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3 years ago

What is the equivalent resistance of a circuit that contains two 50.0 resistors connected in parallel with a 12.0 V battery?

Physics

2 answers:

Pachacha [2.7K]3 years ago

8 0

1/Re=1/R1 +1/R2= 2x1/R1=2/R1=2/R2=2/50=1/25
1/Re=1/25, so the answer is Re= 25

the answer is B. 25.0

sineoko [7]3 years ago

8 0

Answer:

the other person is right, the answer is 25.0

Explanation:

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An electron has a kinetic energy of 3.00 ev. find its wavelength. (b) what if? a photon has energy 3.00 ev. find its wavelength.

Harman [31]

(a) The electron kinetic energy is
$K=3.00 eV$
which can be converted into Joule by keeping in mind that
$1 eV=1.6 \cdot 10^{-19}eV$
So that we find
$K=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The kinetic energy of the electron is related to its momentum p by:
$K= \frac{p^2}{2m}$
where m is the electron mass. Re-arranging the equation, we find
$p= \sqrt{ 2Km}= \sqrt{ 2 ( 4.8 \cdot 10^{-19} J)(9.1 \cdot 10^{-31} kg) } =9.35 \cdot 10^{-25} kgm/s$

And now we can use De Broglie's relationship to find its wavelength:
$\lambda= \frac{h}{p}= \frac{6.6 \cdot 10^{-34} Js}{9.35 \cdot 10^{-25} kg m/s} =7.06 \cdot 10^{-10}m$
where h is the Planck constant.

(b) By using the same procedure of part (a), we can convert the photon energy into Joules:
$E=3.00 eV \cdot 1.6 \cdot 10^{-19} eV/J =4.8 \cdot 10^{-19}J$

The energy of a photon is related to its frequency f by:
$E=hf$
where h is the Planck constant. Re-arranging the equation, we find
$f= \frac{E}{h}= \frac{4.8 \cdot 10^{-19} J}{6.6 \cdot 10^{-34}Js} =7.27 \cdot 10^{14}Hz$

And now we can use the relationship between frequency f, speed of light c and wavelength $\lambda$ of a photon, to find its wavelength:
$\lambda= \frac{c}{f}= \frac{3 \cdot 10^8 m/s}{7.27 \cdot 10^{14} Hz} =4.13 \cdot 10^{-7} m$

8 0

3 years ago

True/False: The points where the potential is the same (in three-dimensional space) lie on a surface.

nikdorinn [45]

1. True
2. False

Does that answer your question?

5 0

3 years ago

Calculate the temperature of the air mass when it has risen to a level at which atmospheric pressure is only 8.00×104 Pa . Assum

cestrela7 [59]

Answer:

$T_{2}=278.80 K$

Explanation:

Let's use the equation that relate the temperatures and volumes of an adiabatic process in a ideal gas.

$(\frac{V_{1}}{V_{2}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$ .

Now, let's use the ideal gas equation to the initial and the final state:

$\frac{p_{1} V_{1}}{T_{1}} = \frac{p_{2} V_{2}}{T_{2}}$

Let's recall that the term nR is a constant. That is why we can match these equations.

We can find a relation between the volumes of the initial and the final state.

$\frac{V_{1}}{V_{2}}=\frac{T_{1}p_{2}}{T_{2}p_{1}}$

Combining this equation with the first equation we have:

$(\frac{T_{1}p_{2}}{T_{2}p_{1}})^{\gamma -1} = \frac{T_{2}}{T_{1}}$

$(\frac{p_{2}}{p_{1}})^{\gamma -1} = \frac{T_{2}^{\gamma}}{T_{1}^{\gamma}}$

Now, we just need to solve this equation for T₂.

$T_{1}\cdot (\frac{p_{2}}{p_{1}})^{\frac{\gamma - 1}{\gamma}} = T_{2}$

Let's assume the initial temperature and pressure as 25 °C = 298 K and 1 atm = 1.01 * 10⁵ Pa, in a normal conditions.

Here,

$p_{2}=8.00\cdot 10^{4} Pa \\p_{1}=1.01\cdot 10^{5} Pa\\ T_{1}=298 K\\ \gamma=1.40$

Finally, T2 will be:

$T_{2}=278.80 K$

6 0

3 years ago

Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

nignag [31]

Answer:

The refractive index of glass, $\mu_{g} = 1.367$

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

$tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}$ (1)