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5

You need to take density of the air into the question. And weight of the balloon itself. The Archimedes' law says: Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object. Therefore a balloon can support the same weight as equal volume of air has (reducing gravitational ...

4

There's a sign error in your equations of motion. The Lagrangian of the system will be $$L=T-U= \frac{m}{2} \left( \dot{x_1}^2 + \dot{x_2}^2 \right)-\frac{k}{2} \left( L + x_1 - x_2 \right)^2$$ So the equation of motion for $x_1$ is: $$\frac{d}{dt}\frac{\partial L}{\partial \dot{x}_1}-\frac{\partial L}{\partial x_1}=0 \\ m\ddot{x}_1+k( L + x_1 - x_2 )=0 \\ ... 3 The net upward force is, according to Wiki buoancy:$$F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-m_\mathrm{balloon} \cdot g$$For helium, the m_\mathrm{balloon}=\rho_\mathrm{helium}V_\mathrm{disp}, thus$$F_\mathrm{net}=\rho_\mathrm{air}V_\mathrm{disp}g-\rho_\mathrm{helium}V_\mathrm{disp}\cdot ...

2

If you want the vector velocity, it's straightforward to use kinematics - if you know the velocity at the beginning, the horizontal component is constant throughout the parabola, $$v_x(t) = v_x^0$$ and the vertical component is given by $$v_y(t) = v_y^0 - g\ t$$ where $v_x^0$ and $v_y^0$ are the initial values of the horizontal and vertical components of ...

2

The integral without delta square would converge to 1, but squaring the function somehow breaks it? Yes. Recall the sifting property: $$\int_{-\infty}^{\infty}f(x)\delta(x - a)dx = \int_{-\infty}^{\infty}f(a)\delta(x - a)dx = f(a)$$ Then, it follows formally that $$\int_{-\infty}^{\infty}\delta^2(x - a)dx = \int_{-\infty}^{\infty}\delta(x - ... 2 What happens when you integrate a function multiplied by a delta function? You get:$$\int f(x) \delta (x-y) dx=f(y)$$(Because the delta function is zero everywhere, except at zero where its integral gives 1.) So, when integrating over \delta^2 (x-y) we get:$$\int \delta^2 (x-y) dx=\int \delta (x-y) \delta (x-y)dx = \delta (y-y) = \infty$$2 The Dirac delta is defined by the equation $$\int_a^b\mathrm{d}x \,\delta(x)f(x) = f(0)$$ for a < 0 < b. By direct application of this definition we get $$\int \mathrm{d}x\, \delta(x - y) \delta(x - y^\prime) = \delta(y - y^\prime)$$ If we let a < c < 0 we can write ... 2 The \delta function has the following property:$$ \int \text{d}x\; f(x)\delta(a-x)=f(a) $$This actually answers both of your questions. First, the non-square-integrability:$$ \int \text{d}x\;\delta(x-y)\delta(x-y)=\delta(y-y)=\delta(0)=\infty $$according to the rule above if you choose one of the \delta's to be the f. Your second question is the ... 2 Your approach is pretty much correct. First you calculate the buoyancy generated by 1L of Helium. This can be calculated by the difference in the mass of 1L of Helium and 1L of air, which you can look up on the internet. The answer is ~0.001kg, as you got. Next you estimate the volume of a balloon. A sensible assumption, followed by some questionable ... 2 How do I identify which ones are parallel or series? If all of the current leaving one resistor enters another resistor, the two resistors are in series. The resistances of series connected resistors can be added together to find the equivalent resistance of a single resistor, e.g.,$$R_{eq} = R_1 + R_2  If all of the voltage across one resistor is ...

1

As you can see, $\rho$ only depends on a single variable: $r$. Thus, it should be intuitive that one can do this problem by integrating only over the variable $r$. To see what you are supposed to do, consider what happens if you fix $r$: You obtain a spherical shell (as was pointed out in the comments). The moment of inertia of a spherical shell is quite ...

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