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3 years ago

Which of the following scenarios will cause the current to stop flowing within a circuit?

2 answers:

8 0

A. The switch within the closed circuit is opened.

If the switch is opened current will stop flowing, as the current will meet a gap, that is not connected by a conductor.

n200080 [17]3 years ago

6 0

On page 39 it explains that conducting materials. If some part of the external circuit is composed of an insulating material— for example, if there’s an air gap in the circuit—it’s an open circuit. An electric current will flow through a circuit only if the circuit is closed. A common method of controlling the flow of current through a circuit is to include a device called a switch. Current will flow through the circuit when the switch is closed and will stop flowing when the switch is open.

So your answer is A. The switch within the closed circuit is opened

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Mining runoff is _____ water that comes from the mining of rock.

PolarNik [594]

A CONTAMINATED WATER

3 0

3 years ago

Read 2 more answers

A thin uniform rod of mass M and length L is bent at its center so that the two segments are now perpendicular to each other. Fi

Tatiana [17]

Answer:

(a) I_A=1/12ML²

(b) I_B=1/3ML²

Explanation:

We know that the moment of inertia of a rod of mass M and lenght L about its center is 1/12ML².

(a) If the rod is bent exactly at its center, the distance from every point of the rod to the axis doesn't change. Since the moment of inertia depends on the distance of every mass to this axis, the moment of inertia remains the same. In other words, I_A=1/12ML².

(b) The two ends and the point where the two segments meet form an isorrectangle triangle. So the distance between the ends d can be calculated using the Pythagorean Theorem:

$d=\sqrt{(\frac{1}{2}L) ^{2}+(\frac{1}{2}L) ^{2} } =\sqrt{\frac{1}{2}L^{2} } =\frac{1}{\sqrt{2} } L=\frac{\sqrt{2} }{2} L$

Next, the point where the two segments meet, the midpoint of the line connecting the two ends of the rod, and an end of the rod form another rectangle triangle, so we can calculate the distance between the two axis x using Pythagorean Theorem again:

$x=\sqrt{(\frac{1}{2}L)^{2}-(\frac{\sqrt{2}}{4}L) ^{2} } =\sqrt{\frac{1}{8} L^{2} } =\frac{1}{2\sqrt{2}} L=\frac{\sqrt{2}}{4} L$

Finally, using the Parallel Axis Theorem, we calculate I_B:

$I_B=I_A+Mx^{2} \\\\I_B=\frac{1}{12} ML^{2} +\frac{1}{4} ML^{2} =\frac{1}{3} ML^{2}$

5 0

3 years ago

The quantity that can be zero if body remains in motion for some time

SpyIntel [72]

Answer:

ok I think sooo

Explanation:

Velocity and displacement

3 0

2 years ago

What is an example of frequency

emmasim [6.3K]

'Frequency' is a word that often confuses some people ... for no good reason.
It just means "frequent-ness" or "often-ness" ... how often something happens.

The SI unit of frequency is the Hertz (Hz). Hz means 'per second'.
So " 13 Hz " means 13 per second.

Here are examples of frequency:

-- 780 kilohertz (on your AM radio dial)
-- 98.7 Megahertz (on your FM dial)
-- 5.8 Gigahertz
-- twice a day
-- three per week
-- every 6 months

6 0

3 years ago

Read 2 more answers

A truck is traveling at 27 m/s down the interstate highway where you are changing a flat tire. frequency of 185 Hz.

Reil [10]

Answer:

(a) the observed frequency is 200 Hz

(b) the observed frequency is 188 Hz.

Explanation:

speed of the truck, Vs = 27 m/s

frequency of the truck as it approaches, Fs = 185 Hz

(a) Apply Doppler effect to determine the frequency you will hear.

As the truck approaches you, the observed frequency will be higher than the source frequency because of decrease in distance.

$F_s = F_o [\frac{V}{V_S + V} ]$

Where;

Fo is the observed frequency which is the frequency you will hear.

V is speed of sound in air

$F_s = F_o [\frac{V}{V_S + V} ]\\\\185 = F_o [\frac{340}{27 + 340} ]\\\\185 = F_o (0.926)\\\\F_o = \frac{185}{0.926}\\\\F_o = 199.78 \ Hz$

$F_o = 200 \ Hz$

(b) Apply the following formula for a moving observer and a moving source;

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )$

The observed frequency is negative since you are driving away from the truck and the source frequency is also negative since it is driving towards you.

$F_o = F_s[\frac{V-V_o}{V} ](\frac{V}{V-V_S} )\\\\F_o = 185[\frac{340-22}{340} ](\frac{340}{340-27} )\\\\F_o = 185(0.9353)(1.0863)\\\\F_o = 188 \ Hz$

5 0

3 years ago