A saturated air parcel is at a temperature of 10 degrees Celsius. It is forced to rise over a mountain that i?
You are describing what is called the Foehn wind, also known as the Chinook wind in the USA and Canada. It works like this: Since the air is saturated, it will cool down at the saturated adiabatic lapse rate of 0.5 C per 100 meters. If it rises 5,000 meters, it will cool down 25 C degrees. At the top of the mountain, the temperature will then be 10 - 25 = -15 C. Since moisture has gone out of the air as it condenses into precipitations, when it comes down on the lee side of the mountain, it will be at the dry adiabatic lapse rate of 1 C per 100 meters (actually, I think it is 0.97 C but I round it to 1 C). When it will reach again the sea level (or the base of the mountain, assuming it is also 5000 meters down) it will warm up by 50 degrees and the temperature will be -15 + 50 = 35 C. Of course, this is a theoretical question. Usually, on the weather side of the mountain, it will take a while before the temperature sinks to dew point. Then on the lee side, some moisture will be left and it may take some time before it warms up at the dry adiabatic lapse rate. Then ... a 5000 meter high mountain is ... a high mountain! It is 200 meters higher than the highest mountain of Europe, the Mont Blanc. The air moving over the Mount Blanc is not much since it is a peak and most of the wind simply moves around and not over the peak.
A parcel of air is forced to rise up and over a 5000 meter high mountain. The initial temperature of the parc?
Given those factors, the air will cool to the dew point at a level of 3000m, since that amount of rise will cause the air parcel to cool 24C (8C x 3km @ 8 degrees of cooling per km) to 12C. Now condensation will initiate which reduces the cooling to 4C per 1km rise. Since the air will rise another 2km, it will cool another 8C, thus giving a final temperature of 4C.
Why is the air “thinner” in the mountains than at sea level?
As the others have mentioned, the air at a certain height is compressed by the air above it. You can imagine a column of air, with its bottom at sea level, rising to the top of the atmosphere. At sea level, the air is being compressed by all of the air in the column. That is a lot of weight, about 10,000 kg per square meter on the surface of the Earth. A significant fraction of that is also sitting on top of you, by the way. All that weight, pushes down hard on the air and compresses it. When you compress a gas, it becomes denser.As you go up in elevation, the air at the bottom of the column goes below you and the column of air above you, pushing down, gets shorter. Moreover, because the densest air is at the bottom, the weight of the column above you decreases rapidly.So, as you increase in elevation there is less force on the air molecules around you, and the pressure is lower. Lower pressure means lower density.For example, Mount Everest is about 8,848 meters above sea level. From the above graph, you can see that the air pressure at that height is less than one third of the pressure at sea level. Therefore, the density of air is also less that one third of the density at sea level. That means every breath draws in less than one third of the oxygen that most people are comfortable with. This is why it is absolutely necessary to wear an oxygen mask when climbing that high:
A hot air balloon is moving vertically upwards with a constant speed of 3.00 m/s. A sandbag is dropped from the balloon. It takes 5.00 seconds for the sandbag to fall to the ground. What was the height of the balloon when the sandbag was released?
Easy peasy. All you need is 2 dist-time equations, and a basic understanding of the object trajectory.Now, you’re on top of that hot-air balloon, you drop the sand-bag. But wait!Your frame of reference (hot-air-balloon) also has an upward velocity of 3 m/sec, meaning the initial velocity of the sandbag when dropped is 3 m/sec upwards. The difference in trajectories of the sandbag and the balloon is now caused by the gravitational pull of the Earth causing a deceleration of ~10 m/sec^2 .i). Understanding the basic trajectory :Sandbag has an initial upward velocity of 3 m/sec; and a deceleration due to gravity of ~10 m/sec^2. It soon loses all its upward velocity and peaks at a point of zero velocity. It is free fall from here.At the original dropping height, the sandbag has an equal but opposite velocity of the dropping value (i.e - 3 m/sec, or 3 m/sec downwards). Let us assume the sandbag took time t to reach its original dropping height. It then falls for time 5 - t sec. The dist (height) is easily calculable.ii). Using the first equation :So, let’s calculate the time taken for the sandbag to fall back to its original height (where it is released). Use equation : v = u + a*t; where u = 3 m/s (initial velocity). v = -3 m/s (final velocity), a = -10 m/sec^2 (acceleration downwards), and t = time taken (which we find here).This equation gives us t = 0.6 sec; this is the time taken for the sandbag to reach the height at which it was dropped.iii). Using the second equation :Now that you have the time taken to reach the drop height, you can calculate the remaining time of free fall from the dropping height as 5 - t = 5 - 0.6 = 4.4 sec (since it takes a total of 5 sec for this whole drama to unfold).We now have the time of free fall from drop position (t = 4.4 sec), we have the downward acceleration ( a = 10 m/sec^2 , no need to take a negative val here, we’ll be taking a positive velocity as well), and a drop velocity of u = 3 m/sec.Using equation s = u*t + 1/2*a*t^2, we calculate the ‘s’, i.e dist traveled (height) after drop until it reaches the ground. Plugging in the above values of u,t and a; we get s = 110 m, which is the height from which the sandbag was dropped.P.S : You’ll get a more accurate answer by plugging in the value 9.8 m/sec^2 for acceleration due to gravity.
