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nikklg [1K]
2 years ago
5

How do magnetics ruin electronic devices

Physics
2 answers:
attashe74 [19]2 years ago
5 0
Your inducing a current
<span>Stick any computer component inside an MRI machine and you can kiss it bye-bye</span>
yulyashka [42]2 years ago
3 0
IT could ruin your data if it has a strong enough charge causing it to not work as efficiently as it is designed to.
You might be interested in
Solve the inequality 2(n+3) – 4&lt;6. Then graph<br> the solution.
Aloiza [94]
The solution is 22 2(n+3)-4&6
6 0
3 years ago
A weight of 30.0 N is suspended from a spring that has a force constant of 220 N/m. The system is undamped and is subjected to a
Nimfa-mama [501]

Answer:

F_0 = 393 N

Explanation:

As we know that amplitude of forced oscillation is given as

A = \frac{F_0}{ m(\omega^2 - \omega_0^2)}

here we know that natural frequency of the oscillation is given as

\omega_0 = \sqrt{\frac{k}{m}}

here mass of the object is given as

m = \frac{W}{g}

\omega_0 = \sqrt{\frac{220}{\frac{30}{9.81}}}

\omega_0 = 8.48 rad/s

angular frequency of applied force is given as

\omega = 2\pi f

\omega = 2\pi(10.5) = 65.97 rad/s

now we have

0.03 = \frac{F_0}{3.06(65.97^2 - 8.48^2)}

F_0 = 393 N

6 0
3 years ago
WILL GIVE BRAINLIEST! QUICK PLEASE HELP!!!
LenKa [72]

Answer:Weight is the result of gravity . The gravitational field strength of Earth is 10 N/kg (ten newtons per kilogram). This means an object with a mass of 1 kg would be attracted towards the centre of Earth by a force of 10 N. We feel forces like this as weight.

Explanation:

3 0
2 years ago
Read 2 more answers
wo fixed charges, A and B are located at x axis. A is at x = 0 m, B is at x = 4 m. QA = +4.0 μC and QB = -5.0 μC. Calculate the
lys-0071 [83]

Answer:

10250 N/C leftwards

Explanation:

QA = 4 micro Coulomb

QB = - 5 micro Coulomb

AP = 6 m

BP = 2 m

A is origin, B is at 4 m and P is at 6 m .

The electric field due to charge QA at P is EA rightwards

E_{A}=\frac{KQ_{A}}{AP^{2}}=\frac{9\times10^{9}\times4\times10^{-6}}{6^{2}}=1000 N/C (rightwards)

The electric field due to charge QB at P is EB leftwards

E_{B}=\frac{KQ_{B}}{BP^{2}}=\frac{9\times10^{9}\times5\times10^{-6}}{2^{2}}=11250 N/C (leftwards)

The resultant electric field at P due the charges is given by

E = EB - EA

E = 11250 - 1000 = 10250 N/C leftwards

5 0
3 years ago
8. Noticing how much food is on your lunch<br> tray is a quantitative observation because
zhenek [66]

Answer:

you are making an observation that uses numbers.

Explanation:

5 0
2 years ago
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