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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 $T(x \rightarrow \infty, t) = ... 5 The simple answer to this question is that the specific heat capacity of a hot solid body you touch really doesn't matter too much for whether you burn yourself; what is much more important is its heat conductivity. The reason for this is that the surface of a hot body with low heat conductivity will rapidly cool down when you touch it (the blood in ... 4 Because what you feel as weight is actually the force you must exert to your surroundings not to collapse, or start falling anywhere. Weightless is just absence of this force. In fact, I can not imagine what should one feel in a free fall but weightless. 3 I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration$a$the box can have as a result of the static ... 2 Mass is not a function of time, so the only thing inside the integral that needs differentiating is the position. Let me try to make this clearer. The integral over$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 Always, always, always start problems like this by drawing a diagram: This make it obvious why cos and sin are used as they are. 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 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 ... 2 Solve for$F_b$from the horizontal braking distance. Assume$F_b$is constant, then during braking kinetic energy has been converted to friction work: $$F_b \Delta x = \frac12 mv^2$$ where$\Delta x=123\:\mathrm{ft}$is the braking distance and$v=60.0\:\mathrm{miles/hour}$. I've not checked the rest of your work. You don't need to invoke friction ... 2 If it's falling only then you have$F_d=+Cv^2$, where up is the positive direction. You said there is a gravitational force$F_g=-mg$. Write a Newton's 2nd Law equation, set$a=\frac{dv}{dt}$, rearrange, to get dv/g(v) = dt, (I'll let you find$g(v)$) and integrate away. The$v$integral is not trivial. Look it up in an integral table, if your teacher will ... 1 $$PV=1\times RT \space For \space O_2$$ $$P_2V=1\times R\times 2\times T \space for\space He2$$ Divide Both$\frac{PV}{P_2V}=\frac{T}{2T}$$$\frac{P}{P_2}=1/2$$ 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 Your system has 2 degrees of freedom, but using$x_1$and$x_2$will not be helpful in determining the effective spring rate. To get the spring rate you need the extension$x$of the connection point with mass$M$and the tilt angle$\theta. Do the substitution: \begin{align} x_1 & = x - a \theta \\ x_2 & = x + b \theta \end{align} The ... 1 Here, in the case of your question, the brass rod and the steel rod are connected in parallel combination in between the hot reservoir and the cold reservoir. So, the thermal resistance will be given by, 1/Rp = (k1A1+k2A2)/l..... And, we know that, Q/t = dT/Rp = (k1A1+k2A2)*dT/l. You have been given the values of Q, t, k1, k2, A1, and dT. So, you can easily ... 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. $$q(t)=10\,p_{0}\,t+q_{0}\qquad p=p_{0}$$ Theq$coordinate flows in time in straight lines, while the$p$coordinate doesn't change in time. So each phase ... 1 You can calculate an average power per square metre, but it's going to be a somewhat meaningless figure because the Sun only shines for part of the day and the strength of the sunlight varies continuously throughout the day. But if you want to go ahead this is how to do it. Suppose the total power (not just the PAR) per square metre is$P$(in watts per ... 1 Your mistake is: since$k$is independent from the total length of the spring, the half long spring still has$k$as force constant. We can define a constant$K$to be the force per relative extension of the spring: $$F = K \frac{x}{L}$$ where$x$is the extension and$L$is the original length. This constant$K$is an intrinsic property of the ... 1 Hint : the block starts from rest 11 cm from equilibrium. You don't give it a push. That means that after one oscillation, it'll come back at the same place and never go further away than 11 cm. That makes the rest of the work become trivial. 1 You almost have it. The last step is $$\Omega = \frac{\rho}{\rho_\text{c}} = \frac{\rho_0\,a^{-3}}{\rho_\text{c,0}}\frac{\rho_\text{c,0}}{\rho_\text{c}} = \Omega_m\,(1+z)^3\frac{H_0^2}{H^2},$$ where we used the critical density $$\rho_\text{c} = \frac{3H^2}{8\pi G},\qquad \rho_\text{c,0} = \frac{3H_0^2}{8\pi G}.$$ The result follows immediately from what ... 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 ...

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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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