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4 years ago

What happens if you reverse the orientation of the permanent magnet? why?

1 answer:

7 0

If one reverse the orientation of a permanent magnet ITS MAGNETIZATION WILL BE PERMANENTLY REVERSED. This is because, the magnetic domains inside the permanent magnet aligned with the new applied field and increase with it while those domains that are anti aligned with that field will shrink.

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Consider a product with three components in series, with reliabilities of 0.90, 0.80, and 0.99 for components A, B, and C, r

Feliz [49]

Answer:0.853

Explanation:

Given

Reliability of A is 0.90

Reliability of B is 0.80

Reliability of C is 0.99

B has a backup with reliability of 0.8

i.e. B has a component in parallel to it.

A,B& C is series.

B actual reliability is

$R_b=1-\left ( 1-0.8\right )\left ( 1-0.8\right )$

$R_b=0.96$

Thus $R_{net}$ is given by

$R_{net}=R_A\times R_B\times R_C$

$R_{net}=0.9\times 0.96\times 0.99$

$R_{net}=0.8553$

5 0

3 years ago

The field inside a charged parallel-plate capacitor is __________.

Bess [88]

<span>Uniform. A parallel plate capacitor is a simple arrangement of electrodes and dielectric to form a capacitor. Here two parallel conductive plates are used as electrodes with a medium or dielectric in between them. Charge separation in a parallel-plate capacitor causes an internal electric field, which is uniform.</span>

5 0

4 years ago

Read 2 more answers

Car and a train move together along straight, parallel paths with the same constant cruising speed v0. At t=0 the car driver not

musickatia [10]

Answer:

Part A: $t = v_0/a_0$

Part B: $t = v_0/a_0$

Part C: $v_0^2/a_0$

Explanation:

Part A:

We will use the following kinematics equation:

$v = v_0 + at\\0 = v_0 - a_0t\\t = \frac{v_0}{a_0}$

Part B:

We will use the same kinematics equation:

$v = v_0 + at\\v_0 = 0 + a_0t\\t = \frac{v_0}{a_0}$

Part C:

The total time takes is 2t.

So the train moves a distance of

$x = v_0(2t) = 2v_0(\frac{v_0}{a_0}) = \frac{2v_0^2}{a_0}$

And the car moves a distance in Part A and in Part B:

$d_A = v_0t + \frac{1}{2}at^2 = v_0(\frac{v_0}{a_0}) - \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{a_0} - \frac{v_0^2}{2a_0} = \frac{v_0^2}{2a_0}\\d_B = v_0t + \frac{1}{2}at^2 = 0 + \frac{1}{2}a_0(\frac{v_0^2}{a_0^2}) = \frac{v_0^2}{2a_0}$