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Bingel [31]
3 years ago
9

Describe the phenomenon of lightning?​

Physics
1 answer:
masha68 [24]3 years ago
3 0

Lightning is a naturally occurring electrostatic discharge during which two electrically charged regions in the atmosphere or ground temporarily equalize themselves, causing the instantaneous release of as much as one gigajoule of energy.
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In which phase is the full illuminated face of the Moon visible on Earth? A) New moon Eliminate B) Full moon C) Waning gibbous D
Nitella [24]
Full moon

New cant be seen

Gibbous is 3/4

Crescent is 1/4
4 0
3 years ago
How much would a pair of 0.5 kg shoes weigh on Earth? (Include units in<br> your answer) *
Stells [14]

Answer:

1.1 lbs

Explanation:

To convert kg to lbs you multiply kilograms by 2.2. So 0.5kg × 2.2 equals to 1.1 lbs

4 0
3 years ago
How can tectonic activity change the climate of the earth?
soldi70 [24.7K]
If you really want to find out the answer just pray about it and it will come to you..
3 0
3 years ago
The state of strain at a point is plane strain with εx = ε0, εy = –2ε0, γxy = 0, where ε0 is a positive constant. What is the no
Marat540 [252]

Answer:

The normal strain along an axis oriented 45° from the positive x axis in the clockwise direction is -ε₀/2

Explanation:

Given that

\epsilon_{x}=\epsilon_{o}\\\\\epsilon_{y}=-2\epsilon_{o}\\\\\gamma_{xy}=0\\\\\theta=-45^{o}\\\\\epsilon_{x_{1}}=?

From equation of normal strain in x direction:

\epsilon_{x_{1}}=\epsilon_{x}cos^{2}\theta+\epsilon_{y}sin^{2}\theta+\gamma_{xy{ sin\theta cos\theta

Substituting the values:

\epsilon_{x_{1}}=\epsilon_{o}cos^{2}(-45)-2\epsilon_{o}sin^{2}(-45)+0\\\\\epsilon_{x_{1}}=\frac{\epsilon_{o}}{2}-2\frac{\epsilon_{o}}{2}\\\\\epsilon_{x_{1}}=-\frac{\epsilon_{o}}{2}

6 0
3 years ago
A 594 Ω resistor, an uncharged 1.3 μF capacitor, and a 6.53 V emf are connected in series. What is the current in milliamps afte
ivanzaharov [21]

Answer:

6.88 mA

Explanation:

Given:

Resistance, R = 594 Ω

Capacitance = 1.3 μF

emf, V = 6.53 V

Time, t = 1 time constant

Now,

The initial current, I₀ = \frac{\textup{V}}{\textup{R}}

or

I₀ = \frac{\textup{6.53}}{\textup{594}}

or

I₀ = 0.0109 A

also,

I = I_0[1-e^{-\frac{t}{\tau}}]

here,

τ = time constant

e = 2.717

on substituting the respective values, we get

I = 0.0109[1-e^{-\frac{\tau}{\tau}}]

or

I = 0.0109[1-2.717^{-1}]

or

I = 0.00688 A

or

I = 6.88 mA

5 0
3 years ago
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