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70

The energy of a bullet is around 735 joules (see bullet details here). This is about the same energy that I have when I'm running at about 4.6 m/s. Would you rather be hit by me or the bullet? The bullet kills you because it concentrates all the energy onto a small impact area while my impact area is rather larger (and sadly getting even larger as ...


67

7 TeV is not that much kinetic energy, that has been covered by your question and previous answers. However, in the context of a proton, with a rest mass of $1.672×10^{−27}~\mathrm {kg}$ (very, very little mass), when a single proton has 7 TeV then it is travelling at a specific speed: $$E= mc^2$$ \begin{align}E& = E_0 + \text{KE}\\ \text{KE}&=E- ...


26

Consolidating some of the points made in the answers to the question you linked, and comments: When constructing a chair, 4 legs is easy when you use traditional (wooden) construction - 90 degree angles, and easy to make stackable. A little bit harder than three legs because you have to make sure they are all the same length (or the chair will wobble). ...


4

You do not need the kinetic energy. Working with the total energy $\gamma m c^2$ produces the same result. Assuming both the total initial energy $\bar E_0 = \gamma_0 m c^2$ and the additional energy $E_i$ are known, write $\gamma_1 mc^2 = \frac{mc^2}{\sqrt{1-\beta_1^2}} = \bar E_0 +E_i$ for $\beta_1 = \frac{v_1}{c}$, then $$ \sqrt{1-\beta_1^2} = ...


3

So, considering that 7 TeV is more or less the same kinetic energy of a mosquito, why is considered to be a great amount of energy in LHC? Like other people said, it's not a great amount of energy. However, it's concentrated in a very tiny space. Just think of how much a mosquito is bigger than a subatomic particle.


3

The difference between an office chair's 5 wheels/supports and a regular chair's 4 legs is that the latter has all of its load going straight down. The legs only need to be strong enough not to shatter. In fact, a chair could easily get away with 3 legs but for the stability. In contrast the office chairs legs support load perpendicular to their ...


3

If the velocity of a mass $m$ at some moment of time is $v$, then the kinetic energy and momentum are: $$\begin{align} E &= \tfrac{1}{2}mv^2 \\ p &= mv \end{align}$$ If the velocity is changing with time, i.e. it is a function of time $v(t)$, then the kinetic energy and momentum will also be functions of time: $$\begin{align} E(t) &= ...


3

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 ...


3

I think that this is a very interesting problem which is conceptually difficult. You do not need to worry about the FBD for the truck. The box should be your main focus. Diagram 1 is the FBD as long as the box does not slide relative to the truck. With the aid of diagram 1 work out the maximum acceleration $a$ the box can have as a result of the static ...


2

As you figured out, the horizontal and the vertical part of the motion are independent. If you throw the bottle upwards, it will go upwards for some time, then turn and fall back. When it reaches the height of your hand, it will have the same velocity as when you threw it, just the opposite direction (downwards) - so this doesnt help as well. Clearly, ...


2

From when I worked among missile engineers, accelerometers were used, along with gyroscopes (mechanical or laser). I don't know of 6th order differential equations. I do know of 3rd order, namely in the steering by swiveling the engine nozzle. Specifically, the engine nozzle angle is off-center by a certain amount, causing an angular acceleration (2nd ...


2

The first postulate of Special Relativity - which is also the Principle of Relativity is "Laws of Physics are invariant in all the inertial frames." Now consider the following scenario: Suppose there are two events which are happening simultaneously in frame $O$ and are spatially separated by distance $l$. From your Equation $1$, the spatial interval ...


2

Boosting means you are changing a frame of reference; boosting frames doesn't imply any actual motion. When you talk about boosting, you are talking about changing the way you are observing something instantaneously. Acceleration on the other hand, is a type of motion inside a frame of reference. When you talk about acceleration, you are talking about a ...


1

Let's say you are doing a typical SR experiment and you have Alice and Bob flying around in spaceships and such and you have each ship feeding you data. Things outside look one way to Alice and a different way to Bob because of their inertial frames. Let's say they're observing two supernovas and trying to determine their timing relative to each other, and ...


1

No need to force $\tilde\alpha_1 = \alpha_1$. After taking advantage of the fact that (1), (2) and (5), (6) must be inverses of each other, your transformation reads, in matrix form, $$ \left(\begin{array}{c} x'\\ t' \end{array}\right) = \alpha_1(v)\left(\begin{array}{cc} 1 & -v\\-\frac{v}{c^2} & 1 \end{array}\right)\left(\begin{array}{c} x\\ t ...


