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3 years ago

What do the magnets in a computer hard drive and in the strip on a credit

Physics

1 answer:

Sonja [21]3 years ago

5 0

Answer:

they store information

Explanation:

Information are encoded on the magnetic part of a hard drive and the strip of a credit card.

On the drive, series of magnetic fields are used to encode information.
The data is stored in form of zeros and ones.
A credit card has a magnetic strip where information is encoded.
The stripe provides a readability function for electronic devices.

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Suppose the minimum uncertainty in the position of a

Anika [276]

Answer:

The minimum uncertainty in its speed is greater than $6.52\times 10^4\ m/s$ .

Explanation:

It is given that,

Speed of the particle, $v=4.1\times 10^5\ m/s$

We need to find the minimum uncertainty in its speed. It can be calculated using uncertainty principle as :

$\Delta x.\Delta p\ge \dfrac{h}{2\pi}$

Since, p = mv

$\Delta x.\Delta (mv)\ge \dfrac{h}{2\pi}$

$(\dfrac{h}{mv}).(m\Delta v)\ge \dfrac{h}{2\pi}$

$\dfrac{\Delta v}{v}\ge \dfrac{h}{2\pi}$

$\Delta v\ge \dfrac{v}{2\pi}$

$\Delta v\ge \dfrac{4.1\times 10^5}{2\pi}$

$\Delta v\ge 6.52\times 10^4\ m/s$

So, the minimum uncertainty in its speed is greater than $6.52\times 10^4\ m/s$ . Hence, this is the required solution.

7 0

3 years ago

A 0.250 kgkg toy is undergoing SHM on the end of a horizontal spring with force constant 300 N/mN/m. When the toy is 0.0120 mm f

konstantin123 [22]

Answer:

(a) The total energy of the object at any point in its motion is 0.0416 J

(b) The amplitude of the motion is 0.0167 m

Explanation:

Given;

mass of the toy, m = 0.25 kg

force constant of the spring, k = 300 N/m

displacement of the toy, x = 0.012 m

speed of the toy, v = 0.4 m/s

(a) The total energy of the object at any point in its motion

E = ¹/₂mv² + ¹/₂kx²

E = ¹/₂ (0.25)(0.4)² + ¹/₂ (300)(0.012)²

E = 0.0416 J

(b) the amplitude of the motion

E = ¹/₂KA²

$A = \sqrt{\frac{2E}{K} } \\\\A = \sqrt{\frac{2*0.0416}{300} } \\\\A = 0.0167 \ m$

$E = \frac{1}{2} mv_{max}^2\\\\v_{max} = \sqrt{\frac{2E}{m} } \\\\v_{max} = \sqrt{\frac{2*0.0416}{0.25} } \\\\v_{max} = 0.577 \ m/s$

8 0

3 years ago

An induced emf is produced inA. a closed loop of wire when it remains at rest in a nonuniform static magnetic field.B. a closed

Kazeer [188]

Answer:

C. a closed loop of wire moving at constant velocity in a nonuniform static magnetic field.

Explanation:

As we know by Faraday's law that rate of change in magnetic flux will induce EMF

so mathematically we can say that

$EMF = \frac{d\phi}{dt}$

so we will have

$\phi = B.A$

so EMF will be induced if the flux linked with the closed loop will change with time.

So here correct answer will be

C. a closed loop of wire moving at constant velocity in a nonuniform static magnetic field.

7 0

4 years ago

What is the bobsled’s speed? 1) 100 m/s <br> 2) 0.25 m/s <br> 3) 2500 m/s <br> 4) 4 m/s

iogann1982 [59]

It should be D. 4 m/s
Hope this helps!

7 0

4 years ago

A 5.80kΩ5.80kΩ resistor in a circuit has two mesh currents flowing through it. The first current measures 3.10mA3.10mA. The seco

77julia77 [94]

Answer:

power = 23.2 mW

Explanation:

given data

resister r = 5.80 kΩ

current measures I1 = 3.10 m A

measure current I2 = 1.10mA

solution

we get here I total

I = current 1 - current 2

I = 3.10 - 1.10

I = 2 mA

so here power will be

power = I²×R .............1

power = 2² × 5.80

power = 23.2 mW

3 0

3 years ago