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shtirl [24]
3 years ago
11

Any individual can become an expert in anything. Please select the best answer from the choices provided. T F

Physics
2 answers:
Cerrena [4.2K]3 years ago
8 0

The answer is FALSE.

arsen [322]3 years ago
6 0

The correct answer is <u><em>F; False. </em></u>

I just did the Practice on Edge and got it right.

Good luck bud <3

You might be interested in
When this current is closed which way does the current flow
Anastaziya [24]
Well, Godess, that's not a simple question, and it doesn't have
a simple answer.

When the switch is closed . . .

"Conventional current" flows out of the ' + ' of the battery, through R₁ ,
then through R₂ , then through R₃ .  It piles up on the right-hand side of
the capacitor (C).  It repels the ' + ' charges on the left side of 'C', and
those flow into the ' - ' side of the battery.  So the flow of current through
this series circuit is completely clockwise, around toward the right. 

That's the way the first experimenters pictured it, that's the way we still
handle it on paper, and that's the way our ammeters display it.

BUT . . .

About 100 years after we thought that we completely understand electricity,
we discovered that the little tiny things that really move through a wire, and
really carry the electric charge, are the electrons, and they carry NEGATIVE
charge.  This turned our whole picture upside down.

But we never changed the picture !  We still do all of our work in terms of
'conventional current'.  But the PHYSICAL current ... the actual motion of
charge in the wire ... is all exactly the other way around.

In your drawing ... When the switch is closed, electrons flow out of the 
' - ' terminal on the bottom of the battery, and pile up on the left plate of
the 'C'.  They repel electrons off of the right-side of 'C', and those then
flow through R₃ , then through R₂ , then through R₁ , and finally into the
' + ' terminal on top of the battery.

Those are the directions of 'conventional' current and 'physical' current
in all circuits.

In the circuit of YOUR picture that you attached, there's more to the story:

Battery current can't flow through a capacitor.  Current flows only until
charges are piled up on the two sides of 'C' facing each other, and then
it stops.

Wait a few seconds after you close the switch in the picture, and there is
no longer any current in the loop.

To be very specific and technical about it . . .

-- The instant you close the switch, the current is

       (battery voltage) / (R₁ + R₂ + R₃)        amperes

but it immediately starts to decrease.

--  Every  (C)/((R₁ + R₂ + R₃)  seconds after that, the current is

                  e⁻¹  =  about  36.8 %

less than it was that same amount of time ago.

Now, are you glad you asked ?
4 0
4 years ago
Consider the magnetic field (B) of a wire with a constant current (I). A Magnetic Field sensor is placed at a radius (r). Will t
iren [92.7K]

Answer:

No, the magnitude of the magnetic field won't change.

Explanation:

The magnetic field produced by a wire with a constant current is circular and its flow is given by the right-hand rule. Since this field is circular with center on the wire the magnitude of the magnetic field around the wire will be given by B = [(\mi_0)*I]/(2\pi*r) where (\mi_0) is a constant, I is the current that goes through the conductor and r is the distance from the wire. If the field sensor will move around the wire with a fixed radius the distance from the wire won't change so the magnitude of the field won't change.

8 0
3 years ago
Sobre un barco, que se mueve en forma rectilínea, y con velocidad constante de 30 [km/h], se mueve un perro en el mismo sentido
almond37 [142]

Answer:

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

Explanation:

Para determinar la velocidad del perro con respecto al observador sentado desde la playa a través del concepto de velocidad relativa, descrito en la siguiente fórmula:

v_{P/B} = v_{P} - v_{B} (1)

Donde:

v_{P/B} - Velocidad del perro relativo al barco, en kilómetros por hora.

v_{P} - Velocidad del perro con respecto al observador, en kilómetros por hora.

v_{B} - Velocidad del barco con respecto al observador, en kilómetros por hora.

Si sabemos que v_{B} = 30\,\frac{km}{h} y v_{P/B} = 10\,\frac{km}{h}, entonces la velocidad del perro con respecto al observador es:

v_{P} = v_{B} + v_{P/B}

v_{P} = 30\,\frac{km}{h} + 10\,\frac{km}{h}

v_{P} = 40\,\frac{km}{h}

El observador verá correr al perro sobre la cubierta del barco a una velocidad de 40 kilómetros por hora.

6 0
3 years ago
A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a po
Archy [21]

Answer:

(a) Since net charge remains same,after immersion Q is same

(b) I. 14.56pF ii. 3.05V

(c) ΔU = 5.204nJ

Explanation:

a)

C = kεA/d

k=1 for air

ε is 8.85x10-12F/m

A = .0025m2

d = .125m

C = 8.85x10-12x.0025/.125 = 1.77x10-13F = 0.177pF

Q = CV = .177pF * 244V = 43.188pC

Since net charge remains same,after immersion Q is same

b)

C = kεA/d, for distilled water k is approx. 80

Cwater = Cair x k

=0.177pF x 80 = 14.16pF

Q is same and C is changed V=Q/c holds. where Q is still 43.188pC and C is now 14.16pF, so V = 43.188pC/14.16pF = 3.05V

c) Change in energy: ΔU = Uwater - Uair

Uwater = Q2/2C = (43.188)2/2x.177pF = 5.27nJ

Uair = Q2/2C = (43.188)2/2x14.16pF = 0.066nJ

ΔU = 5.204nJ

6 0
3 years ago
The diagram shows a wave. The quantity shown is the amplitude. This quantity is directly proportional to which of these other qu
Katen [24]

As we know that frequency and wavelength are dependent on each other

and this is given by

f = \frac{c}{\lambda}

here we know that

\lambda = wavelength

f = frequency

c = speed of wave

while for energy and intensity of wave we know that

Energy = KA^2

here A = amplitude of wave

so energy directly depends on amplitude of wave

so correct answer will be

<em>A) Energy</em>

3 0
3 years ago
Read 2 more answers
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