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grigory [225]
4 years ago
15

A runner achieves a velocity of 12.1 m/s 8.2 sec after he begins. What is

Physics
1 answer:
Karo-lina-s [1.5K]4 years ago
7 0

<u>Given</u><u> </u><u>-</u><u> </u>

  • Velocity (v) = 12.1 m/s
  • Time (t) = 8.2 sec

<u>To</u><u> </u><u>find</u><u> </u><u>-</u>

  • Acceleration

<u>Solu</u><u>tion</u><u> </u><u>-</u><u> </u>

★ Acceleration = velocity/ time

→ A = 12.1/8.2

→ 1.47.... m/s²

<u>There</u><u>fore</u><u>,</u><u> </u><u>the</u><u> </u><u>accele</u><u>ration</u><u> </u><u>of</u><u> </u><u>the</u><u> </u><u>runner</u><u> </u><u>will</u><u> </u><u>be</u><u> </u><u>1</u><u>.</u><u>4</u><u>7</u><u> </u><u>m</u><u>/</u><u>s²</u><u>.</u>

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A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum
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Explanation:

From the given information:

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\rho (r) \  \alpha  \  \delta (r -R)

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Thus;

k * 4 \pi  \int ^{R}_{0}  \delta (r -R) * r^2dr = \sigma \times  R^2

k * \int ^{R}_{0}  \delta (r -R)  r^2dr = \sigma \times  R^2

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k  =   R^2 --- (2)

Hence, from equation (1), if k = \sigma

\mathbf{\rho (r) = \delta* \delta (r -R)  \ \  at   \ \  (r=R)}

\mathbf{\rho (r) =0 \ \  at   \ \  rR}

To verify the units:

\mathbf{\rho (r) =\sigma \ *  \ \delta (r-R)}

↓         ↓            ↓

c/m³    c/m³  ×   1/m            

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The integrated charge Q

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Q = (2 \pi) (2) \int ^R_0 \sigma * \delta (r-R) r^2 \ dr

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\mathbf{Q = 4 \pi R^2  \sigma  }

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3 years ago
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