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3 years ago

7

Which of the following instruments can detect the presence of a voltage, but not the voltage level?

2 answers:

ivanzaharov [21]3 years ago

3 0

A neon 'tester' is just a little neon light bulb that glows orange
when it touches a point of more than maybe 90 volts or so.
When it lights up, you can tell that it's touching a live point, but
you can't tell from the brightness of the glow what the actual
voltage is at that point.

Andreas93 [3]3 years ago

3 0

Neon Circuit Tester(A). Multi tests both, Watt tests the voltage level and clamp is the same as a watt meter

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Consider horizontal parallel plates with a fixed potential difference. The upper plate has a voltage difference of 30 V with the

BlackZzzverrR [31]

relation between potential difference and electric field is given as

$E . d = \Delta V$

so here we know that

d = 3 cm

$\Delta V = 30 V$

$E \times 0.03 = 30$

$E = 1000 N/C$

So now when plates are separated to 4 cm distance carefully

the potential difference between them will change but the electric field between them will remain constant

So at distance of 4 cm also the electric field will be E = 1000 N/C

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3 years ago

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valentina_108 [34]

Three 40w lamp for 6 hours

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4 years ago

Marlon is studying a crab population. He has a large batch of crabs that were captured in the ocean. He places a plastic tag on

tatiyna

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3 years ago

A skateboarder skates down a ramp for 3 seconds. If his initial velocity was 0.8 m/s and · his final velocity was 7 m/s, what wa

Contact [7]

Answer:

The acceleration of the skateboarder is 2.067 m/s²

Explanation:

Given;

initial velocity of the skateboarder, u = 0.8 m/s

final velocity of the skateboarder, v = 7 m/s

time of motion, t = 3 s

The acceleration of the skateboarder is given as;

$a = \frac{v-u}{t}\\\\a = \frac{7-0.8}{3}\\\\a = 2.067 \ m/s^2$

Therefore, the acceleration of the skateboarder is 2.067 m/s²

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3 years ago

Assume the average value of the vertical component of Earth's magnetic field is 42 μT (downward) in some region that has an area

Oliga [24]

Answer:

The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

Explanation:

Given that,

Magnetic field = 42 μT

Area $A=3.71\times10^{5}\ km^2$

$A=3.71\times10^{11}\ m^2$

We need to calculate the flux per unit area

$flux\ per\ unit\ area=\dfrac{42\times10^{-6}}{3.71\times10^{11}}$

$flux\ per\ unit\ area=1.132\times10^{-16}\ T/m^2$

We need to calculate the total earth's surface area

$A'=4\pi r^2$

$A'=4\times\pi\times(6.3781\times10^{6})^2$

$A'=5.1120\times10^{14}\ m^2$

We need to calculate the rest of earth's area

$A''=A-A'$

Put the value into the formula

$A''=5.1120\times10^{14}-3.71\times10^{11}$

$A''=5.10829\times10^{14}\ m^2$

We need to calculate the net magnetic flux through the rest of Earth's surface

$B'=5.10829\times10^{14}\times1.132\times10^{-16}$

$B'=5.782\times10^{-2}\ T$

Hence, The net magnetic flux through the rest of Earth's surface is $5.782\times10^{-2}\ T$

3 0

4 years ago