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3 years ago

Which formula is used to find an objects acceleration

Physics

2 answers:

wlad13 [49]3 years ago

8 0

Answer:

$a=\frac{F}{m}$

$a=\frac{v-u}{t}\\a=2\frac{s-ut}{t^2}\\a=\frac{v^2-u^2}{2s}$

Explanation:

The most common formula from which acceleration of an object can be determined is

F = ma

Where,

F = Force

m = Mass

a = Acceleration

Rearranging the equation we get

$\mathbf{a=\frac{F}{m}}$

There are equations also from which acceleration can be determined they are

$v=u+at\quad\\s=ut+{\tfrac {1}{2}}at^{2}\\v^{2}=u^{2}+2as$

Rearranging the equations we get

$\mathbf{a=\frac{v-u}{t}}$

$\mathbf{a=2\frac{s-ut}{t^2}}$

$\mathbf{a=\frac{v^2-u^2}{2s}}$

Where,

a = Acceleration

u = Initial velocity

v = Final velocity

s = Displacement

t = Time taken

RideAnS [48]3 years ago

7 0

I hope this helps you!

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3 years ago

Points P and Q are located at (0, 2, 4) and (-3, 1,5). Calculate

Akimi4 [234]

Answer:

a) The position vector of P is $\vec P =(0, 2,4)$ .

b) The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

Explanation:

a) The position vector of a point is the vector displacement from the origin to the location of the point. That is:

$\vec P = (0,2,4)-(0,0,0)$

$\vec P = (0-0, 2-0, 4-0)$

$\vec P =(0, 2,4)$

The position vector of P is $\vec P =(0, 2,4)$ .

b) First, we calculate the position vector of point Q:

$\vec Q = (-3,1,5)-(0,0,0)$

$\vec Q = (-3-0,1-0,5-0)$

$\vec Q =(-3,1,5)$

The distance vector from P to Q is define by the following vectorial expression:

$\overrightarrow{PQ} = \vec Q - \vec P$ (1)

$\overrightarrow{PQ} = (-3,1,5)-(0,2,4)$

$\overrightarrow{PQ} =(-3-0,1-2,5-4)$

$\overrightarrow{PQ} = (-3,-1,1)$

The distance vector from P to Q is $\overrightarrow{PQ} = (-3,-1,1)$ .

c) There are two approaches to calculate the distance between P and Q:

First Method - Pythagorean Theorem:

$\|\overrightarrow{PQ}\| = \sqrt{(-3)^{2}+(-1)^{2}+1^{2}}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

Second Method - Dot Product:

$\|\overrightarrow{PQ}\| = \sqrt{\overrightarrow{PQ}\,\bullet\,\overrightarrow{PQ}}$ (2)

$\|\overrightarrow{PQ}\| = \sqrt{(-3,-1,1)\,\bullet (-3,-1,1)}$

$\|\overrightarrow{PQ}\|=\sqrt{11}$

The distance between P and Q is $\|\overrightarrow{PQ}\|=\sqrt{11}$ .

d) To determine a vector parallel to PQ with a given magnitude is determined by the following expression:

$\vec v = \frac{k}{\|\overrightarrow{PQ}\|} \cdot \overrightarrow{PQ}$ (3)

Where $k$ is the scale factor.

If we know that $\overrightarrow{PQ} = (-3,-1,1)$ , $\|\overrightarrow{PQ}\|=\sqrt{11}$ and $k = 10$ , then the vector is:

$\vec v = \frac{10}{\sqrt{11}}\cdot (-3,-1,1)$

$\vec v = \left(-\frac{30}{\sqrt{11}},-\frac{10}{\sqrt{11}},\frac{10}{\sqrt{11}}\right)$

$\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$

A vector parallel to PQ with magnitude of 10 is $\vec v = \left(-\frac{30\sqrt{11}}{11},-\frac{10\sqrt{11}}{11}, \frac{10\sqrt{11}}{11} \right)$ .

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Question #14

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The decrease in energy in the hydrogen molecule is what allows its formation on Earth, but in stars the great energy of the explosion has a kinetic energy so great that electrons cannot bind to another atom, which is why hydrogen has a single atom.

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Learn more about the Hydrogen atom here:

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