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3 years ago

Explain what voltage,current, resistance are and how they relate to Ohms law. Then give examples

1 answer:

3 0

Voltage is the difference in charge between two points.
Current is the rate the charge flows
Resistance is the tendency a material has to resist the flow of charge (current)
Combining voltage resistance and current Ohm developed the formula
V (Voltage)= I (Current) x R (Resistance)

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A train moves uniform velocity of 36km/h for 15 sec.find distance

vesna_86 [32]

Answer:

S = V × t

Explanation:

first convert 36km/h to m/s

36km/h=36*5/18=180/18=10m/s

d =s*t

=10*10=100m

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2 years ago

Suppose a magnetic reversal accoured today .

Brut [27]

They would have the magnetic parts of the rocks oriented differently. This leaves a trace that allow scientists to find out how often this actually happens (as far as I remember, it's on average once in 800 000 years)

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3 years ago

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Explain what differentiates the Earth’s crust and lithosphere.

Allisa [31]

Answer:

What is the difference between the crust and lithosphere? The crust (whether continental or oceanic) is the thin layer of distinctive chemical composition overlying the ultramafic upper mantle. ... The lithosphere is the rigid outer layer of the Earth required by plate tectonic theory.

Explanation:

Hope this helps!

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2 years ago

Read 2 more answers

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

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2 years ago

Describe the water cycle process starting from an afternoon thunderstorm. There are many different variations that could happen.

spin [16.1K]

Explanation:

The water cycle basically involves five steps:

evaporation and transpiration ⇄
condensation, ⇄
precipitation, ⇄
runoff, ⇄
infiltration ⇄

So when a thunderstorm occurs it helps in completing the precipitation process by enabling the release of water vapor stored up in the atmosphere to fall on the ground as rain.

After this, the water runoffs to the surface of the ground, on plants, into rocks, rivers, and lakes.

Next, the Infiltration process enables the water on the ground surface to enter the soil some of which becomes groundwater.

The cycle begins again as the evaporation and transpiration process begins, where the groundwater as a result of heat from the sun is taken back into the atmosphere, while water in plants by means of transpiration goes back into the atmosphere.

It then condenses and falls back as precipitation again.

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2 years ago