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myrzilka [38]
3 years ago
8

The gravitational constant G was first measured accurately by Henry Cavendish in 1798. He used an exquisitely sensitive balance

to measure the force between two lead spheres whose centers were 0.27 m apart. One of the spheres had a mass of 188 kg, while the mass of the other sphere was 0.93 kg.
What was the ratio of the gravitational force between these spheres to the weight of the lighter sphere?
Physics
1 answer:
belka [17]3 years ago
7 0

The gravitational force between the spheres is

F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N

where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².

The weight of the lighter sphere is

F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N

where <em>g</em> = 9.80 m/s².

The ratio between the two forces is then

\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}

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<span>                                                            C = 2πr</span>

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<span>                                                            r = 4.5/2π = 0.716 m</span>

Then, we calculate for the volume through the equation,

<span>                                                            V = πr2h</span>

<span>                                                V = π(0.716 m)2(8m) = 12.9 m3</span>

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3 years ago
Positive Charge Q is distributed uniformly along the x-axis from x=0 to x=a. A positive point charge q is located on the positiv
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Answer:

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Explanation:

You are asked to find the electric field of a continuous charge distribution, so we must use the equation

       

           E = k ∫dp /r²

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We must find the charge differential (dq), let's use that uniformly distributed and create a linear charge density

          λ = q / x

As it is constant, we can write it based on differentials

         λ = dq / dx

         dq = λ dx

We already have all the terms, let's  integrate enter its limits, lower the distance from the left end of the distribution to the test charge (x = r) and the upper limit that is the distance from the left end of distribution to the test load ( x = r - a) where r> a

         E = k ∫ λ dx / x²

         E = k la (- 1 / x)

Let's get the negative sign from the parentheses

         E = - k λ (1 / x)

         E = - k λ (1 /(r-a)  -1 /r) = - k λ [a / r (r-a)]

Let's change the charge density with the value of the total charge λ = Q / a

         E = - k Q/a  [a / r (r-a)]

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b) We calculate the force.  

         F = E qo

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c) the force for charge porbe very far r >> a. In this case we can take r from the parentheses and neglect (a/r)

         F = - k Qqo / r² (1 -  a/r)

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Explanation:

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Answer:

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