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Mumz [18]
3 years ago
15

Static discharges occur when

Physics
2 answers:
vredina [299]3 years ago
8 0
The sudden flow of electricity between two electrically charged objects caused by contact, an electrical short, or dielectric breakdown. A buildup of static electricity can be caused by tribocharging or by electrostatic induction.
Allushta [10]3 years ago
6 0

Answer: the air between two oppositely charged objects suddenly becomes charged

Explanation:

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Ill give brainlist help pleaseee asap
KIM [24]

Answer:

20. This is a transverse wave.

21. a is a Crest

b is the wavelength

c is the amplitude

d is the trough

e isthe amplitude

f is the wavelength

g is time

Explanation:

7 0
3 years ago
An ocean wave is an example of a(n) _____ wave form
bearhunter [10]
An ocean is an example of a wave form
3 0
4 years ago
Read 2 more answers
How much energy is stored in the electric field of a 50-μm-diameter cell with a 7.0-nm-thick cell wall whose dielectric constant
Anastaziya [24]
:<span>  </span><span>Under the assumption that a cell is made up of two concentric spheres you find the surface are of the inside sphere which will be your A. 

You already have your separation and dielectric constant so just use the formula you stated towards the end of your question and you get 8.93x10^-11 Farads which is about 89pF</span>
3 0
3 years ago
Read 2 more answers
​A ​very ​long ​insulating ​cylinder ​of ​charge ​of ​radius ​2.50 ​cm ​carries ​a ​uniform ​linear ​density ​of ​15.0 ​nC/m. ​(
Pavlova-9 [17]

Answer:

r2 = 2.401557  cm

distance = 0.10 cm

Explanation:

given data

​radius ​= 2.50 ​cm

​density ​= ​15.0 ​nC/m

voltmeter ​read =  ​175

solution

we know here potential difference that is express as

ΔV = \frac{\lambda }{2\pi \epsilon _o} ln\frac{r2}{r1}     ...........1

so here

ln\frac{r2}{r1} = 2\pi \epsilon _o \times \frac{\triangle V}{\lambda }  

as here \lambda is linear charge density  

\frac{r2}{r1} = e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}  

r2 = r1 × e^{2\pi \epsilon _o \times \frac{\triangle V}{\lambda }}  

r2 = 2.40 × e^{2\pi 8.85\times 10^{-12} \times \frac{175}{15\times 10^{-6} }}  

r2 = 2.401557  cm

and

here distance above surface will be

distance = r1 - r2

distance =  2.50 - 2.40

distance = 0.10 cm

4 0
3 years ago
Two flat conductors are placed with their inner faces separated by 17 mm. If the surface charge density on one of the inner face
blagie [28]

Answer:

Explanation:

electric field between plates

=  surface charge density / ε₀

= 108 x 10⁻¹² / 8.85 x 10⁻¹² (ε₀ = 8.85 x 10⁻¹² )

= 12.20 V/ m

Electric potential diff

= electric field x distance between plates

=  12.20 x 17 x 10⁻³ V

= 0.207 Volt

4 0
4 years ago
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