Intro to Springs and Hooke’s law

1 answer:

wel3 years ago

6 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.
There is nothing particularly magical about the shape of a coil spring that makes it behave like a spring. The 'springiness', or more correctly, the elasticity is a fundamental property of the wire that the spring is made from. A long straight metal wire also has the ability to ‘spring back’ following a stretching or twisting action. Winding the wire into a spring just allows us to exploit the properties of a long piece of wire in a small space. This is much more convenient for building mechanical devices.

You might be interested in

A bird flies 3.6 km due west and then 1.8 km due north. Another bird flies 1.8 km due west and 3.6 km due north. What is the ang

kondor19780726 [428]

Define unit vectors as follows:
$\hat{i}$ is in the eastern direction.
$\hat{j}$ is in the northern direction.

The position of the first bird is
$\vec{a} = -3.6 \, \hat{i} + 1.8 \, \hat{j}$

The position of the second bird is
$\vec{b} = - 1.8 \, \hat{i} + 3.6 \, \hat{j}$

Let θ = the angle between the net displacement vector for the two birds.
By definition,
$\vec{a} . \vec{b} = |a| |b| cos\theta \\\\ \theta = cos^{-1} ( \frac{\vec{a}.\vec{b}}{|a||b|} )$

$\vec{a}.\vec{b} = (-3.6)(-1.8)+(3.6)(1.8) = 12.96$
$|a| = \sqrt{3.24+12.96} =4.025$
Similarly,
|b| = 4.025

Therefore
$\theta = cos^{-1} \frac{12.96}{4.025^{2}} =36.9^{o}$

Answer: 36.9°

5 0

3 years ago

Identify the prefix that would be used to express<br> 2,000,000,000 bytes of computer memory?

Reika [66]

Answer:

It would be 2 GB

4 0

3 years ago

Read 2 more answers

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

Dmitry [639]

Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$