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3 years ago

An external computer flash drive can hold 1 gigabytes of data how many byes is this

Physics

2 answers:

Mariulka [41]3 years ago

5 0

There are 1000000000 bytes in a gigabyte.

irga5000 [103]3 years ago

4 0

The prefix "giga..." means "billion".

When you're talking about data, though, the actual number
is always a power of 2 .

1 gigabyte = actually 2³⁰ = 1,073,741,824 bytes

but we talk about it as if it's 1,000,000,000 bytes.

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Squid use jet propulsion for rapid escapes. A squid pulls water into its body and then rapidly ejects the water backward to prop

Setler [38]

Answer:

a. FTh = 30 N

b. Fw = 30 N

c. a = 200 m/s2

Explanation:

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3 years ago

Why are so many people using this on thing like a freaking dating app???

Westkost [7]

Answer:

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Explanation:

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3 years ago

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A truck accelerating at 0.0083 meters/secondÆ covers a distance of 5.8 × 10Ê meters. If the truck's mass is 7,000 kilograms, wha

tigry1 [53]

work done = force * distance moved (in direction of the force)

force= mass* acceleration

force=58.1N

58.1*(5.8*10^4)
=3,369,800 J

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3 years ago

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an empty density bottle weighed 23.5 and 48.4g when filled with water .The empty bottle was partially filled with sand until and

gavmur [86]

Answer:

Density of Sand is $2.653g/cm^{3}$ .

Explanation:

Given Empty Density bottle weighs 23.5gm(W=23.5gm)

Weight of bottle when completely filled water=48.4gm

So amount of water required to fill the bottle=Weight of bottle filled with water-W

Amount of water required to fill the bottle( $w_{max}$ )=48.4gm-23.5gm

$w_{max}=24.9$ g

Since we know density of water $d_{w} =1g/cm^{3}$ and $w_{max}$

We can calculate volume of empty space in the bottle(V).

$w_{max}=d_{w}$ $\times$ V

V= $\frac{w_{max} }{d_{w} }$

V= $\frac{24.9}{1}$

V=24.9 $g/cm^{3}$

Now bottle is partially filled with sand,and weight of bottle is ( $w_{s}$ )36.5gm

So,

Amount of sand added ( $m_{s}$ )=36.5-Weight of the bottle

$m_{s}$ =13g

After filling the bottle with water again,the weight of the bottle becomes ( $W_{2}$ =56.5g)

Therefore,

amount of water added to the bottle of sand in grams = $W_{2}$ -36.5gm

amount of water added =56.5g-36.5g

amount of water added =20g

As the density of water = 1g/ $cm^{3}$

Amount of water (in grams )=Volume of water occupied

20=volume of water added

Therfore volume of water added to the sand filled bottle( $V_{w}$ )=20 $cm^{3}$

As we know the total volume of the water bottle(V),

Volume of the sand occupied in the water bottle=V- $V_{w}$

$V_{s}$ =24.9g-20g

$V_{s}$ =4.9g

We know,

Density=Mass/Volume

Therefore,

density of sand = $\frac{m_{s} }{V_{s} }$

density of sand = $\frac{13}{4.9}$

density of sand = $2.653g/cm^{3}$

8 0

3 years ago

john is using a pulley to lift the sail on his sailboat. the sail weighs 150N and he must lift it at 4.0m. how much work must be

gogolik [260]

The work done on the sail is 600 J

Explanation:

The work done to lift the sail is equal to the gain in gravitational potential energy of the sail, therefore is:

$W=mg\Delta h$

where

m is the mass of the sail

g is the acceleration of gravity

(mg) is the weight of the sail

$\Delta h$ is the change in height of the sail

In this problem we have

mg = 150 N (weight)

$\Delta h = 4.0 m$

Substituting, we find the work done:

$W=(150)(4.0)=600 J$

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

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3 years ago