Tag Info

0

Accelerating upwards would initially increase the weight of the mass of the mercury causing it to lower in the tube. Changing air resistance would depend on the shape and design of the barometer. If it were accelerated upward for a large distance other factors would have to be taken into account, such as lower g from increased r from Earth, and lower air ...

0

The acceleration increases the weight $m(g+a)$ of the mercury. If we take the moment when the velocity is just starting to increase from zero, not more changes but this. So the column height decreases. With the gradual increase of velocity, there is a drag force that plays the opposite role. I expect that eventually it overcomes the initial effect of the ...

-1

for sucking earth's atmosphere firstly you need to get rid of gravity so basically you have to pull all the gases and do some work. Assume that you somehow managed to suck all the air. firstly you need a good amount of space to store the gases. and as now you have generated a pressure the pressure outside is zero. So gases will try their best to get rid of ...

5

No. The Earth's atmosphere is held in place entirely by the Earth's gravity. There is no barrier at the top of the atmosphere keeping it from escaping. Running a straw from the ground to deep space would not change the gravitational force holding the air molecules down; it would, in fact, change nothing.

4

You are missing a $-\cos\theta$ factor: this will account for the fact that you are integrating the vertical (upward) components of the force distribution on the sphere. Without this, you are just integrating the magnitude of this force, which isn't very useful.

0

It is not the weight of the He gas that determines the buoyancy but the difference of Helium and air. When He is in a bottle it is usually a very heavy steel one and it is either a liquid at very low temperature or a compressed gas a room temperature. In both cases the Helium weighs more that the air it replaces and in any case even the evacuated bottle will ...

2

Substituting helium for air inside a bottle will make the bottle lighter, but unless the combination of helium plus bottle is lighter than an equal volume of air, the helium-filled bottle will NOT float. Putting helium inside a bottle that is initially totally empty (that is, it initially contains a vacuum) simply increases the total weight of the bottle ...

0

Yes. If something has a net density lower than air, it will float. If the circumstances result in a bottle with helium having a net density lower than air, it will float.

0

It depends on the type of valve. Butterfly valves for example have a approximately linear behaviour only at a few percent of opening. Gate valves are more linear. But for small openings as 10% to 20%, an almost linear behaviour is expected. That means the flow doubles, provided the volume of the reservoir is big enough to have its level practically constant.

0

Consider a vessel of any shape. At any 2 or more points is the vessel that are coplanar, no matter what the shape of the vessel, the pressure will be equal. Just collision of liquid particles? If an object is immersed in the liquid, pressure will act on all directions on it from the liquid's particles. The net pressure (the pressure acting on both sides of ...

0

The fraction submerged will not depend on acceleration; it will depend on velocity. Given constant acceleration beginning at time =0, at the first instant the velocity is zero, so the boat is still half submerged immediately after acceleration starts.

0

So static pressure is not due to any external force when the fluid is at rest but if the fluid is in motion an additional static pressure is created by a part of external force to overcome the container's resultant friction(ie constraint) leading to motion of the fluid .This additional static force may be termed as fluid energy due to the additional static ...

0

The Bernoulli equation (which ignores the effect of friction) expresses conservation of energy density. Changes in pressure are included to account for the work done by pressure between the two points under consideration. If the cross section decreases, the velocity must increase. The pressure does work in increasing the kinetic energy (and can exert less ...

0

Apart from the salty/freshwater discussion. If all floating ice (the future for the Northern ice cap looks dim) is melted, less radiation from the Sun is reflected back in the atmosphere. More radiation is absorbed by the water that takes the place of the ice. This on its turn means that the oceans are slightly warmed. Because the average temperature of the ...

0

There are already great answers to this question. This answer is just a slightly intuitive version of Yashas' answer based on this answer to a related question, Why does ice melting not change the water level in a container? We shall analyse what happens to the water level after the ice melts completely for four different cases, which arise due to the ...

1

There is no rigorous definition of laminar v/s turbulent flow. But here is something that comes close: A turbulent flow is characterised by a wide range of length and time scales, while a laminar flow is not. In practice, you can record a signal from the flow (for e.g. velocity, pressure) and subject it to Fourier transform. If the wavenumber or frequency ...

1

the transition from laminar to turbulent begins to occur when the magnitude of the forces caused by viscosity (which tend to smooth out the flow and damp out instabilities) become smaller than the magnitude of the forces generated by the motion of a parcel of the fluid, as gauged by that parcel's kinetic energy. The fluid flow is thus stable to perturbations ...

2

The change from laminar flow to turbulent flow occurs within a range of flows, not at a particular point. The Moody diagram shows this range, and in that range, it is very difficult to confidently state that the flow is laminar or turbulent. For more info, see en.wikipedia.org/wiki/Moody_chart

4

There is no mathematically rigorous line. You are correct that we use the Reynolds number to define the regions, but there is no strict line in the sand between them. As tough as turbulent flow can be to calculate, its actually easier than calculating behaviors in the transition region between laminar and turbulent flow. In laminar flow there are things ...

3

Your interpretation is correct. Single Bubble Case We have a bubble of vapour surrounded by fluid water, with the vapour having a much lower density than the water. The average gas pressure in the bubble will be equal to the average water pressure around the bubble. But the water pressure changes with the height above the bottom surface, while the gas ...

1

Close. When water is heated, the water which is closest to the flame will start to vibrate rapidly such that the hydrogen bonds between the water molecules will break forming a gaseous vapour. These vapours will form bubbles. Because the density of these bubbles are much lower than the surrounding water, they will feel a buoyant force and therefore rise.

0

This question has been bothering me for a while, and here's what I've come up with. Bernoulli's equation is clearly the starting point to think about this problem: $p + \frac{1}{2} \rho v^2 + \rho g h = E$ where $E$ is a quantity they is conserved throughout the fluid, basically its energy per unit volume. In other words, the sum of the kinetic energy, ...

1

The region to focus upon is not above dotted line AB, but below this line. Since, according to Pascal's law, at a given location in the fluid, pressure is the same in all directions, the pressure at static equilibrium does not vary horizontally; it only varies vertically, according to $$\frac{dp}{dz}=-\rho g$$where z is the elevation. So, if you integrate ...

2

In your diagram, if the pressure were higher on one side then it would push the other side up until pressures were equal. If you have liquids of different densities, that means that their weights will be different for the same volumes. You can think of the U tube as being in balance, you will have the same weight of liquid on each side of the center even if ...

1

IMO there are a number of issues with the wording of the problem. First, the beginning statement says the buckets are of "equal weight". But part 1 asks to calculate the weight of the bucket. It makes no sense. The true weight of the bucket doesn't change inside or outside the water. It's not clear whether the parenthetical comment is yours or in ...

1

The pressure at I and J obviously is not the same, but can be reasonably assumed it is. The pressure difference is $$\Delta p = g \cdot \rho_\mathrm{air} \cdot \Delta h$$ With air density ( about $1.2\ \mathrm{kg/m^3}$ ) roughly 1000 times lower than water density, $100\ \mathrm{mm}$ difference in the level heights makes about $0.1\ \mathrm{mm}$ of the ...

Top 50 recent answers are included