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3 years ago

Can you fill in the blanks for me

Physics

1 answer:

sweet [91]3 years ago

8 0

I’m pretty sure the second blank is solid to a liquid and the third blank is melting point

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As the shuttle bus comes to a sudden stop to avoid hitting a dog, it accelerates uniformly at -4.1 m/s^2 as it slows from 9.0 m/

zhuklara [117]

Acceleration = (change in speed) / (time for the change)

- 4.1 m/s² = (-9 m/s) / (time for the change)

Time for the change = (-9 m/s) / (-4.1 m/s²) = 2.2 seconds

3 0

4 years ago

Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

6 0

3 years ago

A puck of mass 0.70 kg approaches a second, identical puck that is stationary on frictionless ice. The initial speed of the movi

natali 33 [55]

Answer:

$v_1 = \ 5.196 \frac{m}{s}$

$v_2 = 3 \frac{m}{s}$

Explanation:

For this problem, we just need to remember conservation of momentum, as there are no external forces in the horizontal direction:

$\vec{p}_i = \vec{p}_f$

where the suffix i means initial, and the suffix f means final.

The initial momentum will be:

$\vec{p}_i = m_1 \ \vec{v}_{1_i} + m_2 \ \vec{v}_{2_i}$

as the second puck is initially at rest:

$\vec{v}_{2_i} = 0$

Using the unit vector $\vec{i}$ pointing in the original line of motion:

$\vec{v}_{1_i} = 6.0 \frac{m}{s} \hat{i}$

$\vec{p}_i = 0.70 \ kg \ 6.0 \frac{m}{s} \ \hat{i} + 0.70 \ kg \ 0$

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

So:

$\vec{p}_i = 4.2 \ \frac{kg \ m}{s} \ \hat{i} = \vec{p}_f$

$\vec{p}_f = 4.2 \ \frac{kg \ m}{s} \ \hat{i}$

Knowing the magnitude and directions relative to the x axis, we can find Cartesian representation of the vectors using the formula

$\ \vec{A} = | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )$

So, our velocity vectors will be:

$\vec{v}_{1_f} = v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ )$

$\vec{v}_{2_f} = v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

We got

$\vec{p}_f = 0.7 \ kg \ \vec{v}_{1_f} + 0.7 \ kg \ \vec{v}_{2_f}$

$4.2 \ \frac{kg \ m}{s} \ \hat{i} = 0.7 \ kg \ v_1 \ ( \ cos(30 \°) \ , \ sin (30 \°) \ ) + 0.7 \ kg \ v_2 \ ( \ cos(-60 \°) \ , \ sin (-60 \°) \ )$

So, we got the equations:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ v_1 \ cos(30 \°) + 0.7 \ kg \ v_2 \ cos(-60 \°)$

and

$0 = 0.7 \ kg \ v_1 \ sin(30 \°) + 0.7 \ kg \ v_2 \ sin(-60 \°)$ .

From the last one, we get:

$0 = 0.7 \ kg \ ( v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°) )$

$0 = v_1 \ sin(30 \°) + \ v_2 \ sin(-60 \°)$

$v_1 \ sin(30 \°) = - \ v_2 \ sin(-60 \°)$

$v_1 = \ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) }$

and, for the first one:

$4.2 \ \frac{kg \ m}{s} = 0.7 \ kg \ ( v_1 \ cos(30 \°) + v_2 \ cos(60 \°) )$

$\frac{4.2 \ \frac{kg \ m}{s}}{ 0.7 \ kg} = v_1 \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = (\ v_2 \ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + v_2 \ cos(60 \°)$

$6 \ \frac{m}{s} = v_2 (\ \frac{sin(60 \°)}{ sin(30 \°) } ) \ cos(30 \°) + cos(60 \°)$

$6 \ \frac{m}{s} = v_2 * 2$

so:

$v_2 = 6 \ \frac{m}{s} / 2 = 3 \frac{m}{s}$

and

$v_1 = \ 3 \frac{m}{s} \ \frac{sin(60 \°)}{ sin(30 \°) }$

$v_1 = \ 5.196 \frac{m}{s}$

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4 years ago

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Zanzabum

A on that problem :)[email protected]

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False people should always wear gear

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