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2 years ago

11

True or False. A loud noise gives off less energy than a quiet noise.

2 answers:

zvonat [6]2 years ago

5 0

Answer:

true

Explanation:

hope it helped you

Zielflug [23.3K]2 years ago

4 0

Answer:

True

Explanation:

A loud noise cancels out

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Can someone help me please

makvit [3.9K]

Answer:

Time, I believe. Pretty sure it's time lol

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2 years ago

A family communicating using a cell phone at dinner. Using a cell phone is more addicted to technology things than missed opport

tatyana61 [14]

Answer:

The family may be addicted to technology by using their cellphones to communicate is still communicating in a way, but it would be better to use social face- to - face interactions.

Explanation:

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3 years ago

Read 2 more answers

According to Wien's Law, how many times hotter is an object whose blackbody emission spectrum peaks in the blue, at a wave lengt

scZoUnD [109]

Answer:1.55 times

Explanation:

Given

First wavelength $(\lambda _1)=450 nm$

Second wavelength $(\lambda _2)=700 nm$

According wien's diplacement law

$\lambda T=constant$

where $\lambda =wavelength$

T=Temperature

Let $T_1 and T_2$ be the temperatures corresponding to $\lambda _1 & \lambda _2$ respectively.

$\lambda _1\times T_1=\lambda _2\times T_2$

$\frac{T_1}{T_2}=\frac{\lambda _2}{\lambda _1}$

$\frac{T_1}{T_2}=\frac{700}{450}=1.55$

Thus object with $\lambda 450 nm$ is 1.55 times hotter than object with wavelength $\lambda =700 nm$

8 0

3 years ago

At what angle are the electronic and the magnetic wave related in an electromagnetic signal?

VikaD [51]

Answer:

90degrees I'm pretty sure

7 0

2 years ago

A sphere of radius R contains charge Q spread uniformly throughout its volume. Find an expression for the electrostatic energy c

tensa zangetsu [6.8K]

Answer:

$E = \frac{3kQ^2}{5R}$

Explanation:

Let the sphere is uniformly charge to radius "r" and due to this charged sphere the electric potential on its surface is given as

$V = \frac{kq}{r}$

now we can say that

$q = \frac{Q}{\frac{4}{3}\pi R^3} (\frac{4}{3}\pi r^3)$

$q = \frac{Qr^3}{R^3}$

now electric potential is given as

$V = \frac{k\frac{Qr^3}{R^3}}{r}$

$V = \frac{kQr^2}{R^3}$

now work done to bring a small charge from infinite to the surface of this sphere is given as

$dW = V dq$

$dW = \frac{kQr^2}{R^3} dq$

here we know that

$dq = \frac{3Qr^2dr}{R^3}$

now the total energy of the sphere is given as

$E = \int dW$

$E = \int_0^R \frac{kQr^2}{R^3} (\frac{3Qr^2dr}{R^3})$

$E = \frac{3kQ^2}{R^6} (\frac{R^5}{5} - 0)$

$E = \frac{3kQ^2}{5R}$

7 0

3 years ago