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3 years ago

7

Two objects are attracted to each other by electromagnetic forces. Suppose I double the charge of one object. How can I return t

he force to its original value?

1 answer:

nata0808 [166]3 years ago

4 0

Answer:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

Explanation:

Lets assume,

Charge on first object = Q

Charge on second object = q

Distance between them = r

Force between the two charges = F

According to Coulomb's law,

$F = k \frac{Qq}{r^{2}}$

where, k = Coulomb constant

New value of charge on first object = 2Q. Thus the new force(F') will be

$F' = k \frac{2Qq}{r^{2}}$

$F' = 2F$

So, to bring the value of force(F') to original value, there are two options:

1. Reduce the charge on second object by half or

2. Increase the distance between the two charges by a factor of 1.41 (√2).

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A suspension bridge has twin towers that are 1300 feet apart. Each tower extends 180 feet above the road surface. The cables are

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Answer:

$y = 17.04 ft$

Explanation:

Since the cable touches the road at the mid point of two towers

so here we have vertex at that mid point taken to be origin

now the maximum height on the either side is given as

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now we have

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now we need to find the height at distance of 200 ft from center

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3 years ago

Is this true of false <br> Apples and Onions taste the same if you plug your nose?

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3 years ago

A projectile is launched straight up from a height of 960 feet with an initial velocity of 64 ft/sec. Its height at time t is h(

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Answer:

a) $t=2s$

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c) $v_{y}=-256ft/s$

Explanation:

From the exercise we know the initial velocity of the projectile and its initial height

$v_{y}=64ft/s\\h_{o}=960ft\\g=-32ft/s^2$

To find what time does it take to reach maximum height we need to find how high will it go

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Knowing that its velocity is zero at its maximum height

$v_{y}^{2}=v_{o}^{2}+2g(y-y_{o})$

$0=(64ft/s)^2-2(32ft/s^2)(y-960ft)$

$y=\frac{-(64ft/s)^2-2(32ft/s^2)(960ft)}{-2(32ft/s^2)}=1024ft$

So, the projectile goes 1024 ft high

a) From the equation of height we calculate how long does it take to reach maximum point

$h=-16t^2+64t+960$

$1024=-16t^2+64t+960$

$0=-16t^2+64t-64$

Solving the quadratic equation

$t=\frac{-b±\sqrt{b^{2}-4ac}}{2a}$

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So, the projectile reach maximum point at t=2s

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Since the projectile is going down the velocity at the instant it reaches the ground is:

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<em>Since the object is moving with a velocity comparable to the velocity of light with respect to the observer therefore the length will appear shorter according to the theory of relativity.</em>

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