is 14.715 m/s

Learn more; brainly.com/question/13275688

7 0
3 years ago
A jet airliner moving initially at 693 mph (with respect to the ground) to the east moves into a region where the wind is blowin
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Answer:

The new speed of the aircraft with respect to the ground is 1414.3 mph.

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Angle = 37°

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v=v_{a}+v_{w}

We need to calculate the new speed of the aircraft with respect to the ground

Using formula for velocity

v=\sqrt{(v_{a})^2+(v_{w})^2+2v_{a}v_{w}\cos\theta}

Put the value into the formula

v=\sqrt{(693)^2+(798)^2+2\times693\times798\cos37}

v=1414.3\ mph

Hence, The new speed of the aircraft with respect to the ground is 1414.3 mph.

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Answer:

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Explanation

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