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irina1246 [14]
2 years ago
10

On which of saturn’s moons did the cassini-huygens probe land in 2004, providing our first view of the varied and active surface

?
Physics
1 answer:
yawa3891 [41]2 years ago
4 0

On Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

To find the answer, we have to know more about the Cassini-Huygens Mission.

<h3>What is Cassini-Huygens mission?</h3>
  • Before arriving at its final destination of Saturn in 2004 and beginning a series of flybys of Saturn's moons, the spacecraft contributed to studies of Jupiter for six months in 2000.
  • In the same year, it launched the Huygens probe to explore Titan's atmosphere and surface makeup on Saturn's moon.
  • During its second extended mission, Cassini sailed between the rings, entered the planet's atmosphere, and obtained the first measurements of a whole seasonal period for Saturn and its moons.

Thus, we can conclude that, on Titan, the largest moon of of Saturn did the Cassini-Huygens probe land in 2004.

Learn more about the Cassini-Huygens mission here:

brainly.com/question/27907891

#SPJ4

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Freight car A with a gross weight of 200,000 lbs is moving along the horizontal track in a switching yard at 4 mi/hr. Freight ca
zhenek [66]

Answer: a) 4.7 mi/hr.  b) 86,500 lbs. mi²/Hr²

Explanation:

As in any collision, under the assumption that no external forces exist during the very small collision time, momentum must be conserved.

If the collision is fully inelastic, both masses continue coupled each other as a single mass, with a single speed.

So, we can write the following:

p₁ = p₂ ⇒m₁.v₁ + m₂.v₂ = (m₁ + m₂). vf

Replacing by the values, and solving for vf, we get:

vf = (200,000 lbs. 4 mi/hr + 100,000 lbs. 6 mi/hr) / 300,000 lbs = 4.7 mi/hr

If the track is horizontal, this means that thre is no change in gravitational potential energy, so any loss of energy must be kinetic energy.

Before the collision, the total kinetic energy of the system was the following:

K₁ = 1/2 (m₁.v₁² + m₂.v₂²) = 3,400,000 lbs. mi² / hr²

After the collision, total kinetic energy is as follows:

K₂ = 1/2 ((m₁ + m₂) vf²) = 3,313,500 lbs. mi²/hr²

So we have an Energy loss, equal to the difference between initial kinetic energy and final kinetic energy, as follows:

DE = K₁ - K₂ = 86,500 lbs. mi² / hr²

This loss is due to the impact, and is represented by the work done by friction forces (internal) during the impact.

8 0
3 years ago
Chemistry The iron atom (Fe) has 26 protons, 30 neutrons, and 26 electrons. The diameter of the atom is approximately 1.0 × 10 −
Alisiya [41]
Don’t mind this i just need to answer under something because i just signed up !
4 0
3 years ago
Q2. You push a crate up a ramp with a force of 10 N. Despite your pushing, the crate slides down the ramp 4 m. How much work did
Ksivusya [100]

Answer:

40 J

Explanation:

From the question given above, the following data were obtained:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Work done is simply defined as the product of force and distance moved in the direction of the force. Mathematically, we can express the Workdone as:

Workdone = force × distance

Wd = F × s

With the above formula, we can obtain the workdone as follow:

Force (F) = 10 N

Distance (s) = 4 m

Workdone (Wd) =?

Wd = F × s

Wd = 10 × 4

Wd = 40 J

Thus, 40 J of work was done.

5 0
3 years ago
An automobile of mass 2000 kg moving at 30 m/s is braked suddenly with a constant braking force of 10000 N. How far does the car
saveliy_v [14]

Answer:

The car traveled the distance before stopping is 90 m.

Explanation:

Given that,

Mass of automobile = 2000 kg

speed = 30 m/s

Braking force = 10000 N

For, The acceleration is

Using newton's formula

F = ma

Where, f = force

m= mass

a = acceleration

Put the value of F and m into the formula

-10000 =2000\times a

Negative sing shows the braking force.

It shows the direction of force is opposite of the motion.

a = -\dfrac{10000}{2000}

a=-5\ m/s^2

For the distance,

Using third equation of motion

v^2-u^2=2as

Where, v= final velocity

u = initial velocity

a = acceleration

s = stopping distance of car

Put the value in the equation

0-30^2=2\times(-5)\times s

s = 90\ m

Hence, The car traveled the distance before stopping is 90 m.

6 0
3 years ago
HELP PLS MARKING BRANLIST 100 pts TAKING TEST RN
AlladinOne [14]

Answer:

15 m/s^2 The first thing to calculate is the difference between the final and initial velocities. So 180 m/s - 120 m/s = 60 m/s So the plane changed velocity by a total of 60 m/s. Now divide that change in velocity by the amount of time taken to cause that change in velocity, giving 60 m/s / 4.0 s = 15.0 m/s^2 Since you only have 2 significaant figures, round the result to 2 significant figures giving 15 m/s^2

Explanation:

8 0
3 years ago
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