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3 years ago

Where do subduction zones often occur

Physics

1 answer:

vodka [1.7K]3 years ago

4 0

Subductions occur at convergent boundaries. Its where one tectonic plate moves under another.

-Steel jelly

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A JFET has a drain current of 5mA. If IDSS = 10mA and VGS ( off )= -6 v. find The Value Of

levacccp [35]

$\underline {\huge \boxed{ \sf \color{skyblue}Answer : }}$

Given :

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{D} = 5mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: I_{DSS} = 10mA$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS(off)} = -6V$

$\: \:$

$\tt \large {\color{purple} ↬ } \: \: \: \: \: V_{GS} = {?}$

$\: \: \:$

Let's Slove :

$\tt \large I_{D} = I_{(DSS)} (1 - \frac {V_{GS}}{V_{GS(off)}} )^{2}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{I_D} }{ \sqrt{I_{DSS}} } ) \times V_{GS(off)}$

$\: \: \:$

$\tt \large \: V_{GS} = (1 - \frac{ \sqrt{5m} }{ \sqrt{10m} } ) \times { - 6}$

$\: \:$

$\underline \color{red} {\tt \large \boxed {\tt V_{GS} = 1.75 ✓}}$

3 0

1 year ago

Can we travel back in time?

KATRIN_1 [288]

Answer:

I do not believe so.

Explanation:

We have not advanced that far yet in our society.

3 0

3 years ago

What's the noun in "she is my favorite english teacher"

OleMash [197]

Answer:

i think it is teacher

Explanation:

because it is noun

7 0

3 years ago

Read 2 more answers

You step onto a hot beach with your bare feet. A generated in your foot, travels through your nervous system at an average speed

madreJ [45]

Answer: 12.44 millisecond

Explanation:

Based on the information given, to know the time taken for the impulse, which travels a distance of 1.58m to reach the brain will be:

Time taken = Distance/Speed

Time taken = 1.58/127

Time taken = 0.01244

Time taken = 12.44 millisecond.

7 0

3 years ago

An electron travels 0.01 m across an electric potential difference of 6 x 10^6 V. What is the force exerted on the electron? Typ

JulijaS [17]

Answer:

$9.6\cdot 10^{-11}N$

Explanation:

The force experienced by a charged particle immersed in an electric field is

$F=qE$

where

q is the charge of the particle

E is the strength of the electric field

For a uniform field, we can write

$E=\frac{\Delta V}{d}$

where

$\Delta V$ is the potential difference

d is the distance travelled

So we can rewrite the force as

$F=\frac{q\Delta V}{d}$

In this problem:

$q=1.6\cdot 10^{-19}C$ is the charge of the electron

$\Delta V=6\cdot 10^6 V$ is the potential difference

d = 0.01 m is the distance covered

Substituting, we find the force:

$F=\frac{(1.6\cdot 10^{-19})(6\cdot 10^6)}{0.01}=9.6\cdot 10^{-11}N$

3 0

3 years ago