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14

To work out the force needed to move a ship there are two considerations: the mass of the ship the hydrodynamic drag due to the ship's motion through the water At low velocities the force is likely to be dominated by the mass of the ship because the drag is roughly proportional to velocity. Newton's second law tells us that the acceleration of the ship ...

2

I will give a closed form for the integral in Chris Cundy's answer. Doing the substitution $u=nx$, we get $$\sum_{n=1}^{\infty} \int_{x_{min} \cdot n}^{\infty} \frac{1}{n}\left(\frac{u}{n}\right)^3e^{-u}\mathrm{d}u$$ $$\sum_{n=1}^{\infty} \frac{1}{n^4} \Gamma(4,x_{min}\cdot n)$$ where $\Gamma$ is the upper complete gamma function. We write $a=x_{min}$ ...

40

The formula you want is called Planck's Law. Copying Wikipedia: The spectral radiance of a body, $B_{\nu}$, describes the amount of energy it gives off as radiation of different frequencies. It is measured in terms of the power emitted per unit area of the body, per unit solid angle that the radiation is measured over, per unit frequency. $$... 11 The wavelengths of light emitted can be calculated using planks law and the temperature of the object. For your average 100W incandescent light bulb, the filament is 2823 kelvin according to google. The spectral radiance, B, is equal to$$\frac{1.2\cdot10^{52}}{\mathrm{wavelength}^{5}\cdot ...

2

According to Google the mass of the earth is around $5.972 \times 10^{24}$ kg (presumably that figure includes its 7 billion human inhabitants plus everything else were meant to be sharing with). If we take the average weight for a human as given by WolframAlpha as 62 kg this gives humanity a combined weight of $4.34 \times 10^{11}$ kg - so our share in the ...

0

Is it not enough to move the earth from its position (even a bit)? For all practical purposes, nope. Think of it in the following manner: The (Earth + All the Humans) can be viewed as one system, and in this case, the gravitational force exchanged between any one person (e.g. you), and the Earth, is an internal force of this system. And internal forces ...

1

good strategy! Hubble Constant = Ho = [d/s]/[d/1] in basic variables, thus it reduces to 1/s which is time inversed, that is, the inverse of the age of the universe at whatever Ho constant value used and further adjustments from the Standard Model equations. The big adjustment is U. acceleration [ and significant deceleration during first 2 billion years]. ...

4

If an object accelerates over a distance $d_1$ with a constant force $F_1$, then decelerates over a distance $d_2$ with a constant force $F_2$, then conservation of energy would give us that $$F_1 d_1 = F_2 d_2$$ from which it would follow that $$F_2 = F_1 \frac{d_1}{d_2}$$ In other words - if the cat is able to absorb the shock of his fall over a ...

2

When the cat leaves the upper shelf, it will be accelerated by the Earth's gravitational field toward the lower shelf at a rate of 9.8 meters per second, per second (9.8 m/sec^2). Assuming its initial velocity was zero, at the time the cat lands on the lower shelf its velocity will be 19.6 m/sec. The cat's momentum at the lower shelf will be 6.5 Kg * ...

0

You will have to estimate how quickly the cat goes from full speed to completely stopped. Then, given the cat's initial velocity as it gets to the shelf, use the impulse-momentum theorem. The equation is: $$F = \frac {\Delta p}{\Delta t}$$ Is that enough info, or do you need an additional equation to calculate the initial velocity of the cat?

2

If you have $L$ and you have $T$, then nothing more complicated than Stefan's law is required. If $T$ is the effective temperature of the star then this gives an exact answer. $$R = \left(\frac{L}{4\pi \sigma_B T^4}\right)^{1/2}$$, where $\sigma_B = 5.67\times 10^{-8}$ in SI units. If on the other hand you are trying to solve the structure from first ...

1

I can see that this question has been downvoted but I think it still deserves a proper answer. First, it is the Chandrasekhar (one word) limit, named after the Indian-American astrophysicist Subrahmanyan Chandrasekhar. Second, the Chandrasekhar limit does not mean that an object cannot be more massive than 1.4 times the mass of the Sun. There are plenty of ...

4

For a 6 year old, you might want to focus on thickness instead of length, as the numbers get too big with length. A ream of paper (500 sheets) is a bit over an inch thick, say $3.5 \, \text{cm}$, so one sheet is $3.5/50 \, \text{mm}$, or $.07 \, \text{mm}$, which is $7 \times 10^{-5} \text{m}$. An atom has diameter $0.1 \, \text{nm}$ to $0.5 \, \text{nm}$ ...

5

A typical atom is roughly a few times $10^{-10} \text{m}$ wide. A piece of paper is say $(1/4) \text{m}$ wide. Therefore the ratio of the width of an atom to the width of a piece of paper is around $10^9$. A piece of paper is roughly the same width as a human, so $10^9$ is also a rough guess for the ratio of the width of a human to the width of an atom. The ...

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