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NikAS [45]
3 years ago
6

In which application is a magnet used to store information?

Physics
2 answers:
galina1969 [7]3 years ago
6 0

Answer:

d the disk in a computer hard drive

Explanation:

because hard drive use magnetic memory tostore giga- and terabytes of data in computer.

tatiyna3 years ago
3 0

Answer:

D. The disk in a computer hard drive

<em>The</em><em> </em><em>disk</em><em> </em><em>has</em><em> </em><em>a</em><em> </em><em>magnetic</em><em> </em><em>coating</em><em> </em><em>in</em><em> </em><em>which</em><em> </em><em>the</em><em> </em><em>infor</em><em>mation</em><em> </em><em>is</em><em> </em><em>stored </em><em>on</em><em> </em><em>the</em><em> </em><em>magnetic</em><em> </em><em>strips</em><em>.</em>

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Resistance = (voltage) / (current)

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Según el diagrama anterior, ¿cómo se forman las rocas metamórficas?
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3 years ago
A student carried out an experiment adding different weights to a spring and recording the results. Look at the table of results
MAXImum [283]

Answer:

0.25 m.

Explanation:

We'll begin by calculating the spring constant of the spring.

From the diagram, we shall used any of the weight with the corresponding extention to determine the spring constant. This is illustrated below:

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Extention (e) = 0.125 m

Spring constant (K) =?

F = Ke

0.1 = K x 0.125

Divide both side by 0.125

K = 0.1/0.125

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Therefore, the force constant, K of spring is 0.8 N/m

Now, we can obtain the number in gap 1 in the diagram above as follow:

Force (F) = 0.2 N

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5 0
3 years ago
Estimate the final temperature of a mole of gas at 200.0 atm and 19.0°C as it is forced through a porous plug to a final pressur
Yanka [14]

Answer : The final temperature of gas is 266.12 K

Explanation :

According to the Joule-Thomson experiment, it states that when a gas is expanded adiabatically from higher pressure region to lower pressure region, the change in temperature with respect to change in pressure at constant enthalpy is known as Joule-Thomson coefficient.

The formula will be:

\mu_{J,T}=(\frac{dT}{dP})_H

or,

\mu_{J,T}=(\frac{dT}{dP})_H\approx \frac{\Delta T}{\Delta P}

As per question the formula will be:

\mu_{J,T}=\frac{T_2-T_1}{P_2-P_1}   .........(1)

where,

\mu_{J,T} = Joule-Thomson coefficient of the gas = 0.13K/atm

T_1 = initial temperature = 19.0^oC=273+19.0=292.0K

T_2 = final temperature = ?

P_1 = initial pressure = 200.0 atm

P_2 = final pressure = 0.95 atm

Now put all the given values in the above equation 1, we get:

0.13K/atm=\frac{T_2-292.0K}{(0.95-200.0)atm}

T_2=266.12K

Therefore, the final temperature of gas is 266.12 K

5 0
3 years ago
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