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3 years ago

Brainly app · Installed

Physics

1 answer:

Sauron [17]3 years ago

6 0

I need a picture plz I don’t know what to answer.

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A light wave moves from a medium where particles of matter are closely packed to a medium made of very little matter. How do you

shutvik [7]

The wavelength will chang along with the frequency, I think. I don't recall why it would lose energy...

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3 years ago

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Positive charge Q is distributed uniformly along the x-axis from x = 0 to x = a. A positive point charge q is located on the pos

ANTONII [103]

Answer:

(a): $k\dfrac{Q}{r(a+r)}.$ .

(b): 0

(c): $k\dfrac{qQ}{r(a+r)}.$ .

(d): Along positive x axis.

Explanation:

Given that the charge is distributed uniformly on the rod, extended from $x=0$ to $x=a$ and having total charge $Q$ .

Let $\lambda$ be the linear charge density of the rod, such that, $\lambda = \dfrac Qa.$

The electric field due to a charge $q$ at a point $r$ distance away is given by

$E=\dfrac{kq}{r^2}$

where $k$ is the Coulomb's constant whose value is $9\times 10^9] Nm^2/C^2$ .

Now consider a small line element of the rod of length $dx$ at distance $r$ from $x =a+r$ .

The electric field at point $x = a + r$ due to this element is given by

$dE = \dfrac{k\ dq}{x^2}$

$dq$ is the charge on the line charge, then,

$\lambda = \dfrac{dq}{dx}\\\Rightarrow dq=\lambda dx$

Using this value,

$dE = \dfrac{k\lambda\ dx}{x^2}.$

The electric field due to the whole rod is given by

$E=\int dE\\=\int\limits_{r}^{a+r} \dfrac{k\lambda\ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} \dfrac{ dx}{x^2}\\=k\lambda\int\limits^{a+r}_{r} x^{-2}\ dx\\=k\lambda \left (-x^{-1} \right )\limits^r_{a+r}\ dt\\=-k\lambda \left ( \dfrac1{a+r}-\dfrac 1r\right ) \\=-k\lambda \left ( \dfrac{r-(a+r)}{r(a+r)}\right ) \\=k\lambda \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{a} \left ( \dfrac{a}{r(a+r)}\right )\\=k\dfrac{Q}{r(a+r)}.$

This electric is along the x axis only.

(a):

The x component of the electric field at this point = $k\dfrac{Q}{r(a+r)}.$ .

(b):

The y component of the electric field at this point = 0.

(c):

The electric field and the electric force are related as

$F=qE.$

The magnitude of the force that this rod exerts on the charge q = $k\dfrac{qQ}{r(a+r)}.$ .

(d):

The direction of this force is along the positive x axis.

5 0

4 years ago

Part 1

xz_007 [3.2K]

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ .

<h3>Determination of the coordinates of the center of gravity</h3>

Let suppose that each square has an <em>uniform</em> density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the <em>entire</em> system is found by applying definition of <em>weighted</em> averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$ , $W_{2}$ , $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$ . $\blacksquare$

To learn more on center of gravity, we kindly invite to check this verified question: brainly.com/question/20662119

6 0

2 years ago

What is all matter made up of?

aliya0001 [1]

All matter is made up of atoms

3 0

4 years ago

Read 2 more answers

Cart A, with a mass of 0.20 kg, travels on a horizontal air trackat 3.0m/s and hits cart B, which has a mass of 0.40 kg and is i

inysia [295]

Answer:

$\large \boxed{\text{C. 2.3 m/s}}$

Explanation:

Data:

$m_{\text{A} } = \text{0.20 kg};\,v_{\text{Ai}} = \text{3.0 m/s}\\m_{\text{B} } = \text{0.40 kg};\,v_{\text{Bi}} = \text{2.0 m/s}\\$

Calculation:

This is a perfectly inelastic collision. The two carts stick together after the collision and move with a common final velocity.

The conservation of momentum equation is

$\begin{array}{rcl}m_{\text{A}}v_{\text{Ai}} +m_{\text{B}} v_{\text{Bi}}&=&(m_{\text{A}} + m_{\text{B}})v_{\text{f}}\\0.20\times 3.0 + 0.40\times 2.0 & = & (0.20 + 0.40)v_{\text{f}}\\0.60 + 0.80 & = & 0.60v_{\text{f}}\\1.40 & = & 0.60v_{\text{f}}\\v_{\text{f}}&=& \dfrac{1.40}{0.60}\\\\& = & \textbf{2.3 m/s}\\\end{array}\\\text{The centre of mass has a velocity of $\large \boxed{\textbf{2.3 m/s}}$}$

3 0

3 years ago