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Ann [662]
3 years ago
11

Check all choices below that are correct. Increasing the frequency increases the current. Changing the frequency does not affect

the current. Voltage leads current. Current and voltage are in phase. Changing ΔVmax does not affect the current. Changing the resistance does not affect the current. Increasing ΔVmax decreases the current. Increasing the resistance increases the current. Increasing the resistance decreases the current. Increasing ΔVmax increases the current. Current leads voltage. Increasing the frequency decreases the current.
Physics
2 answers:
IgorLugansk [536]3 years ago
6 0

EACH of these statements CAN be true or false.  They ALL depend on the situation, and on the components of the particular circuit.  

For each of these statements, I can give you the diagram of a circuit in which the statement is true, and I can give you the diagram of another circuit in which the same statement is false.

<u>These choices are usually true:</u>

-- Changing the frequency does not affect the current.

-- Increasing the resistance decreases the current.

-- Increasing ΔVmax increases the current.  

<u>These choices are usually false:</u>

-- Increasing the frequency increases the current.

-- Changing ΔVmax does not affect the current.

-- Changing the resistance does not affect the current.  

-- Increasing ΔVmax decreases the current.

-- Increasing the resistance increases the current.

-- Increasing the frequency decreases the current.

<u>These choices can go either way just as easily:</u>

-- Voltage leads current.

-- Current and voltage are in phase.

-- Current leads voltage.

Annette [7]3 years ago
3 0

Answer:

the experyoeyv

Explanation:

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Answer:

Its state is in uniformly accelerated motion

Explanation:

When an object is acted upon the force of gravity only, we said that the object is in free fall.

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4 years ago
Which activities are you always doing when you have a warm-up activity?​
Sati [7]
A jog because it helps to get the muscles moving and your heart pumping blood around your body.

I hope this helps.
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3 years ago
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An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the
Wewaii [24]

Answer:

\frac{dB}{dt} = 3.03 \times 10^6 T/s

Explanation:

As we know that the power emitted by the source is given as

P = 170 W

now we know that

P = \frac{N}{t} (\frac{hc}{\lambda})

now we know that energy density is given as

u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}

now we have

E = B c

u = \frac{B^2}{2\mu_0}

intensity is defined as

I = \frac{P}{A}

now we have

\frac{I}{c} = u = \frac{B^2}{2\mu_0}[/tex]

now we have

\frac{dB}{dt} = \omega B

\frac{dB}{dt} = \frac{2\pi c B}{\lambda}

\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}

here we have

I = \frac{P}{4\pi r^2}

I = \frac{170}{4\pi (410)^2}

I = 8.05 \times 10^{-5}

now we have

\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}

\frac{dB}{dt} = 3.03 \times 10^6 T/s

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3 years ago
A uniform meter stick is hung at its center from a thin wire. It is twisted and oscillates with a period of 5 s. The meter stick
rewona [7]

Answer:

The new time period is  T_2 =  3.8 \  s

Explanation:

From the question we are told that

  The period of oscillation is  T =  5 \ s

   The  new  length is  l_2  =  0.76  \ m

Let assume the original length was l_1 = 1 m

Generally the time period is mathematically represented as

         T  =  2 \pi   \sqrt{ \frac{ I }{ mgh } }

Now  I is the moment of inertia of the stick which is mathematically represented as

           I  =  \frac{m * l^2 }{12 }

So

        T  =  2 \pi   \sqrt{ \frac{  m * l^2 }{12 *   mgh } }

Looking at the above equation we see that

        T  \ \ \  \alpha  \ \ \  l

=>    \frac{ T_2 }{T_1}  =  \frac{l_2}{l_1}

=>    \frac{ T_2}{5} =  \frac{0.76}{1}

=>     T_2 =  3.8 \  s

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gavmur [86]

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Explanation:

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