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svetoff [14.1K]

3 years ago

8

A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on

the xx axis is the net electric field zero? (Be careful to keep track of the direction of the electric field of each particle.)

1 answer:

svet-max [94.6K]3 years ago

5 0

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}$

Put the value into the formula

$\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}$

$\dfrac{27}{x^2}=\dfrac{6}{(1-x)^2}$

$\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}$

$\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}$

$\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1$

$x=0.32\ m$

Hence, The position on the x axis is 0.32 m.

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joja [24]

Answer:

Mechanical advantage of pulleys = 3.47 (Approx)

Explanation:

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4 years ago

A model rocket rises with constant acceleration to a height of 4.2 m, at which point its speed is 27.0 m/s. How much time does i

geniusboy [140]

Answers:

a) $t=0.311 s$

b) $a=86.847 m/s^{2}$

c) $y=1.736 m$

d) $V=17.369 m/s$

Explanation:

For this situation we will use the following equations:

$y=y_{o}+V_{o}t+\frac{1}{2}at^{2}$ (1)

$V=V_{o} + at$ (2)

Where:

$y$ is the <u>height of the model rocket at a given time</u>

$y_{o}=0$ is the i<u>nitial height </u>of the model rocket

$V_{o}=0$ is the<u> initial velocity</u> of the model rocket since it started from rest

$V$ is the <u>velocity of the rocket at a given height and time</u>

$t$ is the <u>time</u> it takes to the model rocket to reach a certain height

$a$ is the <u>constant acceleration</u> due gravity and the rocket's thrust

<h2>a) Time it takes for the rocket to reach the height=4.2 m</h2>

The average velocity of a body moving at a constant acceleration is:

$V=\frac{V_{1}+V_{2}}{2}$ (3)

For this rocket is:

$V=\frac{27 m/s}{2}=13.5 m/s$ (4)

Time is determined by:

$t=\frac{y}{V}$ (5)

$t=\frac{4.2 m}{13.5 m/s}$ (6)

Hence:

$t=0.311 s$ (7)

<h2>b) Magnitude of the rocket's acceleration</h2>

Using equation (1), with initial height and velocity equal to zero:

$y=\frac{1}{2}at^{2}$ (8)

We will use $y=4.2 m$ :

$4.2 m=\frac{1}{2}a(0.311)^{2}$ (9)

Finding $a$ :

$a=86.847 m/s^{2}$ (10)

<h2>c) Height of the rocket 0.20 s after launch</h2>

Using again $y=\frac{1}{2}at^{2}$ but for $t=0.2 s$ :

$y=\frac{1}{2}(86.847 m/s^{2})(0.2 s)^{2}$ (11)

$y=1.736 m$ (12)

<h2>d) Speed of the rocket 0.20 s after launch</h2>

We will use equation (2) remembering the rocket startted from rest:

$V= at$ (13)

$V= (86.847 m/s^{2})(0.2 s)$ (14)

Finally:

$V=17.369 m/s$ (15)

5 0

3 years ago

If gravity did not affect the pain of a horizontally thrown ball the ball would

Drupady [299]

If gravity had no effect on a ball after you threw it ... and there also
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That's the heart and soul of Newton's laws of motion ... any object
keeps moving at the same speed, and in a straight line in the same
direction, until a force acts on it to change its speed or direction.\

If you threw the ball horizontally, then it would keep moving in the
same direction you threw it. But don't forget: The Earth is not flat.
The Earth is a sphere. So, as the ball kept going farther and farther
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4 0

3 years ago

Read 2 more answers

A particle moves with acceleration function a(t) = 5 + 4t - 2t^2. Its initial velocity v(0) = 3 m/s and its initial displacement

eimsori [14]

Answer:

Its position after 4 seconds is 62 meters.

Explanation:

It is given that,

The acceleration of the particle is given by equation :

$a(t)=5+4t-2t^2$

Also, $a=\dfrac{dv}{dt}$

$v=\int\limits {a.dt}$

$v=\int\limits {(5+4t-2t^2).dt}$

$v=5t+2t^2-\dfrac{2}{3}t^3+c$

At t = 0, $v(0)=3\ m/s$ . So, c = 3

$v=5t+2t^2-\dfrac{2}{3}t^3+3$

Also, $v=\dfrac{ds}{dt}$ , s is the position

$s=\int\limits {v.dt}$

$s=\int\limits {(5t+2t^2-\dfrac{2}{3}t^3+3).dt}$

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+c'$

At t = 0, $s(0)=10\ m$ . So, c' = 10

$s=\dfrac{5}{2}t^2+\dfrac{2}{3}t^3-\dfrac{t^4}{6}+3t+10$

At t = 4 s

$s=\dfrac{5}{2}(4)^2+\dfrac{2}{3}(4)^3-\dfrac{(4)^4}{6}+3(4)+10$

s = 62 m

So, at t = 4 seconds the position of the particle is 62 meters. Hence, this is the required solution.

5 0

3 years ago

Help!!!!!!!!!!!!!!!!!!

GREYUIT [131]

I think wind is anther answer

5 0

3 years ago