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Alik [6]
3 years ago
14

You connect five identical resistors in series to a battery whose EMF is 12.0 V and whose internal resistance is negligible. You

measure the current that the circuit draws from the battery and find 0.961 A. What are the resistance of each resistor and the potential difference across each resistor?
Physics
2 answers:
avanturin [10]3 years ago
4 0

Answer:

The resistance of each resistor is 2.5 Ω

The potential difference across each resistor is 2.4 V.

Explanation:

By Ohm's law,

<em>V </em>=<em> IR</em>

where <em>V</em> is the potential difference or voltage across an element, <em>I</em> is the current flowing through it and <em>R</em> is its effective resistance.

For the group of five resistors, let their combined resistance be <em>R</em>.

Then

(12.0 V) = (0.961 A)(<em>R</em>)

R = \dfrac{12}{0.961}\,\Omega

Because they are in series, <em>R</em> is the arithmetic sum of their individual resistances. Because they are all identical, the resistance of each resistor is

= \dfrac{12}{5\times0.961}\,\Omega = 2.5\,\Omega

Also, because they are in series and are equal, the EMF is distributed across them equally. Therefore, the potential difference across each resistor is

\dfrac{12.0\text{ V}}{5} = 2.40\text{ V}

Oksanka [162]3 years ago
3 0

Answer:

R = 2.4974 Ω

V = 2.4 V

Explanation:

When Five identical resistors are connected in series,

R' = R+R+R+R+R

R' = 5R............................ Equation 1

Where R' = Combined resistance of the five resistors connected in series, R = Resistance of each of the resistor.

make R the subject of the equation

R = R'/5.................... Equation 2

Using

E = I(R'+r).................. Equation 3

Where E = Emf of the battery, I = current, r = internal resistance of the battery

Given: E = 12 V, I = 0.961 A, r = 0 Ω (negligible)

Substitute into equation 3

12 = 0.961(R')

R' = 12/0.961

R' = 12.487 Ω

Substitute into equation 2

R = 12.487/5

R = 2.4974 Ω

using ohm's law,

V = IR....................... Equation 4

Where V = potential difference across each resistance.

Given: I = 0.961 A, R = 2.4974 Ω

Substitute into equation 4

V = 0.961(2.4974)

V = 2.4 V.

Hence the resistance and potential difference across each resistor = 2.4974 Ω  and 2.4 V

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pogonyaev

Complete Question

Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcseconds in an arcminute).Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolution with 700 nm light

Answer:

The diameter is  D = 0.59 \  m    

Explanation:

From the question we are told that

      The best resolution is  \theta  =  0.3 \  arcsecond

       The  wavelength is  \lambda  =  700 \  nm =  700 *10^{-9 } \  m

       

Generally the

         1 arcminute  = >  60 arcseconds

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So

       x =  \frac{0.3}{60 }

=>    x = 0.005 \  arcminutes

Now

         60 arcminutes  =>  1 degree

          0.005 arcminutes = >  z degrees  

=>       z =  \frac{0.005}{60 }

=>      z =  8.333 *10^{-5}  \ degree

Converting to radian  

           \theta  = z =  8.333 *10^{-5}  * 0.01745 = 1.454 *10^{-6} \  radian

Generally the resolution is mathematically represented as

            \theta  =  \frac{1.22 *  \lambda  }{ D}

=>    D =  \frac{1.22 * \lambda }{\theta }

=>     D =  \frac{1.22 * 700 *10^{-9} }{ 1.454 *10^{-6} }    

=>     D = 0.59 \  m    

4 0
3 years ago
Part A A uniform solid disk of radius 1.60 m and mass 2.30 kg rolls without slipping to the bottom of an inclined plane. If the
kifflom [539]

Answer:

Height will be 3.8971 m        

Explanation:

We have given that radius of the solid r = 1.60 m

Mass of the solid disk m = 2.30 kg

Angular velocity \omega =4.46rad/sec

Moment of inertia is given by I=\frac{1}{2}mr^2

Transnational Kinetic energy is given by KE=\frac{1}{2}mv^2 as we know that v = v=\omega r

So KE=\frac{1}{2}m(\omega r)^2

Rotational kinetic energy is given by KE_{ROTATIONAL}=\frac{1}{2}I\omega ^2=\frac{1}{2}\left ( \frac{1}{2}mr^2 \right )\omega ^2=\frac{1}{4}m(r\omega )^2

Potential energy is given by mgh

According to energy conservation

mgh=\frac{1}{2}m(\omega r)^2+\frac{1}{4}m(\omega r)^2

h=\frac{3r^2\omega ^2}{4g}=\frac{3\times 1.60^2\times 4.46^2}{4\times 9.8}=3.8971m

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What do astronomers use in addition to parallax to find the actual distance of stars that are close to Earth?
IgorLugansk [536]

Answer:

trigonometry (guessing)

Explanation:

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periapsis : shortest distance between something like the moon and the planet its orbiting around like the earth

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https://science.howstuffworks.com/question224.htm

By comparing the intrinsic brightness to the star's apparent brightness we can calculate the distance of stars

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https://www.space.com/30417-parallax.html

alternative distance measurement for stars used by most astronomers is the parsec. A star with a parallax angle of 1 arcsecond has a distance of 1 parsec, or 1 parsec per arcsecond of parallax, which is about 3.26 light years

blossoms.mit.edu

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7 0
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Light of wavelength 476.1 nm falls on two slits spaced 0.29 mm apart. What is the required distance from the slits to the screen
stich3 [128]

Answer:

The distance is D  =  2.6 \ m

Explanation:

From the question we are told that

    The wavelength of the light is  \lambda  =  476.1 \ nm  =  476.1 *10^{-9} \ m

      The  distance between the slit is  d =  0.29 \  mm  =  0.29 *10^{-3} \ m

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Generally  fringe width is mathematically represented as

       y  =  \frac{\lambda * D }{d}

Where D is the distance of the slit to the screen

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        D  =  \frac{y *  d}{\lambda }

substituting values

       D  =  \frac{ 4.2 *10^{-3} *   0.29 *10^{-3}}{ 476.1 *10^{-9} }

        D  =  2.6 \ m

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Alex777 [14]

Answer:

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