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2 years ago

The amplitude of a mechanical wave shows What?

Physics

1 answer:

Anuta_ua [19.1K]2 years ago

5 0

I’m not sure what you are asking

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I cant figure this out for so long. thank you in advance whoever helps me

kvasek [131]

Answer:

a magnet is a material with a North and South pole

8 0

2 years ago

An experiment on the earth's magnetic field is being carried out 1.00 m from an electric cable. What is the maximum allowable cu

Whitepunk [10]

Answer:

The allowed current in the cable is 1.15 A.

Explanation:

Given that,

Distance = 1.00 m

Suppose the magnetic field is $2.3\times10^{-5}\ T$ and if the experiment is to be accurate to 1.0 %

We need to calculate the current

Using formula of magnetic field

$B=\dfrac{\mu_{0}I}{2\pi r}$

$I=\dfrac{B\times2\pi r}{\mu_{0}}$

Put the value into the formula

$I=\dfrac{2.3\times10^{-5}\times2\times\pi\times1.00}{4\pi\times10^{-7}}$

$I=115\ A$

If the experiment is to be accurate to 1.0%

Then,

We need to calculate the allowed current in the cable

$I'=(1.0%)\times I$

$I'=\dfrac{1.0}{100}\times115$

$I'=1.15\ A$

Hence, The allowed current in the cable is 1.15 A.

4 0

3 years ago

Suppose the tourist in question #1 instead threw the rock with an initial velocity of 8.0

Harman [31]

Answer:

Explanation:u=8m/s, a=8m/s*s, t=4s

Using v= u + at

v=8+-8*4

v=-24m/s

Displacement=s

Using s=½(u+v)t

s=½(8-24)4

s=½*-16*4

s=-8*4

s=32m( you can solve without observing the negative sign)

8 0

3 years ago

It takes 3.0 eV of energy to excite an electron in a 1-dimensional infinite well from the ground state to the first excited stat

valentina_108 [34]

Answer:

0.614 nm

Explanation:

Energy of the nth state of one dimensional infinite wall is,

$E=\frac{n^{2} h^{2} }{8mL^{2} }$

Given the energy to excite an electron from ground state to first excited state is,

$\Delta E=3eV\\\Delta E=3(1.6\times10^{-19})J$

And the Plank's constant is, $h=6.626\times10^{-34}Js$

Mass of electron, $m=9.1\times10^{-31}kg$

Now the energy will of a 1 dimensional infinite wall which excite an electron from ground state to first excited state will be,

$\Delta E=\frac{2^{2} h^{2} }{8mL^{2} } -\frac{1^{2} h^{2} }{8mL^{2} }$

Put all the variables in above equation and rearrange it for L.

$L^{2} =\frac{3((6.626\times10^{-34}Js)^{2} )}{8\times9.1\times10^{-31}kg\times3(1.6\times10^{-19})J} \\L^{2} =0.376922012\times10^{-18} m^{2}\\ L=0.6139\times10^{-9}m\\ L=0.614nm$

Therefore the width of the box is 0.614 nm.

6 0

3 years ago

What two things must a measurement include?

astraxan [27]

The number and units. An example would be 3cm or 4kg.

7 0

2 years ago

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