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3 years ago

Is it possible for lightning to be used as everyday electricity? Is it safe, and or stable? Can lightning be multiplied?

Physics

2 answers:

Anettt [7]3 years ago

6 0

Answer

What the guy above said

Explanation:

PolarNik [594]3 years ago

4 0

Answer:

Though lightning can be harvested in a safe manor, it is not very practical in many areas. Not all places get lightning storms in high amounts, so those locations would not apply, but those who do see frequent lightning can indeed use it to power. Using tall metal beams can attract the lightning and can be escorted into a power converter much like the ones used for nuclear energy.

Explanation:

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3 years ago

A sound wave travels at 379 m/sec and has a wavelength of 8 meters. Calculate its frequency and period.

olga nikolaevna [1]

$\qquad\qquad\huge\underline{{\sf Answer}}$

Wavelength, Wavespeed and frequency depends on each other in the following way :

$\qquad \sf \dashrightarrow \: frequency = \dfrac{wave \: \: velocity}{wavelength}$

$\qquad \sf \dashrightarrow \: f = \dfrac{389}{8}$

$\qquad \sf \dashrightarrow \: f \approx 48.62 \: \: hertz$

And we know the Reciprocal relationship between frequency and period ~

So, let's find it's period •

$\sf{\qquad \sf \dashrightarrow \: t = \dfrac{8}{389}}$

$\sf{\qquad \sf \dashrightarrow \: t \approx 0.02 sec }$

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2 years ago

What is single kind of organism that can reproduce on its own?

Iteru [2.4K]

A cell will reproduce on its own

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3 years ago

Read 2 more answers

A machine

In-s [12.5K]

Answer:

Power_input = 85.71 [W]

Explanation:

To be able to solve this problem we must first find the work done. Work is defined as the product of force by distance.

$W = F*d$

where:

W = work [J] (units of Joules)

F = force [N] (units of Newton)

d = distance [m]

We need to bear in mind that the force can be calculated by multiplying the mass by the gravity acceleration.

Now replacing:

$W = (80*10)*3\\W = 2400 [J]$

Power is defined as the work done over a certain time. In this way by means of the following formula, we can calculate the required power.

$P=\frac{W}{t}$

where:

P = power [W] (units of watts)

W = work [J]

t = time = 40 [s]

$P = 2400/40\\P = 60 [W]$

The calculated power is the required power. Now as we have the efficiency of the machine, we can calculate the power that is introduced, to be able to do that work.

$Effic=0.7\\Effic=P_{required}/P_{introduced}\\P_{introduced}=60/0.7\\P_{introduced}=85.71[W]$

3 0

3 years ago