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brilliants [131]
3 years ago
15

What is the acceleration of an object in free fall near Earth's surface?

Physics
2 answers:
Vitek1552 [10]3 years ago
8 0

Answer:

D 9.8 m/s^2

Explanation:

The force of gravitational gravity on earth is 9.8 m/s^2

Licemer1 [7]3 years ago
8 0
D) 9.8 m/(s^2) is the approximate acceleration of a free falling object near Earth’s surface
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The work done by an external force to move a -6.70 μc charge from point a to point b is 1.20×10−3 j .
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Answer:

108.7 V

Explanation:

Two forces are acting on the particle:

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- The force of the electric field, whose work is equal to the change in electric potential energy of the charge: W_e=q\Delta V

where

q is the charge

\Delta V is the potential difference

The variation of kinetic energy of the charge is equal to the sum of the work done by the two forces:

K_f - K_i = W + W_e = W+q\Delta V

and since the charge starts from rest, K_i = 0, so the formula becomes

K_f = W+q\Delta V

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W=1.20 \cdot 10^{-3}J is the work done by the external force

q=-6.70 \mu C=-6.7\cdot 10^{-6}C is the charge

K_f = 4.72\cdot 10^{-4}J is the final kinetic energy

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\Delta V=\frac{K_f-W}{q}=\frac{4.72\cdot 10^{-4}J-1.2\cdot 10^{-3} J}{-6.7\cdot 10^{-6}C}=108.7 V

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