The atmosphere of Venus is mostly carbon dioxide, and the atmosphere of the Earth has water vapor. Why are these two gases absent in the atmosphere of the satellite around Saturn called Titan

Answers

Answer 1

Answer:

if these gases existed in Titan, this n in solid form, by which its absence in the atmosphere is understood.

Explanation:

Each gas and chemical compound has a defined temperature for changes of state, specifically for the change from gaseous to liquid and from liquid to solid state we have

gas                gas → liquid              liquid → solid

                         (ºC)                            (ºC)

H₂0 (Water)      100                                0

CO₂                 -56.6 (P> = 5.2 bar)     <-56.6

The temperature of the Titan satellite is - 180ºC

From the above, if these gases existed in Titan, this n in solid form, by which its absence in the atmosphere is understood.


Related Questions

A heavier person will have a lower blood level due to a greater amount of ___ in their

Answers

Answer:

Blood alcohol content (BAC) is a measure of the amount of alcohol that is present in the blood.

A heavier person will have a lower blood alcohol level due to a greater amount of water in their body. The more a person weighs, the more water he/she tends to have in his/her body. The water has a diluting effect on the alcohol so the BAC will be lower in comparison with a person who is slimmer  and has less water in the body.

Explanation:

An electrostatic paint sprayer has a 0.17 m-diameter metal sphere at a potential of 25.0 kV that repels charged paint droplets onto the object to be painted, which is grounded (electric potential of 0).

Answers

Answer:

[tex]q=0.236uC[/tex]

Explanation:

From the question we are told that:

Diameter [tex]d=0.17m[/tex]

Radius [tex]r=0.17/2=>0.085[/tex]

Potential [tex]E=25.0kV[/tex]

Generally the equation for Potential on spere is mathematically given by

[tex]E=\frac{1}{4 \pi e_0}*\frac{q}{r}[/tex]

Therefore

[tex]q=\frac{25*10^3*0.085}{\frac{1}{4 \pi e_0}}[/tex]

Where

[tex]\frac{1}{4 \pi e_0}=9*10^9[/tex]

Therefore

[tex]q=\frac{25*10^3*0.085}{(9*10^9}}[/tex]

[tex]q=0.236uC[/tex]

3. Thekinetic energy of an object of
mass, m moving with a velocity
of 5 ms-t is 25 J. What will be its
Toinetic energy when its velocity
is doubled? What will be its
Icinetic energy ruhen its velocity
is increased three times?​

Answers

Answer:

1. 100 J

2. 225 J

Explanation:

We'll begin by calculating the mass of the object. This can be obtained as follow:

Velocity (v) = 5 ms¯¹

Kinetic energy (KE) = 25 J

Mass (m) =?

KE = ½mv²

25 = ½ × m × 5²

25 = ½ × m × 25

25 = 25m / 2

Cross multiply

25m = 25 × 2

25m = 50

Divide both side by 25

m = 50 / 25

m = 2 Kg

1. Determination of the kinetic energy when the velocity is doubled.

Mass (m) = 2 Kg

Velocity (v) = double the initial velocity

= 2 × 5 ms¯¹

= 10 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 10²

KE = ½ × 2 × 100

KE = 100 J

2. Determination of the kinetic energy when the velocity increased three times.

Mass (m) = 2 Kg

Velocity (v) = three times the initial velocity

= 3 × 5 ms¯¹

= 15 ms¯¹

Kinetic energy (KE) =?

KE = ½mv²

KE = ½ × 2 × 15²

KE = ½ × 2 × 225

KE = 225 J

1.The distance moved by objects in a given interval of time can help us to decide which one is faster or slower. Do you agree with this statement?

Answers

Answer:

The rate of change of distance is defined as speed.

Explanation:

The speed is defined as the rate of change of distance.

Speed = distance/ time

When we know the distance and the time, we get the value of speed. So, e know that who is moving fast or slow.

hen a graph is pltted beteen the distance and time, the slope of the graph gives the value of speed. So, by checking the slopes, hoseslope ismore, the speed is more and thusit is moving faster.

So, i agree with the statement.  

(e) Find the height and orbital velocity of the geostationary satellite above the earth assuming earth as a sphere of radius 6370 km. [Radius of earth, R = 6370km, Acceleration due to gravity
= 10ms) ​

Answers

Answer:

I don't really know this one trust me am not lieing

An ice skater pushes harder with her legs and begins to move faster. Which two laws best describes this

Answers

Answer:

Newton' second law  and third law describes the situation.

