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stealth61 [152]
2 years ago
6

I have to write something here, so like hello and please help​

Physics
1 answer:
ExtremeBDS [4]2 years ago
5 0

Answer:

the extension would be less the new extension might be 3 cm

Explanation:

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List three devices that use electric current from batteries and three that use regular house current.
Bond [772]
Charger
counsels
TV

Fan
light
car
4 0
3 years ago
How many protons, neutrons, and electrons does a neutral atom of this element have? (round atomic mass to nearest whole number)
adelina 88 [10]

Answer:

The element "AI"  has:

Protons: 13 Neutrons: 14 Electrons: 13

Have a great day.

8 0
3 years ago
Read 2 more answers
Derive Isothermal process through ideal gas.(anyone plzz!!)​
zysi [14]

Answer:

Hey, bro here is the explanation....

Explanation:

Hope it helps...

8 0
3 years ago
A certain electromagnetic wave traveling in seawater was observed to have an amplitude of 98.02 (V/m) at a depth of 10 m, and an
maksim [4K]

Answer:

The  value is   \alpha =  0.002 Np/m

Explanation:

From the question we are told that

  The first amplitude of the wave is  E_{max}1 =  98.02 \  V/m

  The first  depth  is  D_1 =  10 \  m

   The second amplitude is  E_{max}2 =  81.87 \  (V/m)

   The second depth is D_2 = 100 \ m

Generally from the spatial wave equation we have

   v(x) =  Ae^{-\alpha d}cos(\beta x  + \phi_o)

=>       \frac{v(x)}{v(x)} =\frac{  Ae^{-\alpha d}cos(\beta x  + \phi_o)}{ Ae^{-\alpha d}cos(\beta x  + \phi_o)}

So considering the ratio of the equation for the  two depth

\frac{A}{A_S}  =  \frac{e^{-D_1 \alpha }}{e^{-D_2 \alpha }}

=>   \frac{98.02}{81.87}  =  \frac{e^{-10 \alpha }}{e^{-100 \alpha }}

=>   \alpha  =  \frac{0.18}{90}

=>    \alpha =  0.002 Np/m

       

4 0
3 years ago
an op amp in unity gain configuration (buffer) with slew rate of 5v/us is used to amplify a sinusoidal signal with a frequency o
ZanzabumX [31]

Answer:

The maximum amplitude (V_{max}) will be 7.96 V.

Explanation:

We know, for distortion free operation, the slew rate (S) of an OPAMP is written as

S = 2 \pi f V_{max}

where 'f' is the highest frequency signal.

Therefore, from the above equation we can write,

&&5 \frac{V}{\mu s} = 2 \pi 100 kHz \times V_{max}\\&or,& V_{max} = \frac{5V}{10^{-6} s \times 2 \pi 100 \times 10^{3} Hz}\\&or,& V_{max} = \frac{5}{2 \pi \times 10^{-1}} V = 7.96 V

3 0
3 years ago
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