# Tag Info

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This is not a simple problem. The interactions of the earth's atmospheric fluctuations and the rotational patterns are a matter of research: Atmospheric loads (= air pressure), e.g. during a high pressure weather system, can change the shape of the elastic Earth by up to two centimetres and can also alter the Earth's gravitational force. In this ...

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Let's assume that the initial temperature of the sphere is not 1000C, so that, at least, it is not above the critical temperature of water. Let the initial temperature be $T_0$ and let $T_{\infty}$. Let's also assume that the pressure is high enough so that the water close to the sphere does not boil. My goal will be to arrive at an upper bound to the ...

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If this article is to be believed, you would have no problem at all - in fact you could feel where the cities are, let alone the mountains.

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Newton's Law of Cooling actually states: $$\frac{dQ}{dt}=-kA[T(t)-T_a],$$ where $\frac{dQ}{dt}$ is the heat flux, $k$ the heat transfer coefficient, $A$ the sphere's surface area, $T(t)$ the sphere's temperature as a function of time and $T_a$ the ambient temperature (assumed constant). When the sphere drops in temperature by $dT(t)$ then it loses an ...

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The equation for your curve is given by: $$\frac{dv}{dt} = \frac{F(v)}{m}$$ where $F(v)$ is the net force on the car, which is a function of the velocity. we solve the equation by integrating to get: $$\int \frac{dv}{F(v)} = \frac{t}{m}$$ The trouble is that the net force $F(v)$ is a complicated function that doesn't generally have a simple analytic ...

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You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the ...

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I worked this out a little while back in order to check something said on one of these Nova or other science show specials. I wanted to know how much energy would be required to remove the entire atmosphere of the Earth and whether a supernova (or other astronomical event) could possibly do this. Earth's Atmosphere Let's assume the following quantities: ...

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The first is, is my understanding described above correct? Basically, yes. The second is, how does this affect us healthwise? We are stuck with these debris of cesium-137 and strontium-90 for the next 30 or more years, but what effect do they have on us? Stochastic health effects (induced tumors, cancers, leukemia,…) of ionizing radiation are ...

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The acceleration of the points $B$ and $C$ will be same assuming that the Earth is not breaking apart, i.e., the acceleration of all points on the Earth will be same. If you think of the Earth as a very strong rod, only the Tension in the rod varies so that the entire rod accelerates with the same acceleration as that of the center of mass of the rod. So, ...

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The other answers are adequate, answering not a thing to the first part of the question in your title: Would a neutrino bomb do anything? But questions in titles are important, so I will reply to the second part, Or can weak force kill you? : Of course the weak force can be lethal. The simplest example is the decay of neutrons , it is a weak decay , but ...

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The Earth is much more massive than its human population. If the Earth is transparent to neutrinos from this device, so are the people on it. In supernova explosions the neutrino flux is large enough to have an important effect on fluid transport. (Kip Thorne discusses this in "Black Holes and Time Warps.") Here is an estimate that to receive a lethal dose ...

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There was also a science fiction story where somebody invented a neutrino bomb. It was claimed that such a bomb would turn all the matter in it to neutrinos, which would escape without damaging anything. The first part doesn't work (think baryon conservation) but the second does. It pointed out that a vacuum would be left, so air would rush in with a ...

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let's talk about efficiency bounds The flux of solar neutrinos at the earth's surface is on the order of $10^{11}$ per cm²/s. Even the largest detectors detect less than a few hundreds neutrinos by day. 7Be Solar Neutrino Measurement with KamLAND Let's assume the incredible facts that such detector has only 10 cm² area and that all the ...

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The Reynolds number gives a ratio between forces of inertial origin and those of viscous origin. For a given geometry of the problem, increasing the Reynolds number will lead to turbulent flow from a certain threshold. However, this threshold is strongly dependent on the geometry: this is actually common knowledge, a better design (more "aerodynamic" we say ...

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If the super nova happens in any system it is not safe if you in it. As far a a blast radius You take the class of star to a habitable planet can differ. You would die of radiation poisoning if the Earth survived. The sun is around 150,000,000,000 meters, 1,368 W/m2, 1.7×1017 J from Earth. A super giant star is around 2,100 times bigger than the sun. So ...

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According to Phil Plait and others, anything over 100 light years (and probably a fair bit closer) should be safe. There aren't any known supernova candidates that close. http://earthsky.org/space/supernove-distance https://twitter.com/BadAstronomer/status/201708339904778240 SN 1987A isn't even in our galaxy. It's over 150,000 light years distant.

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Between points A and C The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or Acceleration(a) - Acceleration(c) = 1

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Use the Reynolds number equation: $Re={vL \over \nu}$ where $\nu\approx 1.5\times10^{-5}m^2/s$ is the kinematic viscosity for air. If you enter this into the equation, you end up with $Re\approx 67000{v\over{m/s}}{L\over m}$ i.e. for a race car traveling at 40m/s and with a length of 4m it comes out to be around 10 million, which is certainly ...

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The number of tides stays just about the same, as the earth turns under the tidal bulges. On a water earth we would have two tides per day. Local landforms impact that greatly, varying from place to place. The tidal force is the difference of the moon's gravity on the near and far sides of the earth. It falls off as $\frac 1{r^3}$, so the if the moon ...

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The tides will get smaller as the moon moves away having less gravitational effect. As long as there are tides there will be two per day because this is based on the Earths rotation not the moons distance. As the Earth spins the tides rise on the side facing the moon and the opposite side because of gravity. Its these tidal forces along with the Earths spin ...

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Just imagine there would be no moon at all. Obviously there would be no tides due to the moon. As Dr. Chuck pointed out in the comments, there would still be tides due to the sun, but the tidal forces of the Sun are only about 46% of the moons tidal forces (taken from german Wikipedia about Tides: https://de.wikipedia.org/wiki/Gezeiten). So the farer the ...

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All materials have a resonant frequency Well, sort of. In general, complicated structures will have many resonant frequencies where the amplitudes of any oscillations will have local maxima. However, one of the jobs of structural engineers, and I would assume this would apply to aeroplanes too, is to find these frequencies and make sure that either (a) ...

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You say: All materials have a resonant frequency but this is at best an oversimplification. Any system has a set of normal modes and if you apply driving force at a frequency that matches a normal mode then you will get a resonance. However for any system significantly more complicated than a tuning fork there are many normal modes and non-linearities ...

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69.8 kg That's assuming that your original human weighed 70 kg. If that doesn't sound like much got burnt up, consider that the energy released in this reaction is equivalent to that in 3 millions tonnes of TNT exploding: $14.4 \times 10^{15}J$. The maths In order to work this out, let's see what humans are made of. According to Wikipedia, 61% of a human ...

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Short answer - almost all of it. Your body is a mixture of chemical elements. By number, most of the nuclei are hydrogen, by mass its mostly oxygen. In cremation, most of this oxygen, along with most of the other combustible things like carbon, hydrogen ends up going up the chimney. Hence the low residual mass. Energy (including rest mass energy) is ...

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If we idealize the scenario enough, this is a simple exercise in differential equations, so let's get to work. First, we know that it's initial speed is $150 \text{ m/s}$, but that is by no means its final speed - obviously, the bb slows down as it travels through air! Let's suppose that the moment the bb exits the barrel, it is no longer being pushed (as ...

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Your car consumes gas because of rolling resistance and air resistance. The first one scales with the total weight (at least approximately), the second one mainly with velocity squared. Furthermore, you accelerate with the car, and this energy consumption also scales with the weight. The amount of weight you save is $20/1285=1.6\%$. So the amount of money ...

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