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Since radioactivity is a random process, you'd expect some fluctuation in the number of decays, i.e. if you wait for the half life, there's no guarantee that exactly half of the nuclei will have decayed! Based on your 3 estimates of the half life, you could just take the mean and go with that. And how to find the isotope? There's lists for that. I just ...

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Why don't you use energy conservation? Since this is a 1-dimensional task in potential field, it will be enough $$E/m = 0 - \frac{GM}{r(0)} = \frac{v(t)^2}{2} - \frac{GM}{r(t)}$$ For your assumption that the motion is strictly radial and downwards you have $v(t) = dr(t)/dt < 0$ so you can solve for $dr(t)/dt$ and get an ordinary first order ...

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This problem has a recursive flavor that we'll not try to avoid. Conservation of momentum tells us that $$m v_0 + (p+n-1)m v(n-1) = (p+n)m v(n).$$ Imposing the boundary condition $v(0)=0$ we find $$v(n) = \frac{n}{n+p}v_0$$ as claimed. Let $a_n$ be the time at which the $n$th bullet strike occurs. We have $a_1=x_0/v_0$ and $$v_0 (a_n - T) = v_0 ... 2 why does the voltmeter not form a closed loop with the circuit and hence cause energy to be dissipated by internal resistance It does, but the voltmeter probably has an impedance of 10,000,000 ohms - as most multimeters do - so the current flowing is negligibly small (0.6 microAmps) and therefore the voltage drop due to battery internal resistance is ... 2 The question you have formulated is not an easy one to answer (correctly). But the question you've formulated isn't quite the question that I see. The good news is that the text of the question you've posted implies a much simpler question; it's just asking for the energy change. You can probably assume that the acceleration is dominated by the circular ... 2 Your answer would have been correct if, for example, the spheres were non-conducting and if the charges were distributed uniformly over their surfaces. However, since the spheres are conducting, the surface charge distribution on each sphere will be altered because of the repulsion from the charges on the other sphere. In particular, the charges on each ... 2 The block is accelerating at 1\frac{m}{s^2} up the incline, since it is stationary with respect to the conveyor belt. What force is causing the block to accelerate? It can't be the normal force (which acts perpendicular to the motion). It must be the frictional force, which counteracts the component of gravity parallel to the incline. Using Newton's ... 2 Your angular velocity vector is$$ \vec{\omega} = \Omega \frac{ \vec{r}_D - \vec{r}_A }{|\vec{r}_D - \vec{r}_A|} $$where \vec{r}_A = (0,0.2,0.12), \vec{r}_D = (0.3,0,0), \vec{r}_B = (0.3,0.2,0.12)  in meters and \Omega = 90\;{\rm rad/s}. Your velocity kinematics is$$ \vec{v}_B = \vec{\omega} \times ( \vec{r}_B - \vec{r}_A ) $$And acceleration ... 2 Suppose we have a source of electrical energy, say a battery, that puts out 100 Volts. It is connected through wires with a total resistance of 1 ohm to a heater with a resistance of 99 ohms. The battery sees a total resistance of 100 ohms, and thus pushes 1 Ampere of current through the circuit. The battery is delivering energy at 100 Watts The Power ... 2 The "correct" half-life is the one that best fits all the experimental data available. Start with your original equation,$$N(t)=N_{0}e^{-\lambda t}$$Take natural logarithms of both sides and obtain:$$\ln (N(t))=\ln(N_0)-\lambda \ t$$Note that N(0) is not the same as N_0. The first is an experimental point, with the same sort of random and ... 2 I believe your mistake is with units, and it is the following:$$T=\dfrac{M[c_{rms}]^2}{3R} = \dfrac{\left( 1 \text{amu} \right) [11.2 \frac{km}{s}]^2}{3 \left( 8.3144621 \frac{\mathrm{J}}{\mathrm{\text{mol} K}} \right)} $$This doesn't even cancel out because you're left with a \text{mol} unit. Add avogadro's number.$$T = \dfrac{\left( 1 \text{amu} ...

