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3 years ago

Compare and contrast an electric discharge with an electric current

Physics

1 answer:

vlabodo [156]3 years ago

3 0

The electric current is the number of electrons or charges passing through a conductor per units of time. 1 ampere = 1 coulombs / 1 second. Electric discharge is the consequence of disrupting an isolator (for example, air), thus causing a sudden flow of electricity

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20 points please answer

Anna [14]

Another way to test your question is to build your own miniature buildings. Depending on how in-depth you go, building could get a little pricey, but if you keep it basic there shouldn't be a problem. Decide on a certain number of foundations to test [maybe 3 or so] and try simulating an earthquake.
Hope this helps!

6 0

3 years ago

A man jogs at a speed of 1.6 m/s. His dog

FromTheMoon [43]

I believe it is
1.6x=2.7(x-1.8)
1.1x=2.7*1.8
x~4.4
4.4*1.6
~7.1m

5 0

3 years ago

Two boats leave the same port at the same time, with boat A traveling north at 15 knots (nautical miles per hour) and boat B tra

Mrrafil [7]

Answer:

The chance in distance is 25 knots

Explanation:

The distance between the two particles is given by:

$s^2 = (x_A - x_B)^2+(y_A - y_B)^2$ (1)

Since A is traveling north and B is traveling east we can say that their displacement vector are perpendicular and therefore (1) transformed as:

$s^2 = x_B^2+y_A^2$ (2)

Taking the differential with respect to time:

$\displaystyle{2s\frac{ds}{dt}= 2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt}}$ (3)

where $\displaystyle{\frac{dx_B}{dt}}=v_B$ and $\displaystyle{\frac{dx_A}{dt}}=v_A$ are the respective given velocities of the boats. To find $s$ and $x_B$ we make use of the given position for A, $y_A=30$ , the Pythagoras theorem and the relation between distance and velocity for a movement with constant velocity.

$\displaystyle{y_A = v_A\cdot t\rightarrow t = \frac{y_A}{v_A}=\frac{30}{15}=2 h$

with this time, we know can now calculate the distance at which B is:

$\displaystyle{x_B = v_B\cdot t= 20 \cdot 2 = 40\ nmi$

and applying Pythagoras:

$\displaystyle{s = \sqrt{x_B^2+y_A^2}=\sqrt{30^2 + 40^2}=\sqrt{2500}=50}$

Now substituting all the values in (3) and solving for $\displaystyle{\frac{ds}{dt} }$ we get:

$\displaystyle{\frac{ds}{dt} = \frac{1}{2s}(2x_B\frac{dx_B}{dt}+2y_A\frac{dy_A}{dt})}\\\displaystyle{\frac{ds}{dt} = 25 \ knots}$

4 0

3 years ago

You have discovered a planet that is one-quarter the radius of Earth (Rp = 1/4R⊕) and one-half as massive (Mp = 1/2M⊕). How does

ExtremeBDS [4]

It's 8 times as much as before.

4 0

2 years ago

State the methods of nuclear fission waste disposal and the suitability of each method.

Alexeev081 [22]

Most low-level radioactive waste (LLW) is typically sent to land-based disposal immediately following its packaging for long-term management. This means that for the majority (~90% by volume) of all of the waste types produced by nuclear technologies, a satisfactory disposal means has been developed and is being implemented around the world.


Radioactive wastes are stored so as to avoid any chance of radiation exposure to people, or any pollution.The radioactivity of the wastes decays with time, providing a strong incentive to store high-level waste for about 50 years before disposal.Disposal of low-level waste is straightforward and can be undertaken safely almost anywhere.Storage of used fuel is normally under water for at least five years and then often in dry storage.Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.

I suggest this site on this subject http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/nuclear-wastes/storage-and-dispo...

5 0

3 years ago