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3 years ago

Why audiotape, credit cards and videotape not placed closed to a strong magnet

Physics

2 answers:

Mkey [24]3 years ago

7 0

Answer:

There are many possible causes of magnetic defects in credit cards. But there are two direct causes: 1. Close to a strong magnet 2. Damage from scratches or wear Even if used properly it can cause magnetic defects. But if used incorrectly, it can cause an immediate relapse. If you have never used a credit card due to magnetic defects, check what will cause magnetic defects and for example credit cardMagnetic stripe cards, such as credit cards and cash cards, are more likely to cause magnetic field defects if brought near magnetically generating objects such as magnets. Although exposure to strong magnets does not necessarily lead to failure. However, it should be kept away from objects that may cause magnetic field defects.

Explanation:

e-lub [12.9K]3 years ago

4 0

Audio tapes and credits cards and videotapes have magnetized bars.
Strong magnet can attract them too badly
It will cause harm to them.
They may be destroyed or may not work properly.
So they are not kept closed

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Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

A 25 kh inductor carries a current of 60 ma. how much energy is stored in this system

IgorLugansk [536]

Energy stored= Li^2/2
=45 W

8 0

3 years ago

A resistor has four colored stripes in the following order: orange, orange, brown and silver. What is the resistance of the resi

zubka84 [21]

Answer:

Resistance =330 Ω

Tolerance = 33 Ω

Explanation:

see attached resistor color code table

The first stripe is orange, which means the leftmost digit is a 3.

The second stripe is orange , which means the next digit is a 3.

The third stripe is brown. Since brown is 1, it means add one zero to the right of the first two digits.

The resistance is:

orange-orange-brown= 330 Ω

The tolerance is:

The fourth color band indicates the resistor's tolerance. Tolerance is the percentage of error in the resistor's resistance.

silver is 10%

A 330 Ω resistor has a silver tolerance band.

<em>Tolerance = value of resistor x value of tolerance band </em>

= 330 Ω x 10% = 33 Ω

330 Ω stated resistance +/- 33 Ω tolerance means that the resistor could range in actual value from as much as 363 Ω to as little as 297 Ω.

7 0

3 years ago

If the velocity of a car is 45 km/h west, how far can it travel in 0.5 hours?

Anarel [89]

45km/h * 0.5h= 22.5km

The car can travel 22.5km in 0.5 hours

6 0

3 years ago

Read 2 more answers

Two charged particles are projected into a region where a magnetic field is directed perpendicular to their velocities. if the c

goldfiish [28.3K]

The bearing could be the below:
oppositely charged, same initial direction
same charge, opposite initial direction

You can decide by utilizing your correct hand and put your fingers toward the attractive field (North to South). Thumb toward present or charged molecule. The course of your palm will demonstrate the heading of compelling set on a decidedly charged molecule and the bearing of the back of your hand will demonstrate the bearing of a contrarily charged molecule.

4 0

4 years ago

Why audiotape, credit cards and videotape not placed closed to a strong magnet​

Why audiotape, credit cards and videotape not placed closed to a strong magnet