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Converting units is simple if you simply work with the units as algebraic symbols. Thus start with the equalities $$1\textrm{ cal}=4.18\text{ J},$$ $$1\text{ day}=24\times60\times60\text{ s}=86,400\text{ s, and}$$ $$1\text{ cm}=0.01\text{ m}$$ and just plug the numbers in: $$1\text{ cal cm}^{-2}\text{ day}^{-1} =\frac{4.18\text{ J}}{(0.01\text{ m})^2 \times ... 1 You have to distinguish between the distance the man swims, relative to the water around him, and the total distance the man travels, relative to an observer on the river bank. The total distance relative to an observer on the river bank is the distance the man swims measured relative to the water around him combined with the distance the water moves ... 1 Your calculations are correct, provided the cylinder is indeed ohmic. The constant E you're getting is the difference in electric field between both terminals. As for the current flowing from inside to outside, as you said the cross sectional area will be different, and so will the length. The length L=r_b-r_a, but the cross sectional area is not ... 1 Has your instructor (or your book) mentioned how much bigger a atom is than a atomic nucleus? On order of 10000 times. Moreover, except for the s-shell electrons, most electron never come very close to the center (the p, d, etc shells all have nodes at r=0) so at the moment of fission the nuclei are sitting roughly at the middle of a roughly ... 1 Potential energy is a property of the system, not any one object. Thus there should only be one copy of the typical 1/r potential energy between two charges (plus an analogous gravitational term if that can't be neglected). The easiest way to see this is to start from "infinite" separation. Instead of pushing the two charges together, hold one fixed and ... 1 The Lagrangian density for a Dirac field is$$ \mathcal{L} = i\bar\psi\gamma^\mu\partial_\mu\psi -m \bar\psi\psi $$The Euler-Lagrange equation reads$$ \frac{\partial\mathcal{L}}{\partial\psi} - \frac{\partial}{\partial x^\mu}\left[\frac{\partial\mathcal{L}}{\partial(\partial_\mu\psi)}\right] = 0 $$We treat \psi and \bar\psi as independent dynamical ... 1 When imposing a periodic boundary condition, the amplitude of the wavefunction at coordinate x must match that at coordinate x+L, so we have:$$\Psi(x)=\Psi(x+L)$$In your previous 'particle in a box' scenario, you mention that the general form of the wavefunction is given by a linear combination of sine and cosine with complex coeficients. It might be ... 1 Your approach ignores the body of the slinky and essentially describes two massive particles coupled by a very light spring, which is not allowed to oscillate or show any of the interesting dynamics a real slinky will exhibit. Ideally, you should be using some sort of continuum-mechanics approach to this problem, e.g. treating the slinky as a very elastic ... 1 That line will get Coriolis acceleration$$\vec{a} = -2 \vec{\Omega} \times \vec{v}$$(\Omega is the angular speed of the earth's rotation, with a direction pointing into the ground from the view of the south pole). As it's going across the pole, there's a right angle between \Omega and v and the absolute value will be simply$$a = 2\Omega v$$and the ... 1 For a free particle, the energy/momentum eigenstates are of the form e^{i k x}. Going over to that basis is essentially doing a Fourier transform. Once you do that, you'll have the wavefunction in the momentum basis. After that, time-evolving that should be simple. Hint: The fourier transform of a Gaussian is another Gaussian, but the width inverts, in ... 1 For pushing it up, we have to overcome friction(act downwards) as well as the mg\sin\theta. So,$$3N=f+mg\sin\theta$$Now the block is just slipping , so friction is acting upwards, and so does the force applied externally.So,$$N+f=mg\sin\theta$$Eliminate N and use f=\mu mg\cos\theta. Solve for \mu you get your answer. 1 The ground will provide all of the static friction. Imagine what would happen if the upper block contributed even a tiny amount to the static friction: It would have to move forward due to the reaction force. Having M2 inch along you pull M1 (which stays stationary) would be very strange indeed. Static friction always acts to prevent relative motion. It ... 1 You are halfway there, because you already wrote down the Hadamard gates. The remaining part, I-2\left|0\right>\left<0\right| (I negated it from what you have, since this gives a simpler solution), is diagonal in the computational basis. Write down these diagonal entries and say out loud to yourself what this operator "does". 1 The voltmeter does indeed draw current, but very small. Its internal resistance should be high, unless it's a very cheap voltmeter. Putting the voltmeter's internal resistance, let us say 1MOhm just to make up a value, in series with the battery's internal resistance of a fraction of one Ohm, makes a voltage divider providing practically the entire ... 1 Focus on the integral$$ I_{ij}(k) = \int k_i k_j\ \mathrm{d}\Omega_k.$$This is a rank 2 symmetric tensor which can only depend on \vec{k} through its magnitude k^2, since the direction has been integrated over. So the only possibility is that I_{ij} is proportional to the unit tensor (Kronecker delta):$$ I_{ij}(k) = f(k^2) \delta_{ij}, where ...

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You can prove this by induction. I'll drop the operator hats as they're a pain to write. First step: Suppose $\hat H = \hat q_k$. Then $[ H, p_i] = [q_k, p_i] = i\hbar \delta_{ik} = -\frac{\hbar}{i} \frac{\partial H}{\partial q_i}$ So in the special case that the Hamiltonian has this form, the claim is true! Now suppose that the claim is true for $H = ... 1 Assuming you want to place the$5kg$mass at a point such that entire COM coincides with$(1,2,3)$. Then we can assume a$6kg$mass at$(1,2,3)$and a$5kg$point mass lying at$(-1,3,-2)$. And then the new$5kg$mass should be placed such that$COM$of these two masses is lying at$(1,2,3)$. So the new$5kg$mass should be placed at ... 1 NOTE: This comment was too long so I'll make it an answer. I would assume that the amount that the spring contracts is negligible compared to the distance that the mass has fallen. But anyway, regarding your amplitude problem: consider that when you have a mass on a level surface connected to the spring, there is an interplay between kinetic and spring ... 1 Irrespective of the dimensions, Poisson equation is always true. That is, if$\phi$is the electric potential and$\rho$is the charge density then,$\nabla^2\phi = \rho/\epsilon_0$. The Green's function of this equation satisfies$\nabla^2 G(\vec{x},\vec{x}^\prime) = \delta(\vec{x} - \vec{x}^\prime)$. A Fourier transform of this equation is$k^2G(\vec{k}) ...

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