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3 years ago

An external computer flash drive can hold 1 gigabyte of data. How many bytes is this?

Physics

2 answers:

Fed [463]3 years ago

7 0

Answer: 1 gigabyte of a computer drive can hold of about $10^9$ bytes.

Explanation: We are given an external computer flash drive which can hold a data of about 1 gigabyte.

We know that,

1 gigabyte contains 1000 Megabytes of data

and 1 megabyte of a data can hold of about 1 million bytes.

From the above conversion, we get

$1GB=(1000\times 10^6)bytes=10^9bytes$

LenKa [72]3 years ago

6 0

"gig..." means a billion of something, or 10^9. In binary hardware, it's actually 1,073,741,824 bytes. That's 2^30.

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Calculate the wave length of a water wave with a speed of 20 m/s and a frequency of 2.5 Hz

12345 [234]

Wavelength of the water wave is 8 m

Explanation:

Wavelength measures the distance between two successive crests or troughs of the wave. It is given by the following equation

λ = v/f, where f is the frequency, v is the velocity of the wave

Here, v = 20 m/s and f = 2.5 Hz

⇒ λ = 20/2.5

= 8 m

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3 years ago

A portion of the atmosphere that becomes warmer than surrounding air will ____.

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<span>A portion of the atmosphere that becomes warmer than surrounding air will expand and rise. The warmer atmosphere the more space between the molecules. Therefore, warmer atmosphere </span><span>expands to allow more space for the molecules. Cool air on the other hand, contracts because the molecules in cool air need less space.</span>

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Answer:

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3 years ago

Compute the expected shell-model quadrupole moment of 209Bi () and compare with the experimental value, - 0.37 b

Over [174]

Answer:

0.22 b

Explanation:

Quadrupole moment of the nucleon is,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2}$

And also,

$R^{2}=R^{2} _{0}A^{\frac{2}{3} }$

And, $R _{0}=1.2\times 10^{-15}m$

Now,

$Q=-\frac{2j-1}{2(j+1)}\frac{3}{5}R^{2} _{0}A^{\frac{2}{3} }$

For Bismuth $j=\frac{9}{2}$ and A is 209.

$Q=-\frac{2\frac{9}{2} -1}{2(\frac{9}{2} +1)}\frac{3}{5}(1.2\times 10^{-15}) ^{2}(209)^{\frac{2}{3} }\\Q=0.628\times 35.28\times 10^{-30} \\Q=22.15\times 10^{-30} m^{2} \\Q=0.2215\times 10^{-28} m^{2} \\Q=0.22 barn$

Therefore, the expected value of quadrupole is 0.22 b which is quite related with experimental value which is 0.37 b

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