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Cracks form due to pressure from the nails and propagate through the weak parts of the wood. Wood is a bundle of fibers. These fibers are robust by themselves, but the bonding between individual fibers is weak. This creates weakness planes in wood (Its easy to axe wood along these planes). So if you could pull the fibers in the right direction they tear ...

5

The "$\omega$" in the drawing suggests that the tube is rotating. In that case there will be a pressure gradient in the horizontal part to provide the centripetal acceleration: $$\rho_{\rm air}\omega^2 r = \frac{dP}{dr}$$

3

Let me try to clarify a few key points. $c_p=c_v+R$ is an equality between two state functions (the specific heat depends on the process connected to the heat transfer, but the final quantity is a function f the state). The two specific heats actually correspond to different processes (isobaric and isochoric). It is wrong to think that the equation is valid ...

2

Since the fluid is pushed down by a piston 1 and moves upwards pushing piston 2, then how can we still talk about Pascal's law while the fluid is moving (i.e dynamic pressure would be in presence and thus the pressure is not really equal at pistons based on Bernoulli's principle)? Strictly speaking, you are absolutely correct. Technically while the lift is ...

2

“In an adiabatic process we know there is no change in enthalpy or basically enthalpy change is zero.” This is untrue. Adiabatic compression of a real or ideal gas will increase its enthalpy and temperature. (For an ideal gas, enthalpy is a function of temperature.) People often think enthalpy is the stored thermal energy and only heat flow can increase ...

2

According to Wikipedia, Charles law states: When the pressure on a sample of a dry gas is held constant, the Kelvin temperature and the volume will be in direct proportion Meaning the law doesn't say that when heating something, its pressure doesn't change, but the opposite: if you guarantee that the pressure doesn't change, so a change in temperature will ...

1

The first one is correct. The second one (the way the diagram says it anyway) is incorrect and a very common misconception, or at best misleading and rather thin. In my mind, it's better to say there exist higher and lower pressure regions BECAUSE of the force of the wings on the air. They are two sides of the same coin, but the wing is causing that ...

1

As an explanation of wing lift, I prefer the downward deflection story. This explanation can be explained with other experiences with fluids. A swimmer propels themselves forward by pushing water backwards. An airplane stays up by pushing air down. If you hold a flattened hand out of a moving car and tilt the leading edge upwards, you can feel the wind ...

1

I reasoned that since the water-pressure was higher at the lower faucet... Why is the pressure higher at the lower faucet? It is because of the additional weight of the water pushing on the lower faucet that is not pushing on the upper one. This difference is $\rho g h$. The same reasoning happens on the "pipe" side of the connection as on the &...

1

Let h be the height in cm that the water rises to in the small tank, and 100-h be the distance spanned by the air (in cm). If the gauge pressure at the bottom is 14 psi, the gauge pressure in psi at the air interface is $$p_{ga}=14-\frac{h}{(2.54)(12)}0.4333=14-0.0142h$$where the 0.4333 is the rate of change of water pressure with elevation in psi/ft. So ...

1

There may be several variables. If the extra air is coming from a reciprocating pump, it will come in pulses of varying pressure which are being heated by compression. More predictable is air from another tank at a higher pressure. This cools on leaving the other tank (which may be losing pressure along with the air) and then heats the receiving tank by ...

1

Let $p_i$ be pressure inside the bubble and $p_o$ be the pressure outside. Excess pressure is then $$p = p_i - p_o$$ whether the bubble is in the air or in the water. That is the meaning of excess pressure. Why the expression is given by $\frac{4S}{R}$ as above can be calculated as follows below. The work done by this pressure is force$\times$displacement ...

1

Thank you for clarifying your earlier comment. How much work that is done when reach the final state is dependent on what path we take. This is not true, because (1) we know the initial state (very large atmospheric volume of $V_0$ and zero temperature, in this cartoon idealization) and the final state (atmospheric volume $V_0-V$, where $V$ is the system ...

1

The answer of Thomas is correct; after learning from it I looked into this some more and I am adding this answer to provide more details for anyone interested. The question was two-fold: whether the pressure is uniform in a stationary gas with a temperature gradient, and if so then what is wrong with a simple argument concerning diffusion which suggests the ...

1

I know the pain as i been through the same in understanding why at max pressure the displacement is zero and at max displacement the pressure is normal.but don't worry just imagine what i will tell you and it will be fine. In the particle displacement curve there are points where particles don't move. These are the points where you get max pressure. Treat ...

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