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2 years ago

Identify the prefix that would be used to express 2,000,000,000 bytes of computer memory?

Physics

2 answers:

Reika [66]2 years ago

4 0

Answer:

It would be 2 GB

Fynjy0 [20]2 years ago

4 0

Answer:

2×10^9 = Giga

Explanation:

Because Giga = 10^9

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If you ride your bike 20 miles and it takes you 120 minutes,what is your average speed

Alexus [3.1K]

120 minutes=2 hours
20/2= 10mph

5 0

3 years ago

Which has more thermal energy 1 g if aluminum or 1 g of gold and why

bagirrra123 [75]

gold because it has highest boiling point so it can conduct the heat

8 0

2 years ago

A power plant produces 1000 MW to suply a city 40Km away.Current flows from the power plant on a single wire of resistance0.050

Westkost [7]

Answer:

The current in wire resistance 2Ω

a). 8696 A

b). fraction power 15.1% a 115kV

Explanation:

Resistance

$R=0.05$ Ω/Km*40km

R=2Ω

P=1000 MW

a).

$P=V*I\\I=\frac{P}{V}=\frac{1000x10^{6}W}{115x10^{3}k } =8696.65A$

Using law ohm

b).

$V=I*R\\I=\frac{V}{R}$

$P=I*I*R\\P=I^{2} *R\\P=8696.65^{2}*2\\P=151.228 x10^{6} W$

$e=\frac{151.228x10^{6} }{1000x10^{6} }*100= 15.12$ %

8 0

3 years ago

What will be the force if the particle's charge is tripled and the electric field strength is halved? Give your answer in terms

kirza4 [7]

Answer:

F' = (3/2)F

Explanation:

the formula for the electric field strength is given as follows:

E = F/q

where,

E = Electric Field Strength

F = Force due to the electric field

q = magnitude of charge experiencing the force

Therefore,

F = E q ---------------- equation (1)

Now, if we half the electric field strength and make the magnitude of charge triple its initial value. Then the force will become:

F' = (E/2)(3 q)

F' = (3/2)(E q)

using equation (1)

3 0

3 years ago

In a lab experiment, a student is trying to apply the conservation of momentum. Two identical balls, each with a mass of 1.0 kg,

Studentka2010 [4]

Answer:

Second Trial satisfy principle of conservation of momentum

Explanation:

Given mass of ball A and ball B $=\ 1.0\ Kg.$

Let mass of ball $A$ and $B\ is\ m$

Final velocity of ball $A\ is\ v_1$

Final velocity of ball $B\ is\ v_2$

initial velocity of ball $A\ is\ u_1$

Initial velocity of ball $B\ is\ u_2$

Momentum after collision $=mv_1+mv_2$

Momentum before collision $= mu_1+mu_2$

Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.

Now, $mu_1+mu_2=mv_1+mv_2$

Plugging each trial in this equation we get,

First Trial

$mu_1+mu_2=mv_1+mv_2\\1(1)+1(-2)=1(-2)+1(-1)\\1-2=-2-1\\-1=-3$

momentum before collision $\neq$ moment after collision

Second Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1.5)=1(-.5)+1(-.5)\\.5-1.5=-.5-.5\\-1=-1$

moment before collision $=$ moment after collision

Third Trial

$mu_1+mu_2=mv_1+mv_2\\1(2)+1(1)=1(1)+1(-2)\\2+1=1-2\\3=-1$

momentum before collision $\neq$ moment after collision

Fourth Trial

$mu_1+mu_2=mv_1+mv_2\\1(.5)+1(-1)=1(1.5)+1(-1.5)\\.5-1=1.5-1.5\\-.5=0$

momentum before collision $\neq$ moment after collision

We can see only Trial- 2 shows the conservation of momentum in a closed system.

4 0

3 years ago

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