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44

Is it even possible to hit 350Gs of force to a hard drive? Sure is. Drop it on the floor. You are thinking about sustained forces. 350g sustained won't happen even in rocket launches. But momentary forces can easily peak at this level. Note that the G limit on the drive is for when it's not running. No spinning drive will like 350g, except maybe in ...


16

Here's an application where an ability to withstand high shock is important. Explosions. In the mid 1980s I did work for a mining company's research laboratory (BHP Research, now defunct like all Australian corporate research). We would lower data-logging computers into boreholes to set up a grid of dataloggers, then detonate a charge of known energy at a ...


5

The equivalence principle tells us that gravity and acceleration are locally indistinguishable. So if you are accelerating your glass with some acceleration $\mathbf{a}$ to the right, then this is indistinguishable from a gravitational acceleration acting to the left (the red line): If we add the effective gravitational force to the gravitational ...


4

You are not the first person to ask this question. http://superuser.com/questions/925826/what-would-put-a-hdd-under-350gs-of-force claims that 350 g of force is slightly more than a soccer player kicking a football. What this means is that you basically can kick your case, and it shouldn't brick your hard drive. It might cause other issues though, so don't ...


4

Proper acceleration is acceleration away from following a geodesic. As such, it is $0$ if and only if the object in question is free falling. If there is any net non-gravitational force, then there is proper acceleration. Standing still on the Earth's surface is not free falling. The ground is preventing free fall, and proper acceleration is $g$. Note that ...


2

In general, assume you have some time-varying quantity $f(t)$, that varies from some time $t_{0}$ to some other time $t_{f}$. Now, graph $f(t)$. There will be some area under this curve. The average value of $f(t)$ over this time interval, $\bar f$ will be the height of the square that has the same base as the graph of $f(t)$ and the same area as $f(t)$ ...


2

As the equations of motion are of second order, the higher derivatives give no new information (but follow uniquely from the initial conditions of position and velocity), therefore they usually are not discussed. (Note: As Timaeus pointed out there are specific scenarios, e.g. Norton's dome where intial values for the higher order derivates will change the ...


2

If you really had a "constant power" engine, and all that power was transferred to your rocket which does not lose mass, it would result in a linear increase in the kinetic energy. And since the kinetic energy $E=\frac12 m v^2$, you can find the velocity at a given time from $$P\cdot t = \frac12 m v^2\\ v = \sqrt{\frac{2 \cdot P \cdot t}{m}}$$ If you ...


2

Notice Since I have written the majority of this answer the question changed a lot, and now it's been put on hold. However, I won't erase parts of my answer which are answers for the original (ill posed) question, because other people might find it useful. I have extended it with parts though, which seem (to me) to answer the question the original changed ...


1

In the second law of Newton appears the acceleration $a$. It refers to a generic acceleration due to any phenomenon. $g$ has the same role of $a$, but it refers specifically to the acceleration of gravity (free fall particular case) on the Earth. Usually we approximate $g$ to be constant $\left(9.81\, \mathrm{m}/\mathrm{s}^2\right)$, but in the real case the ...


1

Newton devised a very good law of gravity (until Einstein came along) where the force between the two bodies is scaled by a very small number usually written as a capital G. It's a general law that applies to any two bodies. But if you plug in the mass of the earth, the mass of a test ball, and the distance between the center of earth and the test ball, then ...


1

I am more familiar with graphing in Python than in the language you are using. The following snippet of code produces the graph you are asking for: import numpy as np import matplotlib.pyplot as plt # sci fi velocity graph D = 160934400000 # m v_i = 3070 # m/s total_time = 40*3600 # seconds: 40 hours # to cover half the distance in half the ...


1

The fictitious acceleration due to Earth's spin can be found using centrifugal acceleration, $$ a_{ca} = \omega^2 r, $$ where $\omega$ is Earth's angular velocity (roughly $7.292\cdot10^{-5}\ rad/s$) and $r$ the shortest distance between you and the axis of rotation of Earth. The direction of this acceleration will be aligned with the shortest distance ...


1

Do a free body diagram and you will find for a horizontal plane that $$ F - \mu m g = m a $$ $$ (100) - \mu (50) (9.80665) = (50) (0.1) $$ $$\boxed{ \mu = \frac{(100)-(0.1)(50)}{(50)(9.80665)} = 0.1937\ldots }$$


1

Your last equation is a quadratic in $t$. The $a$ is simply $\sin(\theta).g$. You can then solve it with the usual formula for a quadratic equation. There are two solutions to a quadratic, and that's because if you go into negative time you'd be pulled down by gravity any get to the new X position. This solution, of course, wouldn't apply to your ...


1

You conclude that constant acceleration is appropriate for one of a few reasons: You read in the question text that you should treat it that way. If (1) does not apply, you read in the question text that some physical situation obtains and you know or suspect that this situation is usually well represented by a constant acceleration. If neither (1) nor (2) ...


1

The acceleration is the time derivative of the velocity: $$ a = \frac{dv}{dt} $$ so if the velocity does not change with time the acceleration is necessarily zero. Since in your example the velocity is constant during the interval that means $dv/dt = 0$ and therefore that $a = 0$ during the interval. The velocity doesn't have to be zero. Any constant ...



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