1

Your approach is correct; your ability to read data from a graph is suspect (the divisions are 2 m/s each). The initial velocity is -12 m/s, and at time t=9 s it is up to 18 m/s That should change your answer...


1

There are often lots of ways to solve problems and finding the simplest way is something that comes with experience. The reason your teacher keeps setting you so many problems is to give you that experience. One of the common tricks is choosing a suitable reference frame to do the calculation. As watched by a stationary observer it's a somewhat complicated ...


1

The mistake in your reasoning is to assume that the same force does the same work. That is simply not true - as the force operates over a different distance. To analyze the problem, you have to think in terms of impulse ($F\cdot \Delta t$) or work done ($F\cdot \Delta x$). Let's assume that the same impulse is applied. Then indeed the linear momentum of ...


1

It is because the ball was traveling with you when you threw it. Imagine the following question: I am in a car traveling at 5m/s holding a ball. Where will the ball be relative to me in 10 seconds? Answer in my hand. To be obtuse: ball has velocity 5m/s. After 10s it will have moved forward 10 x 5 meters = 50. I have velocity 5m/s. After 10s it will have ...


1

Newton's third law says the the force on A due to B is equal and opposite to the force on B due to A. This in turn means that the changes of momentum of A and B are the same in magnitude but opposite in direction. This is how the momentum becomes rearranged. B loses some momentum and A gained an equal amount. So when two atoms collide you can think of ...


1

Given the acceleration $a = \sin \left( \frac{\pi t}{T}\right) $, by integration you get $$ v(t) = \int_0^t a\,{\rm d}t = \frac{T}{2} \left(1-\cos\left(\frac{\pi t}{T}\right) \right) $$ $$ s(t) = \int_0^t v\,{\rm d}t = \frac{T}{\pi^2} \left(\pi t - T \sin\left(\frac{\pi t}{T}\right) \right) $$ Since the last one cannot be inverted for $t(s)$, we can ...


1

Your train is travelling east at a speed of 20 m/s, since you walk to the back of your train you are only travellint east at a speed of 18.6 m/s with respect to the ground. Your friend is travelling west at a speed of 28 m/s, or a speed of -28 m/s to the east. From your friends perspective ground is travelling east at a speed of 28 m/s and you have an ...


1

Using the first $2$ statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target.


1

I am giving the solutions of original task (to get the speed at point B). I am not sure if the questions are necessary to perform the task. If you are sure the path taken does not matter (and I will assume that per your statement). So, let us consider a straight line path. Vertical component of F overcomes gravity and causes vertical move. Only horizontal ...


1

To compliment John's answer I'll give you an example: the kinetic energy of a harmonic oscillator. First we need to determine the velocity $ v=\frac{dx}{dt}=\frac{d(Asen(\omega t +\phi_0))}{dt}=A\omega cos(\omega t+\phi_0)$. Because kinetic energy is $\frac{1mv^2}{2}$ we substitute $v$ $KE=\frac{1}{2}mA^2\omega^2cos^2(\omega t+\phi_0)$ Because ...


1

The concept you're after is the dot product between 2 vectors (your displacements). More specifically you want to use $$ \vec{a} \cdot \vec{b} = |a| |b| \cos(\theta) $$ to find $\theta$.


1

You don't need calculus to understand this and I think you are right to be trying to gain a deeper understanding than just memorizing some formulas. During that first second the body accelerates - it starts with 0 velocity and gains linearly giving 9.8 at the end of the first second, so at that point, it hasn't been moving at 9.8m/s for a second, it has ...


1

This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle. $$ \frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx $$ For a constant force and mass $\int F\cdot dx = F (x_A - ...


1

Firstly it's worth noting that such a discontinuity can never be 100 % real. To go from acceleration $a$ to $b$ instantaneously ($\Delta t = 0$) would require an instantaneous change in the net force responsible for the accelerations and that isn't possible in the material world. Secondly, I think you are over-thinking your problem. Just write the ...


1

If you take your final expression $$ x(t) = \underbrace{\left(x_0 + \frac{b - a}{2}\tau^2\right)}_{x_0^*} + \underbrace{\left(v_0 + \tau(a - b)\right)}_{v_0^*} t + \frac{b}{2} t^2, \quad \text{with}\ t>\tau, $$ then $x_0^*$ and $v_0^*$ would be the position and velocity at $t=0$, however this is only meaningful if $\tau<0$ (however in that case $x_0$ ...



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