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3 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]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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For years, the tallest tower in the United States was the Phoenix Shot Tower in Baltimore, Maryland. The shot tower was used fro

serious [3.7K]

Answer:

The answer to your question is:

a) t = 3.81 s

b) vf = 37.4 m/s

Explanation:

Data

height = 71.3 m = 234 feet

t = 0 m/s

vf = ?

vo = 0 m/s

Formula

h = vot + 1/2gt²

vf = vo + gt

Process

h = vot + 1/2gt²

71.3 = 0t + 1/2(9.81)t²

2(71.3) = 9,81t²

t² = 2(71.3)/9.81

t² = 14.53

t = 3.81 s

vf = 0 + (9.81)(3.81)

vf = 37.4 m/s

3 0

3 years ago

Two charges are placed on the x axis. One of the charges (q1 = +7.7 µC) is at x1 = +3.1 cm and the other (q2 = -19 µC) is at x2

Alinara [238K]

Answer:

a) $E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

b) $E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

Explanation:

Given that

q1 = +7.7 µC is at x1 = +3.1 cm

q2 = -19 µC is at x2 = +8.9 cm

We know that electric filed due to a charge given as

$E=K\dfrac{q}{r^2}$

$E_1=K\dfrac{q_1}{r_1^2}$

$E_2=K\dfrac{q_2}{r_2^2}$

Now by putting the va;ues

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.031^2}\ N/C$

$E_1=72.11\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.089^2}\ N/C$

$E_2=21.58\times 10^{6}\ N/C$

The net electric field

$E=E_1-E_2$

$E=50.53\times 10^{6}\ N/C$

The direction will be negative direction.

As we know that electric filed line emerge from positive charge and concentrated at negative charge.

Now

distance for charge 1 will become =5.5 - 3.1 = 2.4 cm

distance for charge 2 will become =8.9-5.5 = 3.4 cm

$E_1=9\times 10^9\times \dfrac{7.7\times 10^{-6}}{0.024^2}\ N/C$

$E_1=120.3\times 10^{6}\ N/C$

$E_2=9\times 10^9\times \dfrac{19\times 10^{-6}}{0.034^2}\ N/C$

$E_2=147.92\times 10^{6}\ N/C$

The net electric field

$E=E_1+E_2$

$E=268.22\times 10^{6}\ N/C$

The direction will be positive direction.

7 0

3 years ago

Which equation is equivalent to Log (26•35) ?

Nady [450]

Answer:

Log (26) + Log (35)

6 0

3 years ago

A 500 kg roller coaster moving at 15 m/s suddenly comes to a stop at the end of the ride. How much work energy was needed to sto

saveliy_v [14]

Answer:

energy required=-energy lost

energy lost=change in kinetic energy

EL=1/2 mv^2

4 0

3 years ago

Suppose you are on an airplane moving at high speed. If you flip a coin straight up it will land in your lap rather than a great

maxonik [38]

It will land in your lap because there's different frames of motion relative to yourself. For example, if you're running at a speed of 6 mph, it doesn't mean you'll run as fast as the Earth spins. Also, since you're on the interior of the plane, any kind of wind or weather on the outside will not affect the coin. A law to back up this claim is Einsteins Special Law of Relativity.

8 0

3 years ago

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