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4 years ago

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It takes about 7000 years for light from the eagle nebula to reach us on earth. using this information, and the speed of light g

iven above, calculate how far the eagle nebula is from us in kilometres
note: the speed of light is 3x10^8

1 answer:

miss Akunina [59]4 years ago

3 0

Speed of light in kilometers is 1,080,000,000

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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

You carry a 50 N backpack for 10000 meters in 500 secs. How much power did<br> you use?

Klio2033 [76]

<h2>for cara a 50+10000=9768m in for 500×9768=586858metres </h2>

<h2>R:1.0000000000.</h2>

5 0

4 years ago

A uniform electric field with a magnitude of 125 000 N/C passes through a rectangle with sides of 2.50 m and 5.00 m. The angle b

jeyben [28]

Answer:

$6.60\cdot 10^5 Nm^2/C$

Explanation:

The electric flux through the rectangle is given by

$\Phi = E A cos \theta$

where

E is the electric field strength

A is the area of the rectange

$\theta$ is the angle between the direction of the electric field and of the vector normal to the plane of the rectangle

In this problem we have

E = 125 000 N/C

The area of the rectangle is

$A=2.50 m \cdot 5.00 m=12.5 m^2$

and the angle is

$\theta=65.0^{\circ}$

so, the electric flux is

$\Phi = (125,000 N/C)(12.5 m^2)(cos 65^{\circ})=6.60\cdot 10^5 Nm^2/C$

8 0

4 years ago

In your lesson guide what is on modification did they list if you do not own or have weights? Question 4 options: Milk jugs fill

Katen [24]

Answer:

I dont have any context, but my best guess is it will be all of the above.

3 0

3 years ago

What force does it take to accelerate a 9.2 kg object 7.0 m/s^2?

Ulleksa [173]

Answer:

<h2>64.4 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question

mass = 9.2 kg

acceleration = 7 m/s²

We have

force = 9.2 × 7 = 64.4

We have the final answer as

<h3>64.4 N</h3>

Hope this helps you

5 0

3 years ago