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oksano4ka [1.4K]

2 years ago

9

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-

directed electric field of magnitude 600 N/C ? Use 9.80 m/s2 for the magnitude of the free-fall acceleration.

1 answer:

sergey [27]2 years ago

8 0

$q = -21 * 10^{-6} C$

<h3>What is Free-fall acceleration?</h3>

The acceleration is constant and equal to the gravitational acceleration g which is 9.8 m/s at sea level on the Earth.

As we know that charge and its sign that remains in equilibrium is under gravity must be such that it will balance the gravitational force by electric force,

mg =qE

$1.45 * 10^{-3}(9.80)=q(600)\\\\q = 21 *10^{6} C\\\\$

and its sign must be negative so that it will have upward electric force

so it is

$q = -21 * 10^{-6} C$

The charge of a particle of mass is $-21 * 10^{-6} C$

Learn more about Free-fall acceleration: brainly.com/question/2165771

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A freshly prepared sample of radioactive isotope has an activity of 10 mCi. After 4 hours, its activity is 8 mCi. Find: (a) the

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Answer:

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Explanation:

Given that,

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(c). We need to calculate the sample's activity

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Put the value intyo the formula

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Hence, (a). The decay constant is $1.55\times10^{-5}\ s^{-1}$

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(b). The value of N₀ is $2.38\times10^{11}\ nuclei$

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