Explanation:

According to the Newton's second law, the force applied on a body is proportional to the rate of change of momentum of the body.

According to the Newton's third law, for every action there is an equal and opposite reaction.

When ice skater pushes harder means more force is applied so he moves fast and more be the action force more be the reaction force.

Thus, Newton' second law  and third law describes the situation.

define regular body in science​

Answers

Answer:

All the sides and inner angles of a regular form must be equal. The sides and angles of an irregular form aren't the same. An equilateral triangle, for example, is a regular form because all of its sides and angles are the same length.

OAmalOHopeO

The electric potential energy of an electric dipole is ______ when the dipole is aligned with an electric field. The electric potential energy of an electric dipole is ______ when the dipole is aligned with an electric field. most positive most negative zero infinite undefined

Answers

Answer:

Most negative

Explanation:

The electric potential energy of a dipole U = -pEcosθ where p = dipole moment, E = electric field and θ = angle between p and E.

When the electric dipole is aligned with the electric field, the angle between p and E is 0°. That is θ = 0°.

So, U = -pEcosθ

U = -pEcos0°

U = -pE

which is the most negative value it can have.

So, the electric potential energy of an electric dipole is most negative when the dipole is aligned with an electric field.

So, most negative is the answer.

Help Meeeeeeeee. Have a nice day:)

Answers

Answer:

01.

Explanation:

Half the acceleration. Its heavier and moves slower. If it moved the same acceleration, the forces would also have to be doubled since the mass was.

an object that has lost its electrons become?​

Answers

Answer: A positively charged ion/ cation

Explanation: When an atom loses all of its electrons, it becomes a positively charged ion. Also referred to as the cation, the object will have a shortage of electrons, leading to a positive charge. Keep in mind that an atom will become positive if it loses electrons, and will become negative if it gains electrons.


Hope this helps! Comment below for more questions.

Imagine you see Mars rising in the east at 6:30 pm. Six hours later what direction would you face (look) to see Mars when it is highest in the sky

Answers

Answer:

The Mars appears in the direction of South.

Explanation:

Mars is rising in the east at 6: 30 PM. The period of rotation of earth is 24 hours.

So, 6 hours is the one fourth of the period of rotation of earth. Earth rotates counter clockwise on its axis, so after 6 hours, we see the Mars in the direction of South.

A train travelling at 20m/s accelerate at 0.5m/s2 for 30 seconds. How far will it travel in this time?

Answers

Answer:

825m

Explanation:

u=20m/s

a=0.5m/(s)^2

s = ut + 1/2a(t)^2

s = 20(30) + 1/2(0.5)(30)^2

s = 600 + 225

s = 825m

Answer:

as we know that

S=ut+1/2(at*t)

S=20*30+1/2(0.5*30*30)

S=600+225

S=825


A hamster in it's ball starts at rest and accelerates to 3ms1 in 6 seconds.
What was the distance the hamster travelled

Answers

Answer:9m

Explanation:

Ball starts from rest . Time taken = 6 seconds. Distance travelled by ball. ∴Distance travelled = 9 m

Hope it helps you

Good luck

Your parallel capacitors are 15 μf and 20 μf. The series capacitors are 10 μf and 12 μf. This circuit is connected to a 14 v battery, also determine the potential energy and the voltage across each capacitor

Answers

Answer:

a. i. 6.608 V ii. 5.507 V iii. 1.89 V iv. 1.89 V

b. i. 0.22 mJ ii. 0.182 mJ iii. 0.027 mJ iv. 0.036 mJ

Explanation:

a. The voltage across each capacitor

Since the 15 μf and 20 μf capacitors are in parallel, their total capacitance is C = 15 μf + 20 μf = 35 μf.

Also, since C is in series with the 10 μf and 12 μf which are in series, their total capacitance, C' is gotten from 1/C' = 1/10 μf + 1/12 μf + 1/35 μf  

1/C' = (12 + 42 + 35)/420 /μf

1/C' = 89/420 /μf

C' = 420/89 μf

C' = 4.72 μf

The total charge in the circuit' is thus Q = C'V where V = voltage = 14 V

So, Q = C'V =  4.72 μf × 14 V = 66.08 μC

Since the 10 μf and 12 μf are in series, Q is the charge flowing through them.

Since Q = CV and V = Q/C

i. The voltage across the 10 capacitor is

V = 66.08 μC/10 μF = 6.608 V

ii. The voltage across the 12 capacitor is

V = 66.08 μC/12 μF = 5.507 V

The voltage across the 15 μF and 20 μF capacitors.

