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Andru [333]
2 years ago
8

A 6.0-v battery maintains the electrical potential difference between two parallel metal plates separated by 1.0 mm. what is the

electric field between the plates?
Physics
1 answer:
Gnoma [55]2 years ago
5 0
The voltage<span> difference between the two plates can be expressed in terms of the </span>work<span> done on a positive test charge q when it moves from the positive to the negative plate.</span><span>
E=V/d
where V is the voltage and d is the distance between the plates.
 
So,

E=6.0V/1mm= 6000 V/m. The electric field between the plates is 6000 V/m.</span>
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Distance is 80 seconds is 8 what is the average speed?
True [87]

Answer:

speed =distance /time

speed =80/8

speed =10m/s

8 0
2 years ago
If the primary of a transformer were connected to a dc power source,
QveST [7]

Answer:

D. only briefly while being connected or disconnected.

Explanation:

As we know that transformer works on the principle of mutual inductance

here we know that as per the principle of mutual inductance when flux linked with the primary coil charges then it will induce EMF in secondary coil

So here when AC source is connected with primary coil then it will give output across secondary coil because AC source will have change in flux with time.

Now when we connect DC source across primary coil then it will not induce any EMF across secondary coil because DC source is a constant voltage source in which flux will remain constant always

So here in DC source the EMF will only induce at the time of connection or disconnection when flux will change in it while rest of the time it will give ZERO output

so correct answer will be

D. only briefly while being connected or disconnected.

8 0
3 years ago
what is the resistance of a car light bulb that conducts 0.025A current when connected to a 12V car accumulator? is the current
Sophie [7]

One form of Ohm's Law says . . . . . Resistance = Voltage / Current .

R = V / I

R = (12 v) / (0.025 A)

R = (12 / 0.025) (V/I)

<em>R = 480 Ohms</em>

I don't know if the current in the bulb is steady, because I don't know what a car's "accumulator" is.  (Floogle isn't sure either.)

If you're referring to the car's battery, then the current is quite steady, because the battery is a purely DC storage container.

If you're referring to the car's "alternator" ... the thing that generates electrical energy in a car to keep the battery charged ... then the current is pulsating DC, because that's the form of the alternator's output.  

7 0
3 years ago
What is the magnitude of the magnetic field at a point midway between them if the top one carries a current of 19.5 A and the bo
Phantasy [73]

Answer:

The magnetic field will be \large{\dfrac{1.4 \times 10^{-4}}{d}} T, '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field (\large{\overrightarrow{B}}) at a distance 'r' due to a current carrying conductor carrying current 'I' is given by

\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}

where '\overrightarrow{dl}' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current 'I', at a distance 'd' is given by

\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}

According to the figure if 'I_{t}' be the current carried by the top wire, 'I_{b}' be the current carried by the bottom wire and '2d' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the \bigotimes symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by \bigodot symbol.

Given \large{I_{t} = 19.5 A} and \large{I_{B} = 12.5 A}

Therefore, the magnetic field (\large{B_{t}}) at 'P' due to the top wire

B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}

and the magnetic field (\large{B_{b}}) at 'P' due to the bottom wire

B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}

Therefore taking the value of \mu_{0} = 4\pi \times 10^{-7} the net magnetic field (\large{B_{M}}) at the midway between the wires will be

\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T

5 0
3 years ago
Which of the following BEST represent a good plan for decreasing your BMI?
Deffense [45]
I’d say b is the better option because d is starving yourself which creates unhealthy habits, c would cause yo yo dieting or binging because you treat food as a reward which is a toxic mindset, and a same explanation. I wish there was an answer to just track what you eat, have daily exercise and enjoy in moderation, though.
3 0
3 years ago
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