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marissa [1.9K]
2 years ago
6

What is the resistance ofa wire made of a material with resistivity of 3.2 x 10^-8 Ω.m if its length is 2.5 m and its diameter i

s 0.50 mm?
Physics
1 answer:
Katarina [22]2 years ago
5 0

R = 0.407Ω.

The resistance  R of a particular conductor is related to the resistivity ρ of the material by  the equation R = ρL/A, where ρ is the material resistivity, L is the length of the material and A is the cross-sectional area of ​​the material.

To calculate the resistance R of a wire made of a material with resistivity of 3.2x10⁻⁸Ω.m, the length of the wire is 2.5m and its diameter is 0.50mm.

We have to use the equation R = ρL/A but first we have to calculate the cross-sectional area of the wire which is a circle. So, the area of a circle is given by A = πr², with r = d/2. The cross-sectional area of the wire is A = πd²/4.  Then:

R =[(3.2x10⁻⁸Ω.m)(2.5m)]/[π(0.5x10⁻³m)²/4]

R = 8x10⁻⁸Ω.m²/1.96x10⁻⁷m²

R = 0.407Ω

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he first excited state of the helium atom lies at an energy 19.82 eV above the ground state. If this excited state is three-fold
bekas [8.4K]

Answer:

Relative population is  2.94 x 10⁻¹⁰.

Explanation:

Let N₁ and N₂ be the number of atoms at ground and first excited state of helium respectively and E₁ and E₂ be the ground and first excited state energy of helium respectively.

The ratio of population of atoms as a function of energy and temperature is known as Boltzmann Equation. The equation is:

\frac{N_{1} }{N_{2} } =  \frac{g_{1}e^{\frac{-E_{1} }{KT} }  }{g_{2}e^{\frac{-E_{2} }{KT} }}

\frac{N_{1} }{N_{2} } = \frac{g_{1}e^{\frac{-(E_{1}-E_{2})  }{KT} }  }{g_{2}}

Here g₁ and g₂ be the degeneracy at two levels, K is Boltzmann constant and T is equilibrium temperature.

Put 1 for g₁, 3 for g₂, -19.82 ev for (E₁ - E₂) and 8.6x10⁵ ev/K for K and 10000 k for T in the above equation.

\frac{N_{1} }{N_{2} } = \frac{1\times e^{\frac{-(-19.82)}{8.6\times 10^{-5}\times 10000} }  }{3}

\frac{N_{1} }{N_{2} } = 3.4 x 10⁹

\frac{N_{2} }{N_{1} } =  2.94 x 10⁻¹⁰

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The graph below shows the sunspot number observed between 1750 and 2000. The graph shows sunspot number on the y axis and years
OlgaM077 [116]
1850 to 1900 because the slope would be 105. It says what is the greatest fall, so the upward slope of 120 wouldn't count.
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A brick and a feather fall to the earth at their respective terminal velocities. which object experiences the greater force of a
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The brick, even though the brick would end up traveling faster, it most likely has a larger surface area therefore it would have more air resistance.
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Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed
olga55 [171]

Initially, the energies are:

U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\
=K_{i}=\frac{1}{2} m v_{0}^{2}

At final point, the energies are:

U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\
K_{f}=\frac{1}{2} m(0)^{2}=0

Using conservation law of energy,

-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\
-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\
\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}

The equation is further simplified as:

r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\
h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\
&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\
& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}

7 0
1 year ago
Which is a correct representation of .000025 in scientific notation?
lidiya [134]

0.000025 → 2.5 × 10⁻⁵ → 2.5E-5

<h3>Further explanation</h3>

Scientific notation represents the precise way scientists handle exceptionally abundant digits or extremely inadequate numbers in the product of a decimal form of number and powers of ten. Put differently, such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent or power. It is a system for expressing extremely broad or exceedingly narrow digits compactly.

Scientific notation should be in the form of  

\boxed{ \ a \times 10^n \ }

where  

\boxed{ \ 1 \leq a \ < 10 \ }

The number 'a' is called 'mantissa' and 'n' the order of magnitude.

From the key question that is being asked, we face the standard form of 0.000025.

\boxed{ \ 0.000025 = \frac{25}{1,000,000} \ }

The coefficient (or mantissa), i.e. 25, is still outside of 1 ≤ a < 10. Both the numerator and denominator are divided by 10.

\boxed{ \ 0.000025 = \frac{2.5}{100,000} \ }

The denominator consists precisely of five zero digits.

Hence, 0,000025 is written in scientific notation as  \boxed{\boxed{ \ 2.5 \times 10^{-5} \ or \ 2.5E - 5 \ }}

The inverse of scientific notation is the standard form. To promptly change scientific notation into standard form, we reverse the process, move the decimal point to the right or left. This expanded form is called the standard form.

<u>A notable example:</u>

\boxed{ \ 3.0 \times 10^{8} \ Hz \ \rightarrow 300,000,000 \ Hz \ or \ 300 \ MHz}

<h3>Learn more</h3>
  1. 0.00069 written in scientific notation brainly.com/question/7263463
  2. Express the pill’s mass in 0.0005 grams using scientific notation or in milligrams brainly.com/question/493592
  3. What 3 digits are in the units period of 4,083,817 brainly.com/question/558692

Keywords: which is, a correct representation, 0.000025, in scientific notation, expanded form, exponent, base, standard form, mantissa, the order of magnitude, power, decimal, very large, small, figures, abundant digits, inadequate

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