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Bumek [7]
3 years ago
11

An electron is released a short distance above earth's surface. a second electron directly below it exerts an electrostatic forc

e on the first electron just great enough to cancel the gravitational force on it. how far below the first electron is the second? the value of coulomb's constant is 8.98755 × 109 n · m2 /c 2 and the acceleration of gravity is 9.81 m/s 2 .
Physics
1 answer:
slamgirl [31]3 years ago
3 0

The mass of an electron is 9.109 x 10⁻³¹ kg

The weight of the electron is (mass) x (g) =  8.926 x 10⁻³⁰ Newton

The charge on an electron is -1.602 x 10⁻¹⁹ Coulomb

The repelling force between the two electrons is (K · q₁ · q₂ / r²) =

(8.98755 x 10⁹ N-m²/C²) x (1.602 x 10⁻¹⁹ C)² / D²

In order for the bottom one to just exactly hold the top one up at a distance 'D', the repelling force has to be exactly equal to the weight of the upper electron.

8.926 x 10⁻³⁰ N = (8.98755 x 10⁹ N-m²/C²)·(1.602 x 10⁻¹⁹ C)² / D²

We have to solve THAT ugly mess for ' D '.

Clean up the units first:

Cancel the C² on the right side, then divide each side by Newton:

8.926 x 10⁻³⁰ = (8.98755 x 10⁹ m²) x (1.602 x 10⁻¹⁹)² / D²

Now, let's multiply both sides by (D² x 10²⁹) :

D² x 8.926 x 10⁻¹ = (8.98755 m²) x (1.602)²

Divide each side by (0.8926):

D² = (8.98755 x 1.602²) / (0.8926)  meter²

D² = 25.84 m²

Take the square root of each side:

<em>D = 5.08 meters</em>

I am shocked, impressed, and amazed !

Are you shocked, impressed, or amazed ?

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A 75 g, 30 cm long rod hangs vertically on a friction less, horizontal axel passing through the center. A 10 g ball of clay trav
kotegsom [21]
Angular momentum is conserved, just before the clay hits and just after; 
<span>mv(L/2) = Iw </span>

<span>I is the combined moment of inertia of the rod, (1/12)ML^2 , and the clay at the tip, m(L/2)^2 ; </span>
<span>I = [(1/12)ML^2 + m(L/2)^2] </span>

<span>Immediately after the collision the kinetic energy of rod + clay swings the rod up so the clay rises to a height "h" above its lowest point, giving it potential energy, mgh. From energy conservation in this phase of the problem; </span>
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<span>Use the "w" found in the conservation of momentum above; and solve for "h" </span>
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<span>Next, get the angle by noting it is related to "h" as; </span>
<span>h = (L/2) - (L/2)Cos() </span>

<span>So finally </span>
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<span>m=mass of clay </span>
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3 years ago
You apply a net force on a soccer ball of 15 N. If the acceleration it has is 5 m/s2 what is the mass of the ball?​
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Answer:

<h2>3 kg </h2>

Explanation:

The mass of the ball can be found by using the formula

m =  \frac{f}{a}  \\

f is the force

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We have

m =  \frac{15}{5}  = 3 \\

We have the final answer as

<h3>3 kg</h3>

Hope this helps you

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irina [24]

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