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3 years ago

What can we conclude from the attractive nature of the force between a positively charged rod and an object?

Physics

1 answer:

Paraphin [41]3 years ago

6 0

Answer:

E; The object is negatively charged

Explanation:

Here, we want to state the conclusion that can be drawn from a positively charged rod being attracted to an object.

Generally as we know, oppositely charged materials attract while the ones with same charges repel each other.

Thus, in this case, for the rod to attract the object, there must have been an opposite charge of negativity on the object

So we conclude that the reason why the rod attracted the object was because of the presence of opposing charges on both of them. And since the rod has taken the positive charge, it is only correct to state that the object is negatively charged

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3 years ago

A piano wire with mass 2.60g and length 84.0 cm is stretched with a tension of 25.0 N. A wave with frequency 120.0 Hz and amplit

likoan [24]

Answer:

Power will be 0.2023 watt

And when amplitude is halved then power will be 0.0505 watt

Explanation:

We have given mass of the Piano wire m = 2.60 gram = 0.0026 kg

Length of wire l = 84 cm = 0.84 m

So mass density $\mu =\frac{m}{l}=\frac{0.0026}{0.84}=0.0031kg/m$

Tension in the wire T = 25 N

Frequency f = 120 Hz

So angular frequency $\omega =2\pi f=2\times 3.14\times 120=753.6rad/sec$

And amplitude A = 1.6 mm = 0.0016 m

We have to find the generated power

Power is given by $P=\frac{1}{2}\sqrt{\mu T}\omega ^2A^2=\frac{1}{2}\times \sqrt{0.0031\times 25}\times 753.6^2\times 0.0016^2=0.2023watt$

From the relation we can see that power $P\ \propto\ A^2$

So if amplitude is halved then power will be $\frac{1}{4}$ times

So power will be equal to $\frac{0.2023}{2}=0.0505watt$

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3 years ago

A wire along the z axis carries a current of 6.8 A in the z direction Find the magnitude and direction of the force exerted on a

Alexeev081 [22]

Answer:

Force is 14.93N along positive y axis.

Explanation:

We know that force 'F' on a current carrying conductor placed in a magnetic field of intensity B is given by

$\overrightarrow{F}=\overrightarrow{Il}\times \overrightarrow{B}$

where L is the length of the conductor

Applying values in the equation we have force F =

$\overrightarrow{F}=6.8\times 6.1\widehat{k}\times 0.36\widehat{i}\\\\\overrightarrow{F}=41.48\widehat{k}\times 0.36\widehat{i}\\\\\therefore \overrightarrow{F}=14.93N\widehat{j}$

Thus force is 14.93N along positive y axis.

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3 years ago