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Dmitry_Shevchenko [17]
3 years ago
5

An alpha particle(the nucleus of a helium atom) has a mass of 6.64*10^-27kg and a charge of +2e. What are the magnitude and dire

ction of the electric field that will balance the gravitational force on it? Mp= 1.67*10^-27kg
Physics
2 answers:
emmasim [6.3K]3 years ago
8 0
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions.

Weight = electric force 
<span>mg = qE </span>

<span>6.64x10^-27 x 9.81 = (2 x 1.60x10^-19) E 
</span>qE =mg, 
<span>E = mg/q = 6.64•10^-27•9.8/2•1.6•10^-19 =2.03•10^-7 V/m</span>
Rainbow [258]3 years ago
3 0

Answer:

Magnitude of electric field E= 2.03×10^-7NC^-1

The charge is positive,the electric force is in the same direction as the electric field according to F= qE

Therefore,the direction of the electric field is upward.

Explanation:

Fg=Fe

Where Fe is the force of the electric field

Fg is force of the gravitational field.

Mg= we

Mg= 2eE

E=mg/2e

E= (6.64×10^-27)(9.8)/2(1.6×10^-19)

E=( 6.50×10^-26)/(3.2×10^-19)

E= 2.03×10^-7NC^-1

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telo118 [61]

Answer:

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Explanation:

The orbital velocity of the vessel around the planet can be found by equalizing the force of gravity between the vessel and the planet and the centripetal force:

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where

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Re-arranging the formula, we find an expression for v:

v=\sqrt{\frac{GM}{r}}

We see that:

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