Name the effect of current in electroplating

1 answer:

4 0

<span>For this, you need a compass needle, an empty matchbox, two electric cells, some wires, a plastic bottle cap and the liquid which is to be tested. Keep the compass needle inside the empty tray of matchbox and wrap a couple of rubber bands around it. Connect the wires to the battery and insert two wires in the liquid which is kept in the plastic bottle cap. You will observe that when the current flows through the wire, there is deflection in the compass needle. This shows that the given liquid conducts electricity. In fact, magnetic compass needle can detect even feeble current. LED can also be used in place of compass needle.</span>

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A football is kicked into the air with a velocity of 32m/s at an angle of 25º. At the very top of the ball’s path, its vertical

Travka [436]

At the "very top" of the ball's path, there's a tiny instant when the ball
is changing from "going up" to "going down". At that exact tiny instant,
its vertical speed is zero.

You can't go from "rising" to "falling" without passing through "zero vertical
speed", at least for an instant. It makes sense, and it feels right, but that's
not good enough in real Math. There's a big, serious, important formal law
in Calculus that says it. I think Newton may have been the one to prove it,
and it's named for him.

By the way ... it doesn't matter what the football's launch angle was,
or how hard it was kicked, or what its speed was off the punter's toe,
or how high it went, or what color it is, or who it belongs to, or even
whether it's full to the correct regulation air pressure. Its vertical speed
is still zero at the very top of its path, as it's turning around and starting
to fall.

7 0

3 years ago

A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximu

Mrac [35]

Answer:

x=±0.026m

Explanation:

In simple harmonic motion the maximum value of the magnitude of velocity

$v_{max}=wA=\sqrt{\frac{k}{m} }A$

The speed as a function of position for simple harmonic oscillator is given by

$v=w\sqrt{A^{2}-x^{2}$

where A is amplitude of motion

Given data

Amplitude A=3 cm =0.03 m

v=(1/2)Vmax

To find

We have asked to find position x does its speed equal half of is maximum speed

Solution

The speed of the particle the maximum speed as:

$v=\frac{V_{max} }{2}\\ w\sqrt{A^{2}-x^{2} }=\frac{wA}{2}\\ A^{2}-x^{2}=\frac{A^{2} }{4}\\ x^{2}=A^{2}- \frac{A^{2} }{4}\\ x^{2}=\frac{3A^{2}}{4} \\x=\sqrt{\frac{3A^{2}}{4}}$