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Korolek [52]
4 years ago
12

Once a ball leaves a table at 10 m/s, how long will it take to hit the floor

Physics
1 answer:
Zarrin [17]4 years ago
6 0

<u>Answer:</u>

The time taken for the ball to hit the floor as 1.02 seconds

<u>Explanation:</u>

As per the given question, the ball leaves at a speed from the table with an initial velocity of 10 m/s, we have the equation

\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at}

where Vf represents the final velocity

 Vi represents the initial velocity  

a represents the acceleration and

t represents the time

\mathrm{V}_{\mathrm{f}}=\mathrm{V}_{\mathrm{i}}+\mathrm{at} after rearranging

 \mathrm{t}=\frac{\mathrm{vf}-\mathrm{vi}}{a}

  \mathrm{t}=\frac{0-10}{9.8}  = 1.022 seconds

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A solid non-conducting sphere of radius R carries a charge Q1 distributed uniformly. The sphere is surrounded by a concentric sp
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Answer:

E = k Q₁ / r²

Explanation:

For this exercise that asks us for the electric field between the sphere and the spherical shell, we can use Gauss's law

           Ф = ∫ E .dA = q_{int} / ε₀

where Ф the electric flow, qint is the charge inside the surface

To solve these problems we must create a Gaussian surface that takes advantage of the symmetry of the problem, in this almost our surface is a sphere of radius r, that this is the sphere of and the shell, bone

           R <r <R_a

for this surface the electric field lines are radial and the radius of the sphere are also, therefore the two are parallel, which reduces the scalar product to the algebraic product.

         E A = q_{int} /ε₀

The charge inside the surface is Q₁, since the other charge Q₂ is outside the Gaussian surface, therefore it does not contribute to the electric field

          q_{int} = Q₁

The surface area is

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we substitute

          E 4π r² = Q₁ /ε₀

          E = 1 / 4πε₀ Q₁ / r²

          k = 1/4πε₀

 

          E = k Q₁ / r²

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object carries a charge of -8.1 µC, while another carries a charge of -2.0 µC. How many electrons must be transferred from the
LenKa [72]

Number of electrons transferred: 1.91\cdot 10^{13}

Explanation:

The charge on the first object is

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while the charge on the 2nd object is

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When they are in contact, the final charge on each object will be

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So, the amount of charge (electrons) transferred from the 1st object to the 2nd object is

\Delta Q = Q_1 - Q = -8.1 -(5.05)=-3.05 \mu C = -3.05\cdot 10^{-6}C

The charge of one electron is

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Learn more about electrons:

brainly.com/question/2757829

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