## anonymous 4 years ago Can anyone show me an example physics numerical which cant be solved just by memorizing formulas and need deep understanding of the concept?Please explain all the steps used. Please explain those deep concepts also

1. anonymous

you know Kirchoff's laws ? if you do, then you know that the closed loop integral of a circuit is always equal to 0, but that also requires you to to take the potential difference of the battery, what if you don't have a battery but instead you have a magnetic field exactly as required for the same amount of current to flow through the circuit.. would Kirchoff's law fail ? if not, why ? if yes, then what is the integral equal to ?

2. anonymous

To paraphrase Poincare: Science(physics) is no more a collection of facts (formulas) than a house is a collection of stones. Formulas are derived from the deep understanding of concepts for instances of special interest to demonstrate the quantitative relationship among the physical quantities involved and for convenience and ease in obtaining desired numerical results for these instances. The deep understanding of concepts is necessary to determine the proper relationship among the physical quantities from which you can choose appropriate formula's to obtain desired information. Most real problems involve more complex situations than can be solved by the simple application of a single equation. Even applying single equation may be problematic in recognizing that the equation is appropriate or in choosing the correct values for the parameters from the known information or knowing how and when to augment the given information with necessary additional information.

3. anonymous

4. anonymous

Consider a tank of water of depth d which you wish to empty with a 5/8 inch diameter garden hose siphon after it passes over an obstruction of height z above the water surface and leads to a level h below the water surface. What is the maximum height of the sIphon loop above the water level before the water column in the sIphon breaks. Use Bernoulli's law because we have a fluid flowing through a tube at different heights. Bernoulli's Law states that at two points in the same streamline for a noncompressible nonviscous fluid in steady flow the sum of the pressure, the kinetic energy per unit volume and the potential energy per unit volume is the same. I'll call this sum the Bernoulli variable. At the top of the siphon the Bernoulli's variable for this point is $P+\frac{ 1 }{ 2 }\rho v ^{2}+\rho gh$. Now at the level of the water the Bernoulli variable has the value of the atmospheric pressure only so we can equate these. So the Bernoulli equation for a point in the loop is $P+\frac{ 1 }{ 2 } \rho v ^{2}+\rho gh=P _{ATM}$ We know that a fluid cannot support a tensile force ( i.e. water cannot pull anything) so when the absolute pressure in the loop drops to 0 the water column will break. So now we have for the maximum z $\rho gz _{\max} =P _{ATM}-\frac{ 1 }{ 2 }\rho v ^{2}$ But what is the velocity v. Since this is a noncompressible fluid the velocity is the same at all points in the siphon. In fact the velocity is determined only by the height of surface of the water above the outlet of the siphon. This is Torricelli's law (derived from Bernoulli's law) given by the equation $v _{out}= \sqrt{2gh}$ So the maximum z equation becomes $z _{\max} = \frac{ P _{MAX} }{ \rho g }- h$ The diameter of the siphon is irrelevant except that its relatively large diameter minimizes viscous forces and surface tension making Bernoulli's law a more valid application to this problem. Reviewing the approach you had to determine the law(s) that are appropriate and the conditions for the breaking of the water column. You had to deduce the Bernoulli variables in the siphon at the water level and top of the siphon. You had to deduce that the velocity at the siphon outlet. And lastly you had to remember the consistent set of units of density, gravitational acceleration, pressure, distance and area. In this case density is in slugs per cubic ft, acceleration of gravity in ft/sec^2, h in feet, and atmospheric pressure in lbs per sq ft. (remember the embarrassment of the company who built a Martian probe when inches and centimeters were interchanged) I think this problem is representative of physics problems in general. Some problems are easier less involved, many are more complex especially in application of physical intuition and/or the development of the mathematical relationships. Generally real problems are not plug and play.

5. anonymous

Well, obviously any numerical problem for which there does not yet exist a formula, but for which a formula can be derived from fundamental principles. Such problems can readily be devised at the graduate student level, but they take a lot of work, so I'm sure not going to do one for your pleasure. Send me \$500 and I might do it. Also, of course, any research problem falls into this category.

6. anonymous

7. anonymous