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2 years ago

Help, please. I am not sure what to do.

Physics

1 answer:

miss Akunina [59]2 years ago

3 0

Answer:

option D) -3m

Explanation:

if 6m is diplaced by -3m then it would be -3+6=3m

feel free to ask if you are confused

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Which accurately explains concave and convex lenses?

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Answer:

its C! I just finished the test on edg :)

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3 years ago

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A spring that is stretched by hanging a 5kg mass. It's equilibrium length was 0.5 meters. Now the length of the spring is 1.6 me

gtnhenbr [62]

45 N/m as attached, please comment

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3 years ago

Charge q1 = +2.00 μC is at -0.500 m along the x axis. Charge q2 = -2.00 μC is at 0.500 m along the x axis. Charge q3 = 2.00 μC i

Kobotan [32]

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons.

<h3>How to calculate the electrical force experimented on a particle</h3>

The vector <em>position</em> of each particle respect to origin are described below:

$\vec r_{1} = (-0.500, 0)\,[m]$

$\vec r_{2} = (+0.500, 0)\,[m]$

$\vec r_{3} = (0, +0.500)\,[m]$

Then, distances of the former two particles particles respect to the latter one are found now:

$\vec r_{13} = (+0.500, +0.500)\,[m]$

$r_{13} = \sqrt{\vec r_{13}\,\bullet\,\vec r_{13}} = \sqrt{(0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{13} =\frac{\sqrt{2}}{2}\,m$

$\vec r_{23} = (-0.500, +0.500)\,[m]$

$r_{23} = \sqrt{\vec r_{23}\,\bullet \,\vec r_{23}} = \sqrt{(-0.500\,m)^{2}+(0.500\,m)^{2}}$

$r_{23} =\frac{\sqrt{2}}{2}\,m$

The resultant force is found by Coulomb's law and principle of superposition:

$\vec R = \vec F_{13}+\vec F_{23}$ (1)

Please notice that particles with charges of <em>same</em> sign attract each other and particles with charges of <em>opposite</em> sign repeal each other.

$\vec R = \frac{k\cdot q_{1}\cdot q_{3}}{r_{13}^{2}}\cdot \vec u_{13} +\frac{k\cdot q_{2}\cdot q_{3}}{r_{23}^{2}}\cdot \vec u_{23}$ (2)

Where:

$k$ - Electrostatic constant, in newton-square meters per square Coulomb.
$q_{1}$ , $q_{2}$ , $q_{3}$ - Electric charges, in Coulombs.
$r_{13}$ , $r_{23}$ - Distances between particles, in meters.
$\vec u_{13}$ , $\vec u_{23}$ - Unit vectors, no unit.

If we know that $k = 8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}}$ , $q_{1} = 2\times 10^{-6}\,C$ , $q_{2} = 2\times 10^{-6}\,C$ , $q_{3} = 2\times 10^{-6}\,C$ , $r_{13} =\frac{\sqrt{2}}{2}\,m$ , $r_{23} =\frac{\sqrt{2}}{2}\,m$ , $\vec u_{13} = \left(-\frac{\sqrt{2}}{2}, - \frac{\sqrt{2}}{2} \right)$ and $\vec u_{23} = \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$ , then the vector force on charge $q_{3}$ is:

$\vec R = \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + \frac{\left(8.988\times 10^{9}\,\frac{N\cdot m^{2}}{C^{2}} \right)\cdot (2\times 10^{-6}\,C)\cdot (2\times 10^{-6}\,C)}{\left(\frac{\sqrt{2}}{2}\,m \right)^{2}} \cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)$

$\vec R = 0.072\cdot \left(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right) + 0.072\cdot \left(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2} \right)\,[N]$

$\vec R = 0.072\cdot \left(0, -\sqrt{2}\right)\,[N]$

And the magnitude of the <em>electrical</em> force on charge $q_{3}$ ( $R$ ), in newtons, due to the others is found by Pythagorean theorem:

$R = 0.102\,N$

The magnitude of <em>electrical</em> force on charge $q_{3}$ due to the others is 0.102 newtons. $\blacksquare$

To learn more on Coulomb's law, we kindly invite to check this verified question: brainly.com/question/506926

8 0

2 years ago

A 20 cm long spring is attached to a wall. The spring stretches to a length of 22 cm when you pull on it with a force of 100 n.

kenny6666 [7]

Answer:

5000 N/m

Explanation:

Hooke's law for the spring is

$F = k \Delta x$

where here we have

F = 100 N is the force applied to the spring

k is the spring constant

$\Delta x = 22 cm - 20 cm = 2 cm = 0.02 m$ is the stretching of the spring with respect to its equilibrium position

Solving the equation for k, we find the spring constant:

$k=\frac{F}{\Delta x}=\frac{100 N}{0.02 m}=5000 N/m$

8 0

4 years ago

The chart shows the times and accelerations for three drivers.

aliina [53]

The lists from greatest to lowest change in velocity.Dustin → Diego → Kira. Option A is correct.

<h3 /><h3>What is acceleration?</h3>

The rate of change of velocity with respect to time is known as acceleration. According to Newton's second law, the eventual effect of all forces applied to a body is its acceleration.

The acceleration is obtained one by one from the given data as the ratio of the velocity to the time.

For the acceleration of each, you can refer to the excel table attached below.

The velocity changes are listed from largest to least. Dustin and Diego Kira.

Hence, option A is the right answer.

To learn more about acceleration, refer to the link;

brainly.com/question/2437624

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6 0

2 years ago