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vesna_86 [32]
3 years ago
11

Which of the following laws relates emf, potential, current, and resistance in a circuit? Ohm’s Law Newton’s Law of Gravity Coul

omb’s Law Conservation of Charge
Physics
2 answers:
lisabon 2012 [21]3 years ago
3 0
I think it’s Ohm’s Law.
joja [24]3 years ago
3 0

Answer:

Ohm's law

Explanation:

As per ohm's law we know that the potential difference across the resistance is applied then the current flows through is given by

i = \frac{V}{R}

here we know that

V = EMF or potential difference across the resistance

R = Resistance

i = electric current

So ohm's law relates the potential difference across the resistance and current flow through the resistance with the resistance of the wire

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To convert minutes per second into kilometre per hour we multiply the speed with​
ankoles [38]

Answer:

To convert m/sec into km/hr, multiply the number by 18 and then divide it by 5.

Explanation:

please mark as brainliest

3 0
3 years ago
If you could shine a very powerful flashlight beam toward the Moon, estimate the diameter of the beam when it reaches the Moon.
grin007 [14]

To develop this problem it is necessary to apply the Rayleigh Criterion (Angular resolution)criterion. This conceptos describes the ability of any image-forming device such as an optical or radio telescope, a microscope, a camera, or an eye, to distinguish small details of an object, thereby making it a major determinant of image resolution.  By definition is defined as:

\theta = 1.22\frac{\lambda}{d}

Where,

\lambda= Wavelength

d = Width of the slit

\theta= Angular resolution

Through the arc length we can find the radius, which would be given according to the length and angle previously described.

The radius of the beam on the moon is

r = l\theta

Relacing \theta

r = l(\frac{1.22\lambda}{d})

r = 1.22\frac{l\lambda}{d}

Replacing with our values we have that,

r = 1.22*(\frac{(384*10^3km)(\frac{1000m}{1km})(550*10^{-9}m)}{7*10^{{-2}}})

r = 3680.91m

Therefore the diameter of the beam on the moon is

d = 2r

d = 2 * (3690.91)

d = 7361.8285m

Hence, the diameter of the beam when it reaches the moon is 7361.82m

8 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
What relationship do you see between a star colour and temperature
Andrei [34K]

Answer:

Stars emit colors of many different wavelengths, but the wavelength of light where a star's emission is concentrated is related to the star's temperature - the hotter the star, the more blue it is; the cooler the star, the more red it is

5 0
3 years ago
Guillaume puts a bottle of soft drink in a refrigerator and leaves it there until its temperature has dropped 15.1 K.
Zarrin [17]

Answer: (a) The magnitude of its temperature change in degrees Celsius is 15.1^{o}C.

(b) The magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is 27.2^{o}F.

Explanation:

(a)  Expression for change in temperature is as follows.

        |\Delta T| = |x - y|K

                         = 15.1 K

                    = |(x - 273.15) - (y - 273.15)|^{o}C

                    = |x - y|^{o}C

                    = 15.1^{o}C

Therefore, the magnitude of its temperature change in degrees Celsius is 15.1^{o}C.

(b)  Change in temperature from Celsius to Fahrenheit is as follows.

           F = 1.8C + 32

          C = \frac{F - 32}{1.8}

Since,   K = C + 273

or,    \Delta K = \Delta C = \frac{\Delta F}{1.8}

         \Delta F = 1.8 \Delta K

                      = 1.8 (15.1)

                      = 27.18^{o}F

or,                  = 27.2^{o}F

Thus, we can conclude that the magnitude of the temperature change (change in T = 15.1 K) in degrees Fahrenheit is 27.2^{o}F.

7 0
3 years ago
Read 2 more answers
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