Why does my bag of chips expand at high altitude?
PRESSURE! PRESSURE! PRESSURE!If there's some air in the bag before the increased altitude it will inflate. Here's why:Atmospheric pressure decreases with increasing altitude. Let's assume you tie up a bag at sea level altitude (with little air inside). That air inside the bag is at 1atm as long as you're at sea level.Now you start going up, your altitude increases and atmospheric pressure decreases (remember what I said in the beginning about atmospheric pressure decreasing with increasing altitude?).Let's say you reach an altitude where the atmospheric pressure is 0.5atm, the force acting inside the bag which is pushing it outward is now greater than the force acting on it which is pushing it inward (1atm>0.5atm). As a result, the air inside the bag will push the sides of the bag outward, thus inflating it to a point where the pressure inside the bag becomes 0.5atm or the bag stops stretching.For the same reason, if you bring the same bag down and take it too deep into the earth (at an altitude lower than sea level) or if you inflate the bag at a very high altitude and bring it down to sea level, it will deflate.
How does a bag of potato chips ("crisps" in British English) become inflated when going up a mountain?
The air in the bag remains equal to the outside pressure, as long as the bag is not under tension. How long that situation lasts depends entirely on the volume change possible in the bag. (Weather balloons are only partly inflated, calculated so they can rise to high altitude before becoming fully inflated at a height determined by the weight they carry) . As you go higher, the pressure inside drops too, both by cooling, and by expansion. (The outside air is normally colder). The expansion and pressure change in the bag stops whenever the bag should become inflated and start to resist further volume change. The bag then increasingly stores more tension energy as you go higher, by very small expansion actually, as the outside pressure drops further, unless it bursts of course. The continued cooling of the bag will lower the internal pressure a little. You can treat temperature as a separate influence. The effect of temperature was to cool the gas and help lower pressure, ie it would not need to expand as much. If you want to heat the gas instead, then the temperature increase will require more expansion still to equalise pressures. It just means you will hit the limit sooner. If the temperature then keeps rising without limit so will the pressure, until it bursts the bag.If you want to calculate, energy can do that for you. Let's look at a bag or balloon with tension in the bag caused by the pressure difference. Energy in surface = pressure energy Tension*area=(Pgas-Patm)*volumeThe gas law at constant volume:Pgas = k Tgas
How far is Jupiter from the moon?
Jupiter is about 669 million kilometers away from earth. Earth is about 385,000 kilometers away from the moon (our moon). The exact distance depends on the moon's orbit, the earth's orbit, and jupiter's orbit, but either way would be about 669 million kilometers.
If a hole is made right through Earth so it reaches the other side, what will happen when one jumps into it?
Earth’s center of mass is pretty much at its core. It is therefore clear that the attractive force exerted by the Earth towards all bodies is directed towards its center.Why is this important? Say the particular hole looks like thisSay the center is at B (it’s not exactly B, but slightly below B). When the person is in the first leg of the journey, i.e. not yet reached the center, he is falling to the center, or basically towards the point of attraction. So, his speed naturally increases.Once he passes B, though, he is now falling away from the center, which is kind of equivalent to jumping up. The force he now experiences is against the direction of his motion, so he now decelerates. Once he completely stops, he falls back towards to the center and the entire cycle repeats.If you’ve noticed, this is basically an SHM (simple harmonic motion). He will keep oscillating about the center with a time period of about 43 minutes (if I’m not wrong).This, however, is assuming that you initially simply dropped into the hole or had low start speed. If you were actively launched into the hole at a high speed, it is possible you may not decelerate fast enough to stop within the hole. You may fly right out the other side. If the speed is high enough, you could even reach space.