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just olya [345]
3 years ago
6

Consider an electron that is 100 m from an alpha particle ( = 3.2 x 10-19 C). (Enter the magnitudes.) (a) What is the electric f

ield (in N/C) due to the alpha particle at the location of the electron? N/C (b) What is the electric field (in N/C) due to the electron at the location of the alpha particle? N/C (c) What is the electric force (in N) on the alpha particle? On the electron? electric force on alpha particle electric force on electron
Physics
1 answer:
scoundrel [369]3 years ago
6 0

Answer:

a)E=2.88*10^{-13}N/C

b)E=1.44*10^{-13}N/C

c)F=4.61*10^{-32}N

Explanation:

The definition of a electric field produced by a point charge is:

E=k*q/r^2

<u>a)Electric Field due to the alpha particle:</u>

E=k*q_{alpha}/r^2=9*10^9*3.2*10^{-19}/(100)^2=2.88*10^{-13}N/C

<u>b)Electric Field  due to the electron:</u>

E=k*q_{electron}/r^2=9*10^9*1.6*10^{-19}/(100})^2=1.44*10^{-13}N/C

<u>c)Electric Force on the alpha particle, on the electron:</u>

The alpha particle and electron feel the same force magnitude but with opposite direction:

F=k*q_{electron}*q_{alpha}/r^2=9*10^9*1.6*10^{-19}*3.2*10^{-19}/(100)^2=4.61*10^{-32}N

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B

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A 3.0-kg cart is rolling across a frictionless, horizontal track toward a 1.3-kg cart that is held initially at rest. The carts
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Answer:

Explanation:

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b )

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3 years ago
A. State whether the acceleration is positive, negative or zero for each section in the speed versus
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Answer:

a=positive

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2 years ago
A positively charged object is brought near but not in contact with the top of an uncharged gold leaf electroscope. The experime
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Answer:

The leaves of the electroscope move further apart.

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7 0
3 years ago
Find the length of a pendulum that oscillates with a frequency of 0.16 hz. the acceleration due to gravity is 9.81 m/s 2 . answe
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The period of the pendulum is the reciprocal of the frequency:
T= \frac{1}{f}= \frac{1}{0.16 Hz}=6.25 s

The period of the pendulum is given by
T=2 \pi  \frac{L}{g}
where L is the length of the pendulum, and g the acceleration of gravity. By re-arranging the formula and using the value of T we found before, we can  calculate the length of the pendulum L:
L=g  \frac{T^2}{(2 \pi)^2}=(9.81 m/s^2)  \frac{(6.25 s)^2}{(2 \pi)^2}=9.71 m
7 0
3 years ago
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