Why is there so much more carbon dioxide in the atmosphere of Venus than in that of Earth? Why so much more carbon dioxide than on Mars?

Answers

Answer 1

Answer:

Explanation:

The reason for the more concentration of carbon dioxide in the atmosphere of Venus than in the Earth -

On the Earth , most amount of the carbon dioxide is in the ocean water and in sea sediments .

Considering Venus , in the planet Venus , there is no Ocean water , hence , carbon dioxide can not get dissolved into the water , hence , it is found in the atmosphere .

So , the escape velocity for carbon dioxide on Mars is smaller than Venus .

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A vertical straight wire carrying an upward 24-A current exerts an attractive force per unit length of 88 X 104N/m on a second parallel wire 7.0 cm away. What current (magnitude and direction) flows in the second wire?
Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.
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Part A (4 pts) Consider light of wavelength λ = 670nm traveling in air. The light is incident at normal incidence upon a thin film of oil with n2 =1.75. On the other side of the thin film is glass with n3 =1.5. What is the minimum non-zero value of the film thickness d that will cause the reflections from both sides of the film to interfere constructively?

A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .

So we get

$B = \cfrac{2\pi * 1.67 * 10^(-27) kg * 9.7 * 10^6 Hz}{1.6 * 10^(-19)C}$

$B =0.636 T$

We have found the initial magnetic field magnitude of 0.636 T

Magnetic field magnitude at increased Hall voltage.

The relation given by Hall voltage with the magnetic field is:

$V_H =\cfrac{R_HI}t B$

Thus if we keep the same current we can write for both cases:

$V_(H1) =\cfrac{R_HI}t B_1\nV_(H2) =\cfrac{R_HI}t B_2$

Thus we can divide the equations by each other to get

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{\cfrac{R_HI}t B_1}{\cfrac{R_HI}t B_2}$

Simplifying

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{ B_1}{ B_2}$

And we can solve for $B_2$

$B_2 =B_1 \cfrac{V_(H2)}{V_(H1)}$

Replacing the given information we get

$B_2= 0.636 T * \left(\cfrac{1.735 mV}{0.549 mV} \right)$

We get

$\boxed{B=2.01\, T}$

Thus when the Hall voltage is 1.735 mV the magnetic field magnitude is 2.01 T

You are given a parallel plate capacitor with plates having a rectangular area of 16.4 cm2 and a separation of 2.2 mm. The space between the plates is filled with a material having a dielectric constant κ = 2.0.Find the capacitance of this system

Answers

Answer: C = 1.319×10^-11 F

Explanation: The formulae that relates the capacitance of a capacitor to the area of the plates, distance between the plates and dielectric constant is given as

C = kε0A/d

Where C = capacitance of plates =?

k = dielectric constant = 2.0

Area of plates = 16.4cm² = 0.00164 m²

d = distance between plates = 2.2 mm = 0.0022m

By substituting the parameters, we have that

C = 2 × 8.85×10 ^-12 ×0.00164/ 0.0022

C = 0.029028 × 10^-12/ 0.0022

C = 13.19× 10^-12

C = 1.319×10^-11 F

Why a switch is connected in phase wire and never is neutral wire?

Answers

If you had the wire connected to a neutral wire, you would never get a charge. You could receive a charge pulse from the phase wire.

Answer:

ifyou have a wre connect you should not have to connected

i think that is the answer

1. Towards the end of a 400m race, Faisal and Edward are leading and are both running at 6m/s. While Faisal is 72m from the finish line Edward is 100m from the finish line. Realising this and to beat Faisal, Edward decides to accelerate uniformly at 0.2 m/s2 until the end of the race while Faisal keeps on the same constant speed. Does Edward succeed in beating Faisal?

Answers

Answer:

Explanation:

Faisal will finish the race in ...

(72 m)/(6 m/s) = 12 s

In order to beat Faisal, Edward's average speed in those 12 seconds must exceed ...

(100 m)/(12 s) = 8 1/3 m/s

To achieve that average speed, Edward's acceleration must be ...

(8 1/3 m/s -6 m/s)/(12 s/2) = 7/18 m/s² ≈ 0.3889 m/s²

Accelerating at only 0.2 m/s², Edward will not beat Faisal.

_____

Additional comment

When acceleration is uniform, the average speed is reached halfway through the period of acceleration.

Based on the measured force between objects that are 10 meters apart, how can you find the force between objects that are any distance apart ?

Answers

The force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

According to the gravitational law, the force acting on an object is directly proportional to the product of their masses and inversely proportional to the square of their distance apart. Mathematically,

$P=(GMm)/(r^2)$

M and m are the masses

r is the distance between the masses

If the force between objects that are 10 meters apart, hence;

$P=(GMm)/(10^2)\nP=(GMm)/(100)\nGMm = 100P$

To find the force between objects that are any distance apart, we will use the same formula above to have;

$P'=(GMm)/(r^2)\n$

Substitute the result above into the expression to have:

$P'=(100P)/(r^2)$

Hence the force between objects that are any distance apart is expressed as $P'=(100P)/(r^2)$

Learn more on gravitational law here: brainly.com/question/11760568

Answer:

F' = 100 F/r²

Explanation:

The gravitational force of attraction between two objects is given by the Newton's Gravitational Formula. The Newton's Gravitational Formula is as follows:

F = Gm₁m₂/r²

where,

F = Force between objects

G = Universal Gravitational Constant

m₁ = mass of first object

m₂ = mass of second object

r = distance between objects = 10 m

Therefore,

F = Gm₁m₂/10²

Gm₁m₂ = 100F --------------------- equation (1)

Now, we consider these objects at any distance r apart. So, the force becomes:

F' = Gm₁m₂/r²

using equation (1), we get:

F' = 100 F/r²

So, if the force (F) between objects 10 m apart is known, we can find it at any distance from the above formula.

A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.What is the speed of ball 1?