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3 years ago

What is an ideal machine?

1 answer:

5 0

An ideal machine is one in which no part of the input energy to the machine gets wasted and the whole input energy is converted into the useful work. The efficiency of such a machine is 100%.

i really hope this helps!

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Why do animals use echolocation? Check all that apply.

jenyasd209 [6]

To find food to find each other and to find their way

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4 years ago

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Buna!! ne pune le fizica intrebari despre legarea mixta!! Va rog , ma puteti ajuta ?? Ptr raspunsuri bune dau coroana!!!!!!

Iteru [2.4K]

Un solid covalent de rețea constă din atomi reținuți împreună de o rețea de legături covalente (perechi de electroni împărțite între atomi de electronegativitate similară) și, prin urmare, pot fi considerați o singură moleculă mare.

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3 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.

x = -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.

x = +x0:

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

What is an example of a wave that is not mechanical and how is it different?

vovikov84 [41]

Answer:

light is an example of a wave that is not mechanical .

it is different as it does not need material medium for its propagation

5 0

3 years ago

A woman who weighs 500 N stands on an 8 m long board that weighs 100 N. The board is supported at each end. The support force at

atroni [7]

Answer:

$1.6\; \rm m$ .

Explanation:

Let $x$ denote the distance (in meters) between the person and the right end of the board.

To keep the calculations simple, consider another unknown: let $y$ denote the support force (in Newtons) on the left end. The support force on the right end of this board would be $3 \, y$ (also in Newtons.)

Now there are two unknowns. At least two equations will be required for finding the exact solutions. For that, consider this board as a lever, but with two possible fulcrums. Refer to the two diagrams attached. (Not to scale.)

In the first diagram, the support at the left end of the board is considered as the fulcrum.
In the second diagram, the support at the right end of the board is considered as the fulcrum.

Calculate the torque in each situation. Note that are four external forces acting on this board at the same time. (Two support forces and two weights.) Why does each of the two diagrams show only three? In particular, why is the support force at each "fulcrum" missing? The reason is that any force acting on the lever at the fulcrum will have no direct impact on the balance between torques elsewhere on the lever. Keep in mind that the torque of each force on a lever is proportional to $r$ , the distance between the starting point and the fulcrum. Since that missing support force starts right at the fulcrum, its $r$ will be zero, and it will have no torque in this context.

Hence, there are three (non-zero) torques acting on the "lever" in each diagram. For example, in the first diagram:

The weight of the board acts at the center of the board, $(1/2) \times 8\; \rm m = 4\; \rm m$ from the fulcrum. This force will exert a torque of $\tau(\text{weight of board}) = 4\; \rm m \times (-100\; \rm N) = (-400\; \rm N \cdot m)$ on this "lever". The negative sign indicates that this torque points downwards.

The weight of the person acts at $x\;\rm m$ from the right end of the board, which is $(8 - x)\; \rm m$ from the fulcrum at the other end of this board. This force will exert a torque of $\tau(\text{weight of person}) = (8 - x)\; {\rm m \times (-500\; \rm N)} = (-500\, \mathnormal{(8 - x)})\; \rm N \cdot m$ on this "lever". This torque also points downwards.

The support on the right end of the board acts at $8\; \rm m$ from the fulcrum (i.e., the left end of this board.) This force will exert a torque of $\tau(\text{support, right}) = 8\; {\rm m} \times (3\, \mathnormal{y})\; {\rm N} = (24\, y)\; \rm N \cdot m$ on the "lever". This torque points upwards.

If the value of $x$ and $y$ are correct, these three torques should add up to zero. That is:

$\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\, (8 - x))}_{\text{person}} + \underbrace{24\, y}_{\text{support}} = 0$ .

That gives the first equation of this system. Similarly, a different equation can be obtained using the second diagram:

$\underbrace{(-400)}_{\text{board}} + \underbrace{(-500\,x)}_{\text{person}} + \underbrace{8\, y}_{\text{support}} = 0$ .

Combine these two equations into a two-by-two system. Solve the system for $x$ and $y$ :

$\left\lbrace\begin{aligned}&x = 1.6\\ &y = 150\end{aligned}\right.$ .

In other words, the person is standing at about $1.6\; \rm m$ from the right end of the board. The support force at the left end of the board is $150\; \rm N$ .

5 0

3 years ago