Since the capacitors are in parallel, the voltage across them is the voltage across their combined capacitance, C

So, V = Q/C = 66.08 μC/35 μF = 1.89 V

iii. The voltage across the 15 μF capacitor is 1.89 V

iv. The voltage across the 20 μF capacitor is 1.89 V

b. The potential energy of each capacitor

i. The potential energy of the 10 μF capacitor

E = 1/2CV² where C = Capacitance = 10 μF = 10 × 10⁻⁶ F and V = voltage across capacitor = 6.608 V

E = 1/2CV²

E = 1/2 × 10 × 10⁻⁶ F(6.608 V)²

E = 5 × 10⁻⁶ F(43.666) V²

E = 218.33 × 10⁻⁶ J

E = 0.21833 × 10⁻³ J

E = 0.21833 mJ

E ≅ 0.22 mJ

ii. The potential energy of the 12 μF capacitor

E = 1/2CV² where C = Capacitance = 12 μF = 12 × 10⁻⁶ F and V = voltage across capacitor = 5.507 V

E = 1/2CV²

E = 1/2 × 12 × 10⁻⁶ F(5.507 V)²

E = 6 × 10⁻⁶ F(30.327) V²

E = 181.96 × 10⁻⁶ J

E = 0.18196 × 10⁻³ J

E = 0.18196 mJ

E ≅ 0.182 mJ

iii. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 15 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 15 × 10⁻⁶ F(1.89 V)²

E = 7.5 × 10⁻⁶ F(3.5721) V²

E = 26.79 × 10⁻⁶ J

E = 0.02679 × 10⁻³ J

E = 0.02679 mJ

E ≅ 0.027 mJ

iv. The potential energy of the 15 μF capacitor

E = 1/2CV² where C = Capacitance = 20 μF = 15 × 10⁻⁶ F and V = voltage across capacitor = 1.89 V

E = 1/2CV²

E = 1/2 × 20 × 10⁻⁶ F(1.89 V)²

E = 10 × 10⁻⁶ F(3.5721) V²

E = 35.721 × 10⁻⁶ J

E = 0.035721 × 10⁻³ J

E = 0.035721 mJ

E ≅ 0.036 mJ

Show that the path followed by a projected horizontally frop top of the tower with uniform speed is a parabola .​

Answers

Answer:

Explained below.

Explanation:

Let the initial velocity from the top of the tower of height(h) be u'

Now, since it is projected horizontally, the horizontal distance covered will be due to having a uniform horizontal velocity (u) while it will possess a vertical distance in the downward direction due to constant acceleration due to gravity (g).

If the time it takes the ball to reach the ground is "t", then we can say the horizontal distance travelled by the ball is denoted as; x(t) = x, while the vertical distance is denoted by; y(t) = y

Now, since it's a projectile the intial velocity (u) will have vertical and horizontal components which are;

u_y and u_x respectively.

Applying kinematic equations, we have;

x = u_x•t + ½at²

Acceleration is zero in the horizontal x direction. Thus;

x = u_x•t

For the vertical y-direction;

y = u_y•t + ½at²

Here since direction is in that of gravity, then a = g.

Also, since the initial velocity has no downward component, then u_y = 0 m/s

Thus;

y = ½gt²

From x = u_x•t, we have;

t = x/u_x

Thus;

y = ½g(x/u_x)²

y = ½gx²/(u_x)²

Let g/(u_x)² be treated as a constant with the letter k.

Thus;

y = kx²

This is the form of a parabolic equation.

Thus, it has been proved that the projectile follows the path of a parabola.

A small car increases its speed from 10 m/s to 15 m/s. Its mass is 1,200 kg. What is the impulse?
A. 6,000 kg-m/s
B. 18,000 kg-m/s
C. 12,000 kg.m/s
D. 1,200 kg-m/s

Answers

Answer:

A

Explanation:

We know that impulse(J) is the force times the change in time OR mass times the change in velocity. Because we're given mass and a final and initial velocity, we'll use the second option written as:

mΔv=J

Since Δ means change in, and Δv= vf-vi, we subtract 10 m/s from 15 m/s to get 5 m/s for Δv.  

Now plugging in Δv and m(1,200 kg)

1,200 kg * 5 m/s =J

6000 kg* m/s = J

To leave the gravitational pull of the Earth, and explore other planets, satellites must have at least:

Answers

Answer:

To explore the other planets, the satellite must have the velocity more than the escape velocity.

