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3 years ago

Someone please help!

Physics

2 answers:

Schach [20]3 years ago

8 0

Answer:

a. 1.81

b. using the lap button feature

Explanation:

a. avg of the times

Rus_ich [418]3 years ago

3 0

I got it I got it I got it I got it I got it yea I don’t got it

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3 years ago

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4 years ago

At what displacement of a sho is the energy half kinetic and half potential? what fraction of the total energy of a sho is kinet

expeople1 [14]

As we know that KE and PE is same at a given position

so we will have as a function of position given as

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

also the PE is given as function of position as

$PE = \frac{1}{2}m\omega^2x^2$

now it is given that

KE = PE

now we will have

$\frac{1}{2}m\omega^2(A^2 - x^2) = \frac{1}{2}m\omega^2x^2$

$A^2 - x^2 = x^2$

$2x^2 = A^2$

$x = \frac{A}{\sqrt2}$

so the position is 0.707 times of amplitude when KE and PE will be same

Part b)

KE of SHO at x = A/3

we can use the formula

$KE = \frac{1}{2}m\omega^2(A^2 - x^2)$

now to find the fraction of kinetic energy

$f = \frac{KE}{TE} = \frac{A^2 - x^2}{A^2}$

$f = \frac{A^2 - (\frac{A}{3})^2}{A^2}$

$f_k = \frac{8}{9}$

now since total energy is sum of KE and PE

so fraction of PE at the same position will be

$f_{PE} = 1 - f_k$

$f_{PE} = 1 - (8/9) = 1/9$

7 0

3 years ago

The histogram below shows the number of downloads of a song over time.

Allisa [31]

Given data:

It is a graphical display where the data is grouped in to ranges

A diagram consists rectangles, whose area is proportional to frequency of a variable and whose width is equal to the class interval.

It is an accurate representation of the distribution of numerical data.

From Figure:

Each box in the graph (small rectangle box) is assumed to be one download. So, in the graph the time between 8 p.m to 9 p.m, the number of downloads are 8.75 approximately (because the last box is incomplete, therefore 8 complete boxes and 9th is more than half).

So, We conclude that the total number of downloads are approximately 9 in the time span of 8 p.m. to 9 p.m.

7 0

3 years ago