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Inessa05 [86]
3 years ago
12

What will happen if you put a usb drive and a magnet too closely together?

Physics
2 answers:
kakasveta [241]3 years ago
6 0
<span>If you put a magnet right next to a USB drive, depending on the strength of the magnet and the amount of steel, nickel or cobalt used in the construction of that particular model of USB drive, the drive would either adhere to, or not adhere to, the magnet. This would cause no other significant effects. The storage of data in solid state form (as in USB drives) is not magnetic in nature, so no deletion or any other damage of the stored data would occur.</span>
choli [55]3 years ago
3 0

Before going to answer this first we have to know the various  types of materials that are present in an USB  drives.

The major materials of the USB drives are copper,silver and to some extent there will be gold plating also.

It contains the alloy like Brass also.

Irrespective of this it contains the integrated circuits[ICs] which are made up of semiconducting materials mostly like silicon.

From the above we get that a USB drive does not contain any type of magnetic materials.The data in the USB drive are stored in the form of electric charges .

Whenever a USB drive is taken towards a magnet and placed gently close to it,the magnetic field can't affect the data stored inside the drive as it is not magnetic in nature.The rest magnet cant not affect the the charge particles present inside the drive.

But if the magnetic filed has great strength and there will be any relative motion between drive and magnet,then it will induce a force on the drive which has still less impact on the drive. A magnet can hamper the stored data if that will be magnetic in nature.


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A 2kg Book is sitting on a table.a 10n Force is pulling to the right.a 3n Force is pulling to the left what is the net force act
salantis [7]


The net force = sum of all forces acting on the body



If we take left side as -ve and right side as +ve,
then,

The net force here would be equal to,
10N + (- 3N)
= 7N.


Therefore, a net force of +7N ( + indicates it's moving towards right) is acting on the book of mass 2kg.


4 0
3 years ago
A body which has surface area 5cm² and temperature of 727°C radiates 300J of energy in one minute. Calculate it's emissivity giv
cestrela7 [59]
<h2>Answer: 0.17</h2>

Explanation:

The Stefan-Boltzmann law establishes that a black body (an ideal body that absorbs or emits all the radiation that incides on it) "emits thermal radiation with a total hemispheric emissive power proportional to the fourth power of its temperature":  

P=\sigma A T^{4} (1)  

Where:  

P=300J/min=5J/s=5W is the energy radiated by a blackbody radiator per second, per unit area (in Watts). Knowing 1W=\frac{1Joule}{second}=1\frac{J}{s}

\sigma=5.6703(10)^{-8}\frac{W}{m^{2} K^{4}} is the Stefan-Boltzmann's constant.  

A=5cm^{2}=0.0005m^{2} is the Surface area of the body  

T=727\°C=1000.15K is the effective temperature of the body (its surface absolute temperature) in Kelvin.

However, there is no ideal black body (ideal radiator) although the radiation of stars like our Sun is quite close.  So, in the case of this body, we will use the Stefan-Boltzmann law for real radiator bodies:

P=\sigma A \epsilon T^{4} (2)  

Where \epsilon is the body's emissivity

(the value we want to find)

Isolating \epsilon from (2):

\epsilon=\frac{P}{\sigma A T^{4}} (3)  

Solving:

\epsilon=\frac{5W}{(5.6703(10)^{-8}\frac{W}{m^{2} K^{4}})(0.0005m^{2})(1000.15K)^{4}} (4)  

Finally:

\epsilon=0.17 (5)  This is the body's emissivity

3 0
3 years ago
Is being a plat 2 in rainbow 6 siege a good rank?
icang [17]

Answer:

In a way it does, but overall, there are many factors that affect your rank. In general, and talking about the average Platinum II, they are pretty decent according to casual player standards.

Explanation:

6 0
3 years ago
A solenoid of radius r 5 1.25 cm and length , 5 30.0 cm has 300 turns and carries 12.0 A. (a) Calculate the flux through the sur
Elina [12.6K]

Answer:

\phi = 7.4 \times 10^{-6} Wb

Explanation:

Magnetic field due to a long solenoid at the center is given by

B = \frac{\mu_o N i}{L}

here we know that

N = 300

L = 30 cm

i = 12 A

now magnetic field is given as

B = \frac{(4\pi \times 10^{-7})(300)(12)}{0.30}

B = 0.015 T

Now magnetic flux through the disc is given as

\phi = B.A

\phi = (0.015)(\pi r^2)

\phi = (0.015)(\pi)(0.0125)^2

\phi = 7.4 \times 10^{-6} Wb

8 0
3 years ago
Please answer for 20 points
Nat2105 [25]
C. Reduce friction between its moving parts
4 0
3 years ago
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