If the distance is halved and the charges of both particles are doubled, the force is ________ as great.

Answers

Answer 1

Answer:

If the distance is halved and the charges of both particles are doubled, the force is 16 times as great.

Coulomb's Law

According to Coulomb's law, the electrostatic force between two charges is given by;

$F=k(q_1 q_2)/(r^2)$

Where 'k' is the Coulomb's constant.

If the charges of both particles are doubled and the distance is halved, the new force will be;

$F\,'=k(2q_1 * 2q_2)/((r/2)^2)=k (4* 4* q_1 q_2)/(r^2) =16\,F$

So, the new force will be 16 times greater than the old force.

Learn more about Coulomb's law here:

brainly.com/question/14049417

Answer 2

Answer:

Answer:

The new force is 16 times of the initial force.

Explanation:

The electric force between charges is given by :

$F=(kq_1q_2)/(r^2)$

If the distance is halved, d' =d/2 and charges are doubles, $q_1'=2q_1\ \text{and}\ q_2'=2q_2$

New force becomes,

$F'=(kq_1'q_2')/(r'^2)\n\nF'=(k(2q_1)(2q_2))/((d/2)^2)\n\nF'=16* (kq_1q_2)/(r^2)\n\nF'=16F$

So, the new force is 16 times of the initial force.

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Answers

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A boat sails along the shore. To an observer, the boat appears to move at a speed of 11 m/s, and a man on the boat walking forward appears to have a speed of 12.4 m/s. How long does it take the man to move 6 m relative to the boat?

Answers

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It therefore equals 4.29 s

Answer:

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Explanation:

Apex

a is a circuit that is energized but dose not allow useful current to flow on the circuit because of a break in the current path

Answers

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Plugging a lamp into an electrical outlet gives the lamp a source ofa. current.
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Answers

Plugging a lamp into an electrical outlet gives the lamp a source of voltage. The answer is letter B. The rest of the choices do not answer the question above.

A runner achieves a velocity of 11.1 m/s 9 seconds after he begins. What is his acceleration? What distance did he cover?

Answers

$Acceleration=\frac { Final\quad Velocity\quad -\quad Initial\quad Velocity }{ Time } \n$

Therefore:

$Acceleration=\frac { 11.1\quad m/s\quad -\quad 0\quad m/s }{ 9\quad seconds } \n \n =\frac { 11.1\quad m/s }{ 9\quad seconds } \n \n =1.23\quad m/{ s }^( 2 )\quad (2\quad decimal\quad places)$

As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.

Explanation:

Acceleration=

Time

FinalVelocity−InitialVelocity

Therefore:

\begin{gathered}Acceleration=\frac { 11.1\quad m/s\quad -\quad 0\quad m/s }{ 9\quad seconds } \\ \\ =\frac { 11.1\quad m/s }{ 9\quad seconds } \\ \\ =1.23\quad m/{ s }^{ 2 }\quad (2\quad decimal\quad places)\end{gathered}

Acceleration=

9seconds

11.1m/s−0m/s

9seconds

11.1m/s

=1.23m/s

(2decimalplaces)

As the runner was travelling for 9 seconds, he covered a distance of 99.9 metres. 9 seconds x 9 seconds = 81 seconds squared, and the runner covers roughly 1.23 metres in distance every second squared.

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I will mark brainliest if you get it right.