Physics, 23.02.2021 01:00 nicjeyr

How do I get through this week?!?!?!

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Answer from: joshualoz5414

lol

Explanation:

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

Answer from: Quest

Answerthe force between the charges will be f/9 (one-ninth of force f)explanationcoulomb's law relates the magnitudes two point charges q₁ and q₂, the distance, r, between the charges, and the electrostatic force, f, between the two charges: $f = \dfrac{k q_1 q_2}{r^2}$ k represents coulomb's constant, k = 8.99×10⁹ n·m²/c².note that if we have r = 3r, then $\begin{aligned} f & = \dfrac{k q_1 q_2}{r^2} \\ f_{3r} & = \dfrac{k q_1 q_2}{(3r)^2} \\ & = \dfrac{k q_1 q_2}{3^2 \cdot r^2} \\ & = \frac{1}{9} \cdot \boxed{\dfrac{k q_1 q_2}{r^2} } \\ & = \frac{1}{9} f & & \left(\text{\footnotesize since $ f = \tfrac{k q_1 q_2}{r^2}$}\right) \end{aligned}$

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