Explanation:

The minimum velocity required by any object to escape from the earth gravitational pull is called the escape velocity.

The escape velocity for any planet depends on the mass of planet and radius of planet. It does not depends on the mass of object. The escape velocity is same for any mass for a particular planet.

So, to explore the other planets, the satellite must have the velocity more than the escape velocity.

A water wave passes by a floating leaf that is made to oscillate up and down two complete cycles each second, which means that the wave's frequency is

Answers

Answer:

2 Hz.

Explanation:

Frequency is simply defined as the number of appearances of a periodic event occurring per time. It is usually measured in cycles/second.

Now, in this question, we are told that there are 2 cycles for each second.

Thus, we can say that the frequency is 2 cycles/1 s = 2 Hz.

A student sects a leaf of length 7.2 cm to draw. Her drawing is 28.8 cm in length. What is the magnification of the drawing?
a) x4
b) x207.36
c) x0.25
d) x36

Answers

Answer:

A) x4

Explanation:

Magnification is equal to image size divided by the actual size, or M = I/A.

The image size is the student's drawing, which is 28.8 cm, and the actual size is 7.2 cm. Divide them, and cancel out the units, and you should get:

28.8 cm/7.2 cm = 4

An instrument rated pilot is planning a flight under IFR on July 10, this year. Before conducting the flight, the pilot must have

Answers

Answer: See explanation

Explanation:

From the information given, we are informed that an instrument rated pilot is planning a flight under IFR on July 10, this year.

It should be noted that before conducting the flight, the pilot must have performed and logged the prescribed tasks and repetitions that are required for instrument currency no earlier than January, 10 for the year.

If the universe is infinite, then how come there are things disappearing from the horizon? I was at a point in this video (TRUE Limits Of Humanity from Kurzgesagt) about true limits of humanity, and the narrator mentioned that (04:12) "...there is a cosmological horizon around us. Everything beyond it, is traveling faster, relative to us, than the speed of light. So everything that passes the horizon is irretrievably out of reach forever and we will never be able to interact with it again." Are they talking about how galaxies passing the horizon are physically absent, or just visually aren't there and we are unable to interact with them?

Answers

Answer:

The big rip theory

Explanation:

      I believe what you are referring to is the big rip theory, in which the universe expands faster than the speed of light Kurzgesagt refers to it as a "horizon" but in reality it's a little more complicated than that. Eventually the expansion of the universe will accelerate far beyond the speed of light creating space between molecules until eventually all matter is fleeting and the entire universe is an endlessly vast cosmic void with not but the occasion molecule left from a time when things weren't so lonely.

A car travelling an unbanked curve of radius 200 ft notices a truckstopped on the road ahead. The driverimmediately applies brakes causing the speed of the carto decrease at the rate of 10 ft/s2. If at that instant, the stationary truckis 100 ft ahead (the distance is measured along the path) and the car is travelling at a speed of 40ft/s, whatis the magnitude of the relative velocity ofthe truck perceived by the driver of the car (i.e. from the moving frame of referenceof the car).

Answers

Answer:

 u = - 40 ft / s

Explanation:

The Galilean relation for the relative velocity is

         v ’= v + u

where u is the speed between the two reference frames, v is the speed of the fixed system and v 'the speed of the mobile system.

In this case the truck has a speed with respect to the ground (fixed system) 0 m / s (it is stopped), the car has a speed with respect to the ground of v = 40 ft / s,

        u = v'- v

         

        u = 0 - 40

        u = - 40 ft / s

the speed perceived by the car if the system is fixed on it is -40 ft / s

A uniform meter rule of mass 10g is balanced on a knife edge placed at 45cm mark. Calculate the distance of a mass 25g from the pivot

Answers

Answer:

2 cm.

Explanation:

Please see attached photo for diagram.

In the attached photo, y is the distance from the pivot to which the 25 g is placed.

The value of y can be obtained as follow:

Clockwise moment = 5 × 10

Anticlock wise moment = y × 25

Anticlock wise moment = Clockwise moment

y × 25 = 5 × 10

y × 25 = 50

Divide both side by 25

y = 50/25

y = 2 cm

Thus, the distance is 2 cm

A plane is flying a circular path at a speed of 55.0 m/ s, with a radius of 18.3 m. The centripetal force needed to maintain this motion is 3000 N. What is the plane's mass?

