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2 years ago

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

Physics

2 answers:

Fed [463]2 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]2 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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A 100 kg boat is floating in water. half of the boat is submerged under water. what is the weight of the boat?

Alex17521 [72]

W=MG
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2 years ago

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3 years ago

the acceleration of a rocket traveling upward is given by a=(6+.02s)m/s^s, where s is in meters. Determine the rocket's velocity

NikAS [45]

Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

where

a is acceleration

s is distance

v is velocity

Given;

a = 6 + 0.02s

so,

$\int\limits^s_0 {a} \, ds = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\$

Remember that

$v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t= (5\sqrt{2} ) ln \frac{| [s + 300 + \sqrt{(s^{2} + 600s)} ] |}{300} .......2$

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

8 0

3 years ago

You drop a rock from rest out of a window on the top floor of a building, 20.0 m above the ground. When the rock has fallen 5.00

Andrej [43]

Answer:

You drop a rock from rest out of a window on the top floor of a building, 30.0 m above the ground. When the rock has fallen 3.00 m, your friend throws a second rock straight down from the same window. You notice that both rocks reach the ground at the exact same time. What was the initial velocity of the ...... rest out of a window on the top floor of a building, 30.0m above the ground. ... You Notice That Both Rocks Reach The Ground At The Exact Same Time. ... You drop a rock from rest out of a window on the top floor of a building, 30.0m ... When the rock has fallen 3.20 m, your friend throws a second rock straight down from ...

4 0

2 years ago

When changing a tire sized 195/75 R15 to a 5 percent lower profile, the correct tire size would be _____

valentinak56 [21]

Answer and explanation:

When you are changing a car tire, the most important thing is to keep the total diameter as equal as possible.

The total car tire diameter can be calculated as:

$D_{tot orig}=\frac{195 \cdot 75}{2540 \cdot 2}+15''=26.5''$

The profile of this tire is 75 (the higher/taller relation), therefore a 5 percent lower profile would be:

pr=0.95·75=71.25

The problem is that the profiles are normalized and the nearest profile available is 70.

If we take a theorical tire with a profile of 71.25:

$D_{tot orig}=D_{new}\\\frac{195 \cdot 75}{2540 \cdot 2}+15''=\frac{X \cdot 71.25}{2540 \cdot 2}+15''\\X=205.26$

The theorical tire size should be 205/71 R15.

If we look for a real tire size, we should look for a tire with a diameter nearest to 26.5'' and a profile of 70.

The best option for real tire size is: Tire 225/70 R14 (wheel diameter of 26.4'') or 205/70 R15 (wheel diameter of 26.3'').

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3 years ago