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ladessa [460]
3 years ago
12

You have a 12 volt battery and placed across a 6 ohm resistor, what will the current be?​

Physics
2 answers:
QveST [7]3 years ago
8 0

Answer:

i hope it help

Explanation:

It's possible to step 12 volts down to 6 volts by incorporating a pair of 10,000-ohm resistors into the circuit. Cut two lengths of wire, and strip each wire of 1/2 inch of insulation at each end. Attach one end of the first wire to the positive terminal on the power supply.

scoray [572]3 years ago
4 0

Answer:

Two

Explanation:

Volt = Current* Resistance

Current = Volt\Resistance

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Light is shining perpendicularly on the surface of the earth with an intensity of 680 W/m^2. Assuming that all the photons in th
andrew11 [14]

Answer:

3.066×10^21  photons/(s.m^2)

Explanation:

The power per area is:

Power/A = (# of photons /t /A)×(energy / photon)

E/photons = h×c/(λ)

photons /t /A = (Power/A)×λ /(h×c)  

photons /t /A = (P/A)×λ/(hc)

photons /t /A = (680)×(678×10^-9)/(6.63×10^-34)×(3×10^-8)

                      = 3.066×10^21

Therefore, the number of photons per second per square meter 3.066×10^21  photons/(s.m^2).

4 0
3 years ago
PLS ANSWER FAST WILL GIVE BRAINLY!!!! <br><br> Why does the look of an atom keep changing?
kotykmax [81]

Answer:Atomic model keep changing because the electrons around the nucleus are not fixed and they keeps rotating or changing their position in valence orbits around the nucleus.

Explanation:

3 0
3 years ago
A cart with a mass of 0.5 kg is at the top of the ramp. The height is 0.40m .
Tju [1.3M]

A=0.05.0M.

B=68.9244GPE.34

C=0

D it would be 79%HIGHER

3 0
3 years ago
Name three categories that are used to classify the elements in the periodic table?
nikitadnepr [17]

Answer:

metals,nonmetals, and inert gases

Explanation:

6 0
3 years ago
Read 2 more answers
to 10 Hz. Superimposed on this signal is 60-Hz noise with an amplitude of 0.1 V. It is desired to attenuate the 60-Hz signal to
givi [52]

Answer:

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

Explanation:

For this case we can use the formula for the Butterworth filter gain given by:

[tec] G = \frac{1}{\sqrt{1 +(\frac{f}{f_c})^{2n}}}[/tex]

Where:

G represent the transfer function and we want that G =0.1 since the desired signal is less than 10% of it's value

f_c = 10 Hz represent the corner frequency

f= 60 Hz represent the original frequency

n represent the filter order and that's the variable that we need to find

G \sqrt{1 +(\frac{f}{f_c})^{2n}} = 1

If we square both sides we got:

G^2 (1+\frac{f}{f_c})^{2n}= 1

We divide both sides by G^2 and we got:

(1+\frac{f}{f_c})^{2n} = \frac{1}{G^2}

Now we can apply log on both sides and we got:

2n ln(1+\frac{f}{f_c}) = ln (\frac{1}{G^2})

And solving for n we got:

n = \frac{ ln (\frac{1}{G^2})}{2ln(1+\frac{f}{f_c})}

And replacing we got:

n = \frac{ln (\frac{1}{0.1^2})}{2ln(1+\frac{60}{10})}

n = \frac{4.60517}{3.8918}=1.18

And since n needs to be an integer the correct answer would be n=2 for the filter order.

7 0
3 years ago
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