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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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At what percentage of their full-grown weight are elephants born?

solmaris [256]

New-born elephants<span> are </span>born<span> with an incredible mass of 77-113 kg. But they </span>weight<span> only 4% of an </span>adult<span> female's </span>weight<span> and only 2% of an </span>adult<span> male's. New-born may consume 11.4 ... </span>

7 0

3 years ago

A big olive (* - 0.50 kg) lies at the origin of an xy coordinate system, and a big BrazlI nut (M - 1.5^kg) lie^s at the point (1

Afina-wow [57]

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ .

<h3>Procedure - Estimation of the displacement of the center of mass of the olive</h3>

In this question we should apply the definition of center of mass and difference between the coordinates for <em>dynamic</em> ( $\vec r$ ) and <em>static</em> conditions ( $\vec r_{o}$ ) to estimate the displacement of the center of mass of the olive ( $\overrightarrow{\Delta r}$ ):

$\vec r - \vec r_{o} = \left[\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot(m_{i}\cdot g + F_{i, x})}{\Sigma \limits_{i =1}^{2}(F_{i,x}+m_{i}\cdot g)} ,\frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot(m_{i}\cdot g + F_{i, y})}{\Sigma \limits_{i =1}^{2}(F_{i,y}+m_{i}\cdot g)} \right]-\left(\frac{\Sigma\limits_{i=1}^{2}r_{i,x}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}, \frac{\Sigma\limits_{i=1}^{2}r_{i,y}\cdot m_{i}\cdot g}{\Sigma \limits_{i= 1}^{2} m_{i}\cdot g}\right)$ (1)

Where:

$r_{i, x}$ - x-Coordinate of the i-th element of the system, in meters.
$r_{i,y}$ - y-Coordinate of the i-th element of the system, in meters.
$F_{i,x}$ - x-Component of the net force applied on the i-th element, in newtons.
$F_{i,y}$ - y-Component of the net force applied on the i-th element, in newtons.
$m_{i}$ - Mass of the i-th element, in kilograms.
$g$ - Gravitational acceleration, in meters per square second.

If we know that $\vec r_{1} = (0, 0)\,[m]$ , $\vec r_{2} = (1, 2)\,[m]$ , $\vec F_{1} = (0, 3)\,[N]$ , $\vec F_{2} = (-3, -2)\,[N]$ , $m_{1} = 0.50\,kg$ , $m_{2} = 1.50\,kg$ and $g = 9.807\,\frac{kg}{s^{2}}$ , then the displacement of the center of mass of the olive is:

<h3>Dynamic condition $\vec{r} = \left[\frac{(0)\cdot (0.50)\cdot (9.807)+(0)\cdot (0) + (1)\cdot (1.50)\cdot (9.807) + (1)\cdot (-3)}{(0.50)\cdot (9.807) + 0 + (1.50)\cdot (9.807)+(-3)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (0)\cdot (3) + (2)\cdot (1.50)\cdot (9.807) +(2) \cdot (-2)}{(0.50)\cdot (9.807) + (3)+(1.50)\cdot (9.807)+(-2)} \right]$ $\vec r = (0,704, 1.233)\,[m]$ </h3>

<h3>Static condition</h3><h3> $\vec{r}_{o} = \left[\frac{(0)\cdot (0.50)\cdot (9.807) + (1)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807) + (1.50)\cdot (9.807)}, \frac{(0)\cdot (0.50)\cdot (9.807) + (2)\cdot (1.50)\cdot (9.807)}{(0.50)\cdot (9.807)+(1.50)\cdot (9.807)} \right]$ </h3><h3> $\vec r_{o} = \left(0.75, 1.50)\,[m]$ </h3><h3 /><h3>Displacement of the center of mass of the olive</h3>

$\overrightarrow{\Delta r} = \vec r - \vec r_{o}$

$\overrightarrow{\Delta r} = (0.704-0.75, 1.233-1.50)\,[m]$

$\overrightarrow{\Delta r} = (-0.046, -0.267)\,[m]$

The <em>estimated</em> displacement of the center of mass of the olive is $\overrightarrow{\Delta r} = -0.046\,\hat{i} -0.267\,\hat{j}\,[m]$ . $\blacksquare$

To learn more on center of mass, we kindly invite to check this verified question: brainly.com/question/8662931

3 0

2 years ago

Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume

Mamont248 [21]

Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 490 km

a = Acceleration

g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s$

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

3 0

4 years ago

Read 2 more answers

Three capacitors with individual capacitances of 2 μF, 5 μF, and 10 μF

rodikova [14]

Answer:

1.25C

Explanation:

When capacitance is in series we add them like this: 1/Ctotal = 1/C1 +1/C2 + 1/C3.....

1/C = 1/2 + 1/5 + 1/10 = 5 + 2 + 1/10 = 8/10

C = 10/8 = 1.25

Capacitance = Charge/potential difference(Q/V)

1.25 = Charge/12

Total charge = 1.25 ×12 =15coulombs

5 0

3 years ago

A rocket is headed away from earth at a speed of 0.8c. the rocket fires a missile at a speed of 0.7c (the missile is aimed away

dsp73

To solve this problem, let us consider that the Earth is the origin, the initial reference point. Therefore the speed of rocket plus the missile would be 0.8 C

Now after the rocket had moved away from Earth, it fired a missile at a speed of 0.7 C. Now the reference made to this is relative to the rocket. We have established that our initial reference point is the Earth, therefore the real speed of the missile with reference to Earth is:

Speed of missile relative to Earth = 0.8 C + 0.7 C

Speed of missile relative to Earth = 1.5 C

Answer is:

8 0

3 years ago