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Tju [1.3M]
3 years ago
6

What are some of the obstacles to exploration of outer planets or even Venus right next door?

Physics
1 answer:
weqwewe [10]3 years ago
5 0

Answer:

Distance and coping with the extreme atmospheres

Explanation:

The outer planets and venus are so far away that they are difficult to visit. Even if they were closer, the extreme atmospheric pressure and weather on these planets would make them  difficult to explore.

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Why don't the planets move exactly along the ecliptic?
AVprozaik [17]
As seen from the Earth, the Sun, Moon, and planets all appear to move along the ecliptic. ... Unlike the Sun, however, the planets don't always move in the same direction along the ecliptic. They usually move in the same direction as the Sun, but from time to time they seem to slow down, stop, and reverse direction!

Because of various events in their (unknown) past history that resulted in deviations from the theoretical orbit. That formed in the plain of the ecliptic.

Capturing a large passing comet or asteroid might do it.
6 0
3 years ago
2. Compare and Contrast A fault cuts through
Elenna [48]

Explanation:

The principle of cross-cutting relationships states that a fault or intrusion is younger than the rocks that it cuts through. The fault labeled "E" cuts through all three sedimentary rock layers (A, B,and C) and also cuts through the intrusion (D). So the fault must be the youngest formation that is seen and known of.

6 0
3 years ago
If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan
andrew-mc [135]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².  

<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>

  • Δt = (√2xgxh)/9,8
  • Δt = (14√10)/9,8
  • Δt ≈ 4,5 s

4 0
3 years ago
A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled
GaryK [48]

The boat's position x relative to its starting point x_0=0 is determined by

x=x_0+v_0t+\dfrac12at^2

where v_0 is its initial velocity, a is its acceleration, and t is time. After t=6\,\mathrm s, the boat has traveled

x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2

\implies x=91\,\mathrm m

3 0
3 years ago
(1 pt) A bucket of water of mass 20 kg is pulled at constant velocity up to a platform 35 meters above the ground. This takes 14
Elenna [48]

Answer:

w = 5832.372 Joules

Explanation:

Mass of water, m = 20 kg

The water was pulled up to a height of 35 meters, i.e. h = 35 m

It takes 14 minutes to pull up the water through the height, 35 m

speed = distance/ time = 35/14 = 2.5 m/min

The bucket's height, y = speed * time = 2.5t meters

6 kg of water drips out of the bucket throughout the 14 minutes

The rate at which the water drips drips out = (6/14) = 0.4286 kg/min

Mass of water that drips out in time, t = 0.4286t kg

The mass of water remaining = (20 - 0.4286t) kg

Change in Workdone, Δw = mgΔy

Δy = 2.5 Δt

Δw = mg *  2.5 Δt

dw =  (20 - 0.4286t)g2.5 dt

integrating both sides

dw = (50g - 1.07gt)dt

w = \int\limits^a_b {(50g-1.07gt)} \, dx where b = 0, a = 14

w = 50gt - 1.07g(t²)/2      g = 9.8 m/s²

w = 490t - 5.243t²

w = (490*14 - 5.243*14²) - (490*0 - 5.243*0²)

w = 6860 - 1027.628

w = 5832.372 Joules

3 0
3 years ago
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