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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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Answer:

Explanation:

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3 years ago

Which factors could be potential sources of error in the experiment? Check all that apply.

mote1985 [20]

Answer:

1. energy lost in the lever due to friction

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Explanation:

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3 years ago

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3 years ago

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

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3 0

3 years ago

A ship leaves the island of Guam and sails a distance 255 km at an angle 49.0 o north of west. Part A: In which direction must i

kiruha [24]

Answer:

Explanation:

We shall represent displacement in vector form .Consider east as x axes and north as Y axes west as - ve x axes and south as - ve Y axes . 255 km can be represented by the following vector

D₁ = - 255 cos 49 i + 255 sin49 j

= - 167.29 i + 192.45 j

Let D₂ be the further displacement which lands him 125 km east . So the resultant displacement is

D = 125 i

D₁ + D₂ = D

- 167.29 i + 192.45 j + D₂ = 125 i

D₂ = 125 i + 167.29 i - 192.45 j

= 292.29 i - 192.45 j

Angle of D₂ with x axes θ

tan θ = -192.45 / 292.29

= - 0.658

θ = 33.33 south of east

Magnitude of D₂

D₂² = ( 192.45)² + ( 292.29)²

D₂ = 350 km approx

Tan

7 0

3 years ago