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3 years ago

6. The star nearest to our Sun is Alpha Centauri, which is 4.3 light years away. A light

Physics

1 answer:

jok3333 [9.3K]3 years ago

4 0

Answer:

273,000 Astronomical Units.

Explanation:

4.3 light years = 40,850,000,000,000 kilometeres, 4.3 x 9,500,000,000,000.

Convert kilometers to astronomical unit:

149,597,870 kilometers = 1 AU, convert unit.

40,850,000,000,000 divided by 149,597,870 = 273,065.385222 or approx. 273,000 AU.

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3 years ago

A 2.0-mm-long, 1.0-mmmm-diameter wire has a variable resistivity given by rho(x)=(2.5×10−6)[1+(x1.0m)2]Ωmrho(x)=(2.5×10−6)[1+(x1

Tresset [83]

Answer:

1.144 A

Explanation:

given that;

the length of the wire = 2.0 mm

the diameter of the wire = 1.0 mm

the variable resistivity R = $\rho (x) =(2.5*10^{-6})[1+(\frac{x}{1.0 \ m})^2]$

Voltage of the battery = 17.0 v

Now; the resistivity of the variable (dR) can be expressed as = $\frac{\rho dx}{A}$

$dR = \frac{(2.5*10^{-6})[1+(\frac{x}{1.0})^2]}{\frac{\pi}{4}(10^{-3})^2}$

Taking the integral of both sides;we have:

$\int\limits^R_0 dR = \int\limits^2_0 3.185 \ [1+x^2] \ dx$

$R = 3.185 [x + \frac {x^3}{3}}]^2__0$

$R = 3.185 [2 + \frac {2^3}{3}}]$

R = 14.863 Ω

Since V = IR

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$I = \frac{17}{14.863}$

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8 0

3 years ago

Two positive charges of 20 micro coulomb and 100 micro coulomb and the distance between them is 150cm.What will be the electrica

Roman55 [17]

Answer:

0.8 N

Explanation:

From coulomb's law,

Formula:

F = kqq'/r²........................ Equation 1

Where F = Force of repulsion, k = coulomb's constant, q = first positive charge, q' = second positive charge, r = distance between the charge.

Given: q = 20 μC = 20×10⁻⁶ C, q' = 100 μC = 100×10⁻⁶ C, r = 150 cm = 1.5 m.

Constant: k = 9×10⁹ Nm²/C²

Substitute these values into equation 1

F = (20×10⁻⁶ )( 100×10⁻⁶)(9×10⁹)/1.5²

F = 1800×10⁻³/2.25

F = 1.8/2.25

F = 0.8 N

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