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3 years ago

Which are ways to improve the design of this experiment? Check all that apply.

Physics

2 answers:

krok68 [10]3 years ago

8 0

Answer:

b d e

Explanation...................................

Arte-miy333 [17]3 years ago

3 0

Answer:

* Experiment with a higher range of materials

* Use a galvanometer.

* Vary in number of coils of the electromagnet

Explanation:

This is an experiment of electricity and magnetism, in general the best way to improve the results are:

* Experiment with a higher range of materials

allowing to know the scope of the initial assumptions

* Use a galvanometer.

The more accurate the readings the error of the derived quantities is the less which will improve the precision of the experiment.

* Vary in number of coils of the electromagnet

Since it allows to have greater magnetic fields and therefore expand the range of measurements

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A cylinder with a mass M and radius r is floating in a pool of liquid. The cylinder is weighted on one end so that the axis of t

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Answer:

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4 years ago

Examine the graph showing the half-life of the radioactive isotope substance x.

kow [346]

Answer:

T_{1/2} = 5,776 10³ years

We see this life time as it is very close to the life time of carbon 14, so it could be used for dating ancient objects

Explanation:

The radioactive decay of described by an equation of the form

N = N₀ $e^{-\lambda t}$

where N is the number of atoms present, N₀ is the number of initial atoms λ is the activity of the material.

The average life time is defined as the time for which the number of remainng atoms is N = N₀ / 2

$T_{1/2}$ = ln 2 /λ

With these expressions, the best method to determine the average life time is to find the activity of the first equation.

For this we look for a point on the graph as accurate as possible,

N₀ = 100, N = 30 and t = 10,000 years

we substitute in the equation

30 = 100 e^{-\lambda 10000}

ln 0.3 = - λ 10000

λ = - (ln 0.3) / 10000

λ = 1.2 10-4

now we can find the average life time

$T_{1/2}$ = ln 2 / 1,2 10-4

T_{1/2} = 5,776 10³ years

We see this life time as it is very close to the life time of carbon 14, so it could be used for dating ancient objects

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4 years ago

As gas rises the temperature gets hotter?<br> True or False

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Answer:

i think its true

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3 years ago

Read 2 more answers

If the f-number of a camera lens is doubled, say from F4.0 to F8.0, that means the diameter of the lens aperture is

yan [13]

Answer:

The diameter is halved or reduced by a factor of 2

Explanation:

Because

f number = F/D

That is f number is inversely proportional to diameter so if it is doubled diameter is halved

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3 years ago

When a low-pressure gas of hydrogen atoms is placed in a tube and a large voltage is applied to the end of the tube, the atoms w

FromTheMoon [43]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The value of n is $n =7$

Explanation:

From the question we are told that

The value of m = 2

For every value of $m, n = m+ 1, m+2,m+3,....$

The modified version of Balmer's formula is $\frac{1}{\lambda} = R [\frac{1}{m^2} - \frac{1}{n^2} ]$

The Rydberg constant has a value of $R = 1.097 *10^{7} m^{-1}$

The objective of this solution is to obtain the value of n for which the wavelength of the Balmer series line is smaller than 400nm

For m = 2 and n =3

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{3^2} ]$

$\lambda = \frac{1}{1523611.1112}$

$\lambda = 656nm$

For m = 2 and n = 4

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{4^2} ]$

$\lambda = \frac{1}{2056875}$

$\lambda = 486nm$

For m = 2 and n = 5

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{5^2} ]$

$\lambda = \frac{1}{2303700}$

$\lambda = 434nm$

For m = 2 and n = 6

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{6^2} ]$

$\lambda = \frac{1}{2422222}$

$\lambda = 410nm$

For m = 2 and n = 7

The wavelength is

$\frac{1}{\lambda } = (1.097 * 10^7)[\frac{1}{2^2} - \frac{1}{7^2} ]$

$\lambda = \frac{1}{2518622}$

$\lambda = 397nm$

So the value of n is 7

7 0

3 years ago