Please help fast due next period!!!

As part of your daily workout, you lie on your back and push with your feet against a platform attached to two stiff springs arr

Nitella [24]

Answer:

<u><em>1. Part A: 648N</em></u>
<u><em></em></u>
<u><em>2. Part B: 324J</em></u>
<u><em></em></u>
<u><em>3. Part C: 1,296N</em></u>

Explanation:

The magnitude of the force is calculated using the Hook's law:

$|F|=k\Delta x$

You know $\Delta x=0.250m$ but you do not have $k$ .

You can calculate it using the equation for the work-energy for a spring.

The work done to compress the springs a distance $\Delta x$ is:

$Work=\Delta PE=(1/2)k(\Delta x)^2$

Where $\Delta PE$ is the change in the elastic potential energy of the "spring".

Here you have two springs, but you can work as if they were one spring.

You know the work (81.0J) and the length the "spring" was compressed (0.250m). Thus, just substitute and solve for k:

$81.0J=(1/2)k(0.250m)^2\\\\k=2,592N/m$

In reallity, the constant of each spring is half of that, but it is not relevant for the calculations and you are safe by assuming that it is just one spring with that constant.

Now calculate the magnitude of the force:

$|F|=k\Delta x=2,592N/m\times 0.250m=648N$

<u><em>2. Part B. How much additional work must you do to move the platform a distance 0.250 m farther?</em></u>

The additional work will be the extra elastic potential energy that the springs earn.

You already know the elastic potential energy when Δx = 0.250m; now you must calculate the elastic potential energy when Δx = 0.250m + 0.250m = 0.500m.

$\Delta E=(1/2)2,592n/m\times(0.500m)^2=324J$

Therefore, you must do 324J of additional work to move the plattarform a distance 0.250 m farther.

<u><em>What maximum force must you apply to move the platform to the position in Part B?</em></u>

The maximum force is when the springs are compressed the maximum and that is 0.500m

Therefore, use Hook's law again, but now the compression length is Δx = 0.500m

$|F|=k\Delta x=2,592N/m\times 0.500m =1,296N$

8 0

3 years ago

A convex mirror of focal length 34 cm forms an image of a soda bottle at a distance of 18 cm behind the mirror. The height of th

Mademuasel [1]

<h2>Answers:</h2>

In convex mirrors the focus is virtual and the focal distance is negative. This is how the reflected rays diverge and only their extensions are cut at a point on the main axis, resulting in a virtual image of the real object .

<h2 /><h2>a) Position of the soda bottle (the object) </h2><h2 />

The Mirror equation is:

$\frac{1}{f}=\frac{1}{u}+\frac{1}{v}$ (1)

Where:

$f$ is the focal distance

$u$ is the distance between the object and the mirror

$v$ is the distance between the image and the mirror

We already know the values of $f$ and $v$ , let's find $u$ from (1):

$u=\frac{v.f}{v-f}$ (2)

Taking into account the explanation at the beginign of this asnswer:

$f=-34cm$ and $v=-18cm$

The <u>negative signs</u> indicate the <u>focal distance</u> and the <u>distance between the image and the mirror are virtual </u>

Then:

$u=\frac{(-18cm)(-34cm)}{-18cm-(-34cm)}$

$u=38.25cm$ (3) >>>>Position of the soda bottle

<h2>b) magnification of the image </h2><h2 />

The magnification $m$ of the image is given by:

$m=-\frac{v}{u}$ (4)

$m=-\frac{(-18cm)}{38.25cm}$

$m=0.47$ (5)>>>the image is 0.47 smaller than the object

<h2>The fact that this value is positive means the image is upright </h2>

<h2>c) Describe the image </h2><h2 />

According to the explanations and results obtained in the prior answers, the correct option is 9:

<h2>The image is <u>virtual, upright, smaller</u> than the object </h2><h2 /><h2>d) Height of the soda bottle (object) </h2>

Another way to find the magnification is by the following formula:

$m=\frac{h_{i}}{h_{o}}$ (6)

Where:

$h_{i}$ is the image height

$h_{o}$ is the object height

We already know the values of $m$ and $h_{i}$ , let's find $h_{o}$ :

$h_{o}=\frac{h_{i}}{m}$ (7)

$h_{o}=\frac{6.8cm}{0.47}$

$h_{o}=14.46cm$ (8) >>>height of thesoda bottle

6 0

3 years ago

What are em waves expain in details

Tresset [83]

Electromagnetic waves or EM waves are waves that are created as a result of vibrations between an electric field and a magnetic field. ... They are hence known as 'electromagnetic' waves. The electric field and magnetic field of an electromagnetic wave are perpendicular (at right angles) to each other.

Hope this helps.
:)

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3 years ago

Which is NOT part of the equation for angular momentum?

Solnce55 [7]

For the answer to the question above asking <span>Which is NOT part of the equation for angular momentum? The answer is the last one among the given choices above which is D. Force
I hope my answer helped you. Have a nice day!</span>

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Why do some appliances have step-down transformers built in?

elena-s [515]

To increase the current and lower the voltage of the appliance to make it safe for use

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