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3 years ago

At its nearest, Venus comes within about 41 million km of Earth. How distant is it at its farthest?

Physics

2 answers:

mariarad [96]3 years ago

5 0

Answer:

261 million km

Explanation:

Venus has an almost perfectly circular orbit with 0.007 eccentricity. Earth however has an elliptical orbit with 0.016 eccentricity. Distance of Earth from Venus at its farthest is 261 million km. So, the difference in the farthest and closest distance is very high almost 220 million km. At the farthest Venus is 108,939,000 km from the sun and 107,477,000 km at its closest from the Sun.

Eva8 [605]3 years ago

4 0

Answer:

162 million km

Explanation:

The nearest distance between Earth and Venus is 41 million kilometers. Whereas the Farthest distance between Earth and Venus is 162 million kilometers. The Venus is the hottest planet of the solar system. It is also called Evening and morning star, as it appears brightest in the sky in morning and Evening after moon.

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A disk of radius a has a total charge Q uniformly distributed over its surface. The disk has negligible thickness and lies in th

sleet_krkn [62]

The electric potential V(z) on the z-axis is : V = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

<u>Given data :</u>

V(z) =2kQ / a²(v(a² + z²) ) -z

<h3>Determine the electric potential V(z) on the z axis and magnitude of the electric field</h3>

Considering a disk with radius R

Charge = dq

Also the distance from the edge to the point on the z-axis = √ [R² + z²].

The surface charge density of the disk ( б ) = dq / dA

Small element charge dq = б( 2πR ) dr

dV $\frac{k.dq}{\sqrt{R^2+z^2} } \\\\= \frac{k(\alpha (2\pi R)dR}{\sqrt{R^2+z^2} }$ ----- ( 1 )

Integrating equation ( 1 ) over for full radius of a

∫dv = $\int\limits^a_o {\frac{k(\alpha (2\pi R)dR)}{\sqrt{R^2+z^2} } } \,$

V = $\pi k\alpha [ (a^2+z^2)^\frac{1}{2} -z ]$

= $\pi k (\frac{Q}{\pi \alpha ^2})[(a^2 +z^2)^{\frac{1}{2} } -z ]$

Therefore the electric potential V(z) = $(\frac{Q}{a^2} ) [ (a^2 + z^2)^{\frac{1}{2} } -z$

Also

The magnitude of the electric field on the z axis is : E = kб 2 $\pi$ ( 1 - [z / √(z² + a² ) ] )

Hence we can conclude that the answers to your question are as listed above.

Learn more about electric potential : brainly.com/question/25923373

7 0

2 years ago

An 80-kg hiker climbs to the top of a tall hill and builds up 470,000 J of gravitational potential energy. How high did the hike

nekit [7.7K]

Answer:

599 meters is the answer rounded to the nearest whole number and 599.489795918 meters is the complete answer

Explanation:

to find gravitational potential energy you multiply mass x acceleration due to gravity (always 9.8 on earth) x hight

since we know the gravitational potential energy and want to find out the hight, we take the gravitational potential energy (470,000) and divide it by the product of acceleration due to gravity x mass (9.8 x 80)

so how high the hiker climbed is equal to 470,000 divided by (9.8 x 80)

hight = 470,000 / (9.8 x 80)

hight = 470,000 / 784

hight = 599.489795918 meters

as for rounding, if the decimal is less than 5 you round "down" and keep the current whole number, if the decimal is 5 or greater you round "up" and add 1 to get your new number

5 0

3 years ago

Read 2 more answers

These graphs show measurements of the density of air as different sound waves pass a single point.

Ymorist [56]

Answer: D!

It is the option with the greatest amplitude.