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Vitek1552 [10]
3 years ago
5

Two electric charges are moved so that they are twice as far apart as they had originally been. Is the force they experience fro

m each other now half the original force? Explain.
Physics
1 answer:
Natasha_Volkova [10]3 years ago
5 0

Answer:

No.

Explanation:

The force that two particle experience is inversely proportional to the sqare of the distance, this is:

F \  \alpha \  \frac{1}{D^{2}} for a distance D

If we move them so that D is doubled:

\frac{1}{2^{2}.D^{2}  }= \frac{1}{4} \eq  \frac{1}{.D^{2}  } \eq

Then the force they experience is one fourth of the original.

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A mass weighting 48 lbs stretches a spring 6 inches. The mass is in a medium that exerts a viscous resistance of 27 lbs when the
Mademuasel [1]

Answer:

a)

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b)

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c)

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Explanation:

The general equation of a damping oscillate motion is given by:

u(t)=u_oe^{-\frac{b}{2m}t}cos(\omega t-\alpha)    (1)

uo: initial position

m: mass of the block

b: damping coefficient

w: angular frequency

α: initial phase

a. With the information given in the statement you replace the values of the parameters in (1). But first, you calculate the constant b by using the information about the viscous resistance force:

|F_{vis}|=bv\\\\b=\frac{|F_{vis}|}{v}\\\\|F_{vis}|=27lbs=27*32.17ft.lb/s^2=868.59ft.lb/s^2\\\\b=\frac{868.59}{6}lb/s=144.76lb/s

Then, you obtain by replacing in (1):

6in = 0.499 ft

u(t)=0.499ft.e^{-\frac{144.76lb/s}{2(48lb)}t}cos(\omega t)\\\\u(t)=0.499ft.e^{-1.5t}cos(\omega t)

b.

mass, m = 48lb

c.

b = 144.76 lb/s

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The ends of a magnet where the force is strongest are magnetic
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Answer:

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Explanation:

Its a magnet that is magnetic

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