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2 years ago

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

2 answers:

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]2 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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According to ohm's law what is the resistance of a light bulb if the applied voltage is 21 volts and the current is 3 amps?

Nady [450]

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2 years ago

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2 years ago

A metal ball has a net charge of 4.5x10-7 C

netineya [11]

a) the number of protons is $2.81\cdot 10^{12}$ more than the electrons

b) $4.69\cdot 10^{-15} kg$

Explanation:

The net electric charge on the ball is

$Q=+4.5\cdot 10^{-7}C$

This electric charge is given by the algebraic sum of the charge of the protons and of the charge of the electrons.

The charge of one proton is:

$q_p =+e= +1.6\cdot 10^{-19}C$

While the charge of one electron is

$q_e = -e=-1.6\cdot 10^{-19}C$

So the net charge on the metal ball will be given by

$Q=N_p q_p + N_e q_e = (N_p -N_e)e$

where

$N_p$ is the number of protons

$N_e$ is the number of electrons

So we find:

$N_p-N_e=\frac{Q}{e}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

This means that the number of protons is $2.81\cdot 10^{12}$ more than the electrons.

In this case, we want to make the ball neautral, so we have to remove a net charge of Q' such that the new charge is zero:

$Q-Q'=0$

This implies that the charge that we must remove is

$Q'=Q=4.5\cdot 10^{-7}C$

To do that (and to make the ball losing mass at the same time), we have to remove protons, since they have positive charge.

The number of protons that must be removed is:

$N_p = \frac{Q'}{q_p}=\frac{4.5\cdot 10^{-7}}{1.6\cdot 10^{-19}}=2.81\cdot 10^{12}$

The mass of one proton is

$m_p = 1.67\cdot 10^{-27}kg$

Therefore, the total mass that must be removed from the ball is

$M=m_p N_p = (1.67\cdot 10^{-27})(2.81\cdot 10^{12})=4.69\cdot 10^{-15} kg$

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2 years ago

The word that identifies the color of an object seen under ordinary daylight is:

IRINA_888 [86]

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2 years ago