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2 years ago

All known frequencies of the visible spectrum are..

Physics

1 answer:

crimeas [40]2 years ago

7 0

Explanation:

$b. \: light$

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How does Physics help you as a student?

Sonja [21]

Answer:

The goal of physics is to understand how things work from first principles. ... Courses in physics reveal the mathematical beauty of the universe at scales ranging from subatomic to cosmological. Studying physics strengthens quantitative reasoning and problem solving skills that are valuable in areas beyond physics

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2 years ago

Read 2 more answers

1. What are three questions you should ask before joining a club?

lara [203]

Answer:

what time does it start.

what do I need to join.

what are your expectations.

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3 years ago

The boundary between tectonic plates that are moving toward each other

Alisiya [41]

Answer:

A divergent boundary is the answer

Explanation:

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3 years ago

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$

Substituting our values:

$F=\frac{(73.14m/s)}{(9.81m/s^2)(30\times10^{-3}s)}W=248.5W N$

Where the average force exerted has been written it terms of the tennis ball's weight W.

8 0

2 years ago

Normalize the equations

tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

$\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k$

Part b)

$\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1$

$\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}$

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3 years ago