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9

TL; DR The material with the greater effusivity will be more likely to burn you upon contact. Analysis of a simplified case First consider the case where your palm comes into contact constant temperature wall. Often, we can consider your palm as a semi-infinite solid. The requirement for the semi-infinite approximation is that $T(x \rightarrow \infty, t) = ...


4

The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason for this is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in ...


4

Because what you feel as weight is actually the force you must exert to your surroundings not to collapse, or start falling anywhere. Weightless is just absence of this force. In fact, I can not imagine what should one feel in a free fall but weightless.


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I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static ...


2

You can't get an exact number for this without some assumptions. However, you can develop a relationship. What we do know is that one full period corresponds to some amount of time (my niave guess would be 1 second but I believe I have heard faster clocks which might be half seconds). So we introduce some constant C which is the amount of time reported on ...


2

You have two basic options: Realize that this is actually just two resistors in series with three parallel resistors, and analyze it using the equivalent resistances of resistors in parallel and series, or Use Kirchoff's laws to derive the equivalent resistance. If you choose option 1, I'll help you out by revealing the parallel resistors in this ...


2

Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over $m$ is really the integral over the volume: $$\int_m r \cdot dm = \int_V r \cdot\rho(r')~ dV$$ Here I deliberately distinguish between $r$, the vector from the origin of the coordinate ...


2

I fear you are overthinking it. I am not sure why the original references, Kubo (1964) and Royer (1976) and their proofs are not adequate for you. Working in Fock space with creation and annihilation operators is a bit self-defeating, unless you were suitably adroit. Sticking to standard phase space operators $\hat{x}, \hat{p}$ and using the standard CBH ...


2

In your last question it is important as to what you mean by WRT the question. If you are trying to find the E-field due to a point charge using Gauss then to make the surface integration easier you choose a surface which has the following properties: the E-field direction is everywhere perpendicular to the surface the E-field has a constant ...


2

You feel weightless while free falling because no normal reaction force is present. While sitting on a chair you feel the normal reaction force that chair exerts on you which is equal to your weight and the force you exert on the chair.


2

Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are.


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Reusing your car example: use the fact that acceleration is "change in velocity". This can be positive (acceleration in the usual/common sense), but everyone knows that velocity can also be decreasing. This is what phsicists call negative acceleration.


2

The inertia matrix for a thin rectangular foil (laying along the xy plane) in body coordinates is $$ I_{body} = \begin{vmatrix} \frac{m}{12} b^2 & 0 & 0 \\ 0 & \frac{m}{12} a^2 & 0 \\ 0& 0 & \frac{m}{12}(a^2+b^2) \end{vmatrix} $$ where $a$ and $b$ are the side dimensions. The inertia matrix in world coordinates, while rotated by ...


1

Positive means speeding up, negative means slowing down. Now this is assuming you are traveling in the positive direction but through an axial change you could always guarantee this. I think this would be a good starting point for a child.


1

I would explain that we can feel acceleration in a car. As our speed increases we are pushed back into the car seats. This is positive acceleration. As we slow down we are pulled back by the edge of the seat/seat bell. If the child is happy that acceleration is a change of speed then they should be able to tell with their eyes closed if the ...


1

It is a method that is generally used for conservative unidimensional problems (problems with only one degree of freedom, here your angle $\theta$ or cartesian coordinate $x$). You'll notice that it is equivalent to using Newton's second law in this case : let us write the total energy $E = \frac{1}{2} m v^2 + V(x)$, $V$ being potential energy. The problem ...


1

No, placing the spheres at he vertices of a cube is not the most efficient packing. The most efficient packing of spheres is in a face-centred-cubic arrangement. The volume taken up is 74%. Computing this value follows a similar procedure to what you used. https://en.wikipedia.org/wiki/Cubic_crystal_system


1

Your solutions are wrong. As $$ \frac{dq}{dt}=\frac{\partial H}{\partial p}\qquad \frac{dp}{dt}=-\frac{\partial H}{\partial q} $$ you get $$ \frac{dq}{dt}=10\,p\qquad \frac{dp}{dt}=0 $$ i.e. $$ q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0} $$ The $q$ coordinate flows in time in straight lines, while the $p$ coordinate doesn't change in time. So each phase ...


1

Potential at center due to +ve sphere is not correct. What you had found is when cavity is at center. However potential due to $-\rho$ is correct. First consider no cavity Potential at center of sphere due to uniformly charged complete sphere $ V = 3kq/2a$ Now, potential due to positive charged sphere $cavity$ at center. $$ V_{1}= \frac{4\rho π ...


1

Using the first $2$ statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target.


1

If you did not have resistor R in the circuit the voltmeter would always give the same reading - the voltage of the battery. A thermistor changes its resistance when a temperature changes. You are not given a resistance meter. All you have is a battery, a resistor R and a voltmeter. The circuit as set up is called a potential divider which means that the ...


1

You have identified the problem as the factor $2$ between your calculation and the book. You have read the problem correctly, so it seems the book is wrong.


1

They aren't. Frictional force down an incline is the coefficient of kinetic or static friction multiplied by mass and the acceleration of gravity and the sine of the angle of elevation. Meanwhile, the component of weight down the incline is all of the above without the coefficient of kinetic or static friction.


1

In such circuit see that all resistance at rim have same potential, since no resistance reduced it. At rim all have same potential as that at $a$. At center all have same potential as that at $b$ So, all have same potential difference $V_{a} - V_{b}$ Same potential difference means that they are connected in parallel.


1

I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task. If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal ...


1

It is true that two electrons can't have identical quantum numbers, but spin itself is a quantum number. That is, the state with quantum numbers 111 can hold two electrons: one of spin up and one of spin down. So when you are finding the ground state, for example, find the four lowest energy eigenstates (ignoring spin), and the ground state of eight ...


1

You don't feel gravity pulling on you, because gravity doesn't actually pull on you or anything else. Gravity bends spacetime. What this means is that your own perceived frame of reference (a system of coordinates to measure space and time), which is all nice and straight, does not match up with the actual shape of spacetime. So even though you think you are ...


1

First of all, as this is a homework question, I can't tell you the complete solution. Choose P as the origin of coordinate system and resolve the forces into x and y component. And as the body is in equilibrium, the net force is zero. So you get these two relations (when the net force on x and y component equated to zero.) $$G\cos\theta=H\cos\phi\\ ...


1

The result you described says that the projection of antisymmetric rank 2 $SO(4)$ tensors onto self-dual and anti-self-dual subspaces commutes with the action of $SO(4)$. This just implies that the space of antisymmetric rank 2 tensors of $SO(4)$ is reducible. To show that the self-dual and anti-self-dual subspaces are themselves irreducible, think of ...


1

The reason you feel weightless is because there is no force pushing against you, since you are not in contact with anything. Gravity is pulling equally on all the particles in your body. This creates a sensation where no forces are acting on you and you feel weightless. It would be the same sensation as if you were floating in space.



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