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1 year ago

The dense atmosphere of venus is , causing most incoming solar radiation to be _.

Physics

1 answer:

ad-work [718]1 year ago

7 0

The dense atmosphere of Venus is light in color, causing most incoming solar radiation to be reflected.

<h3 /><h3>What is the composition of atmosphere of Venus?</h3>

The atmosphere of Venus is composed of carbon dioxide (CO2, 96,5 %) and nitrogen (N2, less than 3,5%) already represent more than 99,9% of the composition. Water Vapor is present in extremely low quantities, making Venus the driest planet of the solar system.

<h3>What is the effect of solar radiation on planet Venus? </h3>

On Venus, only 10% of the incoming solar radiation reaches the surface; most of it is reflected back into space by the planet's dense cloud cover, making Venus appear to shine brightly in the night sky.

Despite the small amount of solar radiation reaching the surface, it is trapped effectively by gases in lower atmosphere, giving rise to the high temperatures.

Scientists think that the Venus was once more like the Earth, but at some point it started heating up uncontrollably. Any water that may have once existed in oceans on the surface evaporated into atmosphere causing further runaway heating.

It’s like Earth gone wrong with a runaway greenhouse effect that some planetary scientists say should be a warning as to what may become of our home planet if human activities – such as burning fossil fuels and destroying forests, which intensifies natural greenhouse effect – continue.

To know more about the planet Venus visit:

brainly.com//18452664

#SPJ4

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The distance from the Earth to the Sun equals 1 AU. Neptune is 30 AU from the Sun. How far is Neptune from the Earth? AU

valina [46]

Answer:

The answer is 29 AU

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3 years ago

Read 2 more answers

A loop of wire is perpendicular to a magnetic field such that the field is coming straight at you through the loop. The field be

Dima020 [189]

Answer:

Inducted Magnetic field will be toward from you

Inducted current direction will be counter clockwise.

Explanation:

Lenz's law states that the direction of the current induced in a wire by a changing magnetic field is such that the magnetic field created by the induced current opposes the initial changing magnetic field.

So if the field begins to decrease, the induced magnetic field would try to stop this, so its direction will be the same as the magnetic field, toward from you.

This induced magnetic field is produced by the current in the wire. If the inducted magnetic field will be toward you, the right hand rule says that the direction from the inducted current will be counter clockwise.

5 0

3 years ago

A coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged con

Tcecarenko [31]

Complete question:

A 50 m length of coaxial cable has a charged inner conductor (with charge +8.5 µC and radius 1.304 mm) and a surrounding oppositely charged conductor (with charge −8.5 µC and radius 9.249 mm).

Required:

What is the magnitude of the electric field halfway between the two cylindrical conductors? The Coulomb constant is 8.98755 × 10^9 N.m^2 . Assume the region between the conductors is air, and neglect end effects. Answer in units of V/m.

Answer:

The magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

Explanation:

Given;

charge of the coaxial capable, Q = 8.5 µC = 8.5 x 10⁻⁶ C

length of the conductor, L = 50 m

inner radius, r₁ = 1.304 mm

outer radius, r₂ = 9.249 mm

The magnitude of the electric field halfway between the two cylindrical conductors is given by;

$E = \frac{\lambda}{2\pi \epsilon_o r} = \frac{Q}{2\pi \epsilon_o r L}$

Where;

λ is linear charge density or charge per unit length

r is the distance halfway between the two cylindrical conductors

$r = r_1 + \frac{1}{2}(r_2-r_1) \\\\r = 1.304 \ mm \ + \ \frac{1}{2}(9.249 \ mm-1.304 \ mm)\\\\r = 1.304 \ mm \ + \ 3.9725 \ mm\\\\r = 5.2765 \ mm$

The magnitude of the electric field is now given as;

$E = \frac{8.5*10^{-6}}{2\pi(8.85*10^{-12})(5.2765*10^{-3})(50)} \\\\E = 5.793*10^5 \ V/m$

Therefore, the magnitude of the electric field halfway between the two cylindrical conductors is 5.793 x 10⁵ V/m

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3 years ago

How does an object's position and velocity change as the object accelerates?”

taurus [48]

Answer:

aerodynamics

Explanation:

if an object like a car is going 200 mph at max speed and then the car gets aerodynamic or smoothed to the point that air can get by the car it could end up going another 20 mph faster

7 0

3 years ago

The period of a simple pendulum is 3.5 s. The length of the pendulum is doubled. What is the period T of the longer pendulum?

Rudik [331]

Answer:

Explanation:

T = 2π√(L/g)

If you increase L to 2L, the period is increased by a factor of √2

T = 3.5√2 ≈ 4.9 s

6 0

3 years ago

The dense atmosphere of venus is ________, causing most incoming solar radiation to be _________.

The dense atmosphere of venus is , causing most incoming solar radiation to be _.