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1 year ago

9

Random errors can be reduced by taking repeated measurements.Error and uncertainty are interchangeable words that describe the s

ame concept.are these statements true or false

1 answer:

fiasKO [112]1 year ago

5 0

repeated mesurement can reduce the error

it is true

if you take any mesurement repeatedly and the average is taken, the error will be less

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Scientist who discovered that the nucleus contains neutrons in addition to protons.

ZanzabumX [31]

Answer: James Chadwick

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2 years ago

Compare between chromosphere, corona and photosphere

gladu [14]

Answer:

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Explanation:

Photosphere: The lowest layer of the sun is called photo sphere . It is about 300 miles thick from the surface. It is the source of solar flares. It is marked by bright bubbling granules of plasma.

chromosphere emits a reddish glow as the super heated hydrogen burns off but the red rim can only be seen during total solar eclipse.

The third layer of the sun atmosphere is Corona. It can also only be seen during during a total solar eclipse. Temperature in corona can reach as high as 3.5 million degree fahrenheit. As the gases cool they become solar winds.

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3 years ago

Which of the following is true about a planet orbiting a star in uniform circular motion? A. The direction of the velocity vecto

Luda [366]

<span>As it is uniform circular motion therefore speed is constant. Therefore we can rule out option B. Also in circular motion the direction of velocity vector changes therefore velocity can't be constant. Therefore option B is incorrect as well. Also centripetal acceleration is always towards the center so option D is wrong as well. That implies option A is correct.</span>

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3 years ago

Read 2 more answers

A stone tumbles into a mine shaft strikes bottom after falling for 4.2 seconds. How deep is the mine shaft

ziro4ka [17]

$d = u \times t \: + \frac{1}{2} \times a \times {t}^{2}$
Since initial velocity is zero hence , u = 0

=> d = 1/2 * a * t2

$d = 0.5 \times 9.8 \times {4.2}^{2}$
on solving we get

d = 86.436 metres

Note ; Here Gravitational Acceleration is take as , g = 9.8 m/s2

3 0

3 years ago

to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to

givi [52]

Answer:

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

$f_c = 10 Hz$ represent the corner frequency

$f= 60 Hz$ represent the original frequency

n represent the filter order and that's the variable that we need to find

$G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1$

If we square both sides we got:

$G^2 (1+\frac{f}{f_c})^{2n}= 1$

We divide both sides by $G^2$ and we got:

$(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}$

Now we can apply log on both sides and we got:

$2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})$

And solving for n we got:

$n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}$

And replacing we got:

$n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}$

$n = \frac{4.60517}{3.8918}=1.18$

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0

3 years ago