Identify which gas had the greatest volume change.

What is a charge? how objects can be charged? short answer pls

nevsk [136]

Answer:

An electrical charge is created when electrons are transferred to or removed from an object. Because electrons have a negative charge, when they are added to an object, it becomes negatively charged. When electrons are removed from an object, it becomes positively charged.

Explanation:

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8 0

2 years ago

Web Spiders and Oscillations All spiders have special organs that make them exquisitely sensitive to vibrations. Web spiders det

chubhunter [2.5K]

Complete Question

The complete quetion is shown on the first uploaded image

Answer:

Explanation:

From the question we are told that

The mass of the fly is $m_f = 11 mg = 11*10^{-3} g = 1.1*10^{-5} \ kg$

The extension of the web is $e= 4.00 \ mm = 0.004 \ m$

The spring constant is mathematically evaluated as

$k = \frac{mg}{e}$

substituting values

$k = \frac{1.1 *10^{-5} *9.8}{0.004}$

$k = 0.027 \ N/m$

The frequency of vibration is

$f = \frac{1}{2 \pi} \sqrt{\frac{k}{m} }$

substituting values

$f = \frac{1}{2 * 3.142 } \sqrt{\frac{0.027}{1.1*10^{-5}} }$

$f = 7.9 Hz$

8 0

3 years ago

Model a hydrogen atom as a three-dimensional potential well with Uo = 0 in the region 0 < x a. 283 eV <br> b. 339 eV <br> c.

denis23 [38]

This question is incomplete, the complete question is;

Model a hydrogen atom as a three-dimensional potential well with U₀ = 0 in the region 0 < x < L, 0 < y < L and 0 < z < L, and infinite otherwise, with L = 1.0 × 10⁻¹⁰ m.

Which of the following is NOT one of the lowest three energy levels of an electron in this model?

a. 283 eV

b. 339 eV

c. 113 eV

d. 226 eV

Answer:

the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

Explanation:

Given the data in the question;

Three dimension cube or particle in a cubic box

the energy value is given by;

$E_{nx,ny,nz$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

where h" = h/2π and h is Planck's constant ( 6.626 × 10⁻³⁴ m² kg / s )

m is mass of electron ( 9.1 × 10⁻³¹ kg )

l is length of side of box ( 1.0 × 10⁻¹⁰ m )

for ground level ( $n_x = n_y = n_z = 1$ )

so

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h"² / 2ml²

since h" = h/2π

$( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

so we substitute

$E_{111$ = ( 1² + 1² + 1² ) × [ π²( 6.626 × 10⁻³⁴ )² ] / [ (2π)² × 2 × 9.1 × 10⁻³¹ kg × ( 1.0 × 10⁻¹⁰)² ]

$E_{111$ = 3 × [ (4.333188779 × 10⁻⁶⁶) / ( 7.185072 × 10⁻⁴⁹ ) ]

$E_{111$ = 3 × [ 6.03082165 × 10⁻¹⁸ ]

Now, we know that electric charge = 1.602 x 10⁻¹⁹

so

$E_{111$ = 3 × [ (6.03082165 × 10⁻¹⁸) / (1.602 x 10⁻¹⁹) ]

$E_{111$ = 3 × [ 37.645578 ]

$E_{111$ = 112.9 ≈ 113 eV

$E_{211$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{211$ = ( 1² + 1² + 2² ) × [ 37.645578 ]

$E_{211$ = 6 × [ 37.645578 ]

$E_{211$ = 225.87 ≈ 226 eV

$E_{221$ = $( n_x^2 + n_y^2 + n_z^2 )$ × π²h² / (2π)²2ml²

we substitute

$E_{221$ = ( 2² + 2² + 1² ) × [ 37.645578 ]

$E_{211$ = 9 × [ 37.645578 ]

$E_{211$ = 338.8 ≈ 339 eV

Therefore, the lowest three energy are; 113 eV, 225 eV, and 339 eV.

Hence Option a) 283 eV is not among the three lowest energy

8 0

2 years ago

A flutist assembles her flute in a room where the speed of sound is 342m/s . When she plays the note A, it is in perfect tune wi

USPshnik [31]

Answer:

a.3Hz

b.0.0034m

Explanation:

First, we know the flute is an open pipe, because open pipe as both end open and a close organ pipe as only one end close.

The formula relating the length and he frequency is giving as

$f=\frac{nv}{2l}\\$ .

a.we first determine the length of the flute at the fundamental frequency i.e when <em>n</em>=1 and when the speed is in the 342m/s

Hence from

$f=\frac{nv}{2l}\\\\l=\frac{342}{2*440}\\ l=0.389m\\$ .

since the value of the length will remain constant, we now use the value to determine the frequency when the air becomes hotter and the speed becomes 345m/s.

$f=\frac{nv}{2l} \\f=\frac{345}{2*0.389}\\f=443.4Hz$

Hence the require beat is

$B=/f_{1}-f_{2}/\\B=/440-443/\\B=3Hz$ .

b. since the length is dependent also on the speed and frequency, we determine the new length when she plays with a fundamental frequency when the speed of sound is 345m/s

using the formula

$L_{new}=\frac{v}{2f}\\\\L_{new}=\frac{345}{2*440}\\\\L_{new}=0.39204$