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3 years ago

State a Newtowns second law

Physics

1 answer:

astraxan [27]3 years ago

5 0

Answer:

the second law states that the force F is the product of an object's mass and its acceleration a: F = m * a. For an external applied force, the change in velocity depends on the mass of the object.

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3 years ago

Select the statement that correctly completes the description of phase difference.

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Answer:

The difference in the phase angle between any two waves at any given position along the waves.

Explanation:

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$\Delta d=d\ sin\theta$

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3 years ago

Read 2 more answers

A particle of charge +2q is placed at the origin and particle of charge -q is placed on the x-axis at x = 2a. Where on the x-axi

zzz [600]

Answer:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Explanation:

We know that the electric force equation is:

$F=k\frac{q_{1}*q_{2}}{r^{2}}$

k is the electric constant $9*10^{9} Nm^{2}/C^{2}$
r is the distance between the particles
q1 and q2 are the particle

Now, we have three particles, the first one at x=0, the second one at x=2a and the third in some place between these two particle.

1. Let's find the electric force between the first particle and the third particle.

$F_{31}=k\frac{q_{3}*q_{1}}{r_{31}^{2}}$

$F_{31}=k\frac{q*2q}{r_{31}^{2}}$

$F_{31}=k\frac{2q^{2}}{r_{31}^{2}}$

r(31) is the distance between 3 and 1

2. Now, let's find the electric force between the third particle and the second particle.

$F_{32}=k\frac{q_{3}*q_{2}}{x_{32}^{2}}$

$F_{32}=k\frac{q*(-q)}{r_{32}^{2}}$

$F_{32}=-k\frac{q^{2}}{r_{32}^{2}}$

r(32) is the distance between 3 and 2.

Now, $r_{31}+r_{32}=2a$ or $r_{32}=2a-r_{31}$

The net force must be zero so:

$F_{31}+F_{32}=0[\tex][tex]k\frac{2q^{2}}{r_{31}^{2}}-k\frac{q^{2}}{r_{32}^{2}}=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{r_{32}^{2}})=0[\tex] [tex]kq^{2}(\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}})=0[\tex] It means that:[tex]\frac{2}{r_{31}^{2}}-\frac{1}{(2a-r_{31})^{2}}$

We just need to solve it for r(31)

$r_{31}^{2}=2(2a-r_{31})^{2}$

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

Therefore the distance from the origin will be:

$r_{31}=\frac{2a\sqrt{2}}{1+\sqrt{2}}$

I hope it helps you!

4 0

4 years ago

State a Newtowns second law​

State a Newtowns second law