Answers

The plane has a centripetal acceleration a of

a = v ²/r

where v is the plane's tangential speed and r is the radius of the circle. By Newton's second law,

F = mv ²/r

Solve for the mass m :

m = Fr/v ² = (3000 N) (18.3 m) / (55.0 m/s)² ≈ 18.1 kg

In both the camera and the __________, light enters a narrow opening and is projected onto a photosensitive surface. Group of answer choices

Answers

Answer: The HUMAN EYE

Explanation:

The human eye is made up of different parts which ranges from controlling the amount of light that enters the eye to the focusing of the image that is formed. The camera is a device which is both mechanically and electronically operated which shares a number of similarities with the eye.

In the human eye, the IRIS helps to regulate the amount of rays passing through the pupil to the lens by either contracting or dilating in light or dark environment respectively. While in the camera, the DIAPHRAGM controls the amount of light entering the camera.

The PUPIL serves as the passage for light into the eye while in the camera, the APERTURE does the same.

The photosensitive surface in the eye is the YELLOW SPOT while in the camera, the photosensitive surface is the PHOTOGRAPHIC FILM.

Refer to the periodic table above a student is asked

Answers

Answer:

there is no context??????

A child is playing on a swing. As long as he does not swing too high the time it takes him to complete one full oscillation will be independent of

Answers

Answer:

We know that for a pendulum of length L, the period  (time for a complete swing) is defined as:

T = 2*pi*√(L/g)

where:

pi = 3.14

L = length of the pendulum

g = gravitational acceleration = 9.8 m/s^2

Now, we can think on the swing as a pendulum, where the child is the mass of the pendulum.

Then the period is independent of:

The mass of the child

The initial angle

Where the restriction of not swing to high is because this model works for small angles, and when the swing is to high the problem becomes more complex.

What happen to the frequency of transverse vibration of a stretched string if its tension is halved and the area of cross section of the string is doubled?

Answers

Answer:

The fundamental frequency of the stretched string is:

[tex]f=[/tex] [tex]\frac{1}{2} \sqrt{\frac{T}{L} }[/tex] [ T = Tension and μ = mass per unit length]

Here,

 μ = [tex]\frac{m}{L} = \frac{Vp}{L} = Ap[/tex]

[tex]f= \frac{1}{2} \sqrt{\frac{T}{Ap} }[/tex]

If T is halved and A is doubled,

[tex]f= \frac{1}{2} \sqrt{\frac{T'}{A'p} } = \sqrt{\frac{1}{2* 2* A* p} } = \frac{1}{2} (\frac{1}{2} \sqrt{\frac{T}{Ap} } = \frac{1}{2} f[/tex]

Thus, the frequency is reduced to half if its tension is halved and the area of cross-section of the string is doubled.

When the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

The frequency of transverse vibration of a stretched string is calculated as follows;

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }[/tex]

where;

T is the tension on the stringμ is the mass per unit length

[tex]f = \frac{1}{2l} \sqrt{\frac{T}{\mu} }\\\\f = \frac{1}{2l} \sqrt{\frac{Tl}{m} } \\\\f = \frac{l^2}{2l} \sqrt{\frac{T}{m} } \\\\f = \frac{A}{2l} \sqrt{\frac{T}{m}}[/tex]

when the tension is halved and the area of cross section is doubled, the frequency is calculated as;

[tex]\frac{f_1}{A_1 \sqrt{T_1} } = \frac{f_2}{A_2\sqrt{T_2} } \\\\f_2 = \frac{f_1 A_2\sqrt{T_2}}{A_1 \sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1\sqrt{0.5T_1} }{\sqrt{T_1} }\\\\f_2 = \frac{f_1 \times 2A_1 \times 0.7071\sqrt{T_1} }{\sqrt{T_1} }\\\\f_2 = 1.414 f_1[/tex]

Thus, when the tension is halved and the area of cross section is doubled, the frequency increases by a factor of 1.414.

Learn more about tension in a string here: https://brainly.com/question/25743940

A car is moving at 30km/h. What is its speed in m/s? Give your answer to 2 significant figures.

Answers

Explanation is in the attachment

hope it is helpful to you

While visiting the beach, you enjoy the warm ocean water, but the sand burns your feet. That night you walk along the beach and notice that the sand is colder than the ocean water. Why?
Group of answer choices

It takes a long time for sand to heat up, but it cools down very quickly. Water takes a short time to heat up and cool down.

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

Water is naturally colder than sand.

Sand is naturally colder than water.

Answers

The answer would be B..

Since sand can heat up quickly, it will also cool off quickly. But water takes a long time to heat up and cool down.

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