# Tag Info

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In a radial field, the center of gravity is closer to the central body than the center of mass. This effect is enhanced in an object of lower density. When measuring the acceleration of the center of mass, the lower density object (of the same total mass) should feel a greater force (assuming no friction).

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In a vacuum the density doesn't affect the acceleration of a falling object. If there were air resistance the density does matter. In air a dense ball bearing would fall with a similar acceleration as in a vacuum, but a less dense object, for example a polystyrene ball, would be slowed down significantly by the air resistance.

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you have to solve those two differential equation numerically for example with MATLAB program. $${\frac {d}{dt}}v \left( t \right) ={\frac {p}{v \left( t \right) }}-b \left( v \left( t \right) \right) ^{2}+c \\{\frac {d}{dt}}x \left( t \right) =v \left( t \right)$$ simulation results initial conditions $~v(0)=10/3.6~$[m/s] $~x(0)=0~$[m] $~p=30~,c=100~$

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You could swap $\frac{dv}{dt}$ to $\frac{dv}{ds}\times \frac{ds}{dt} = v\frac{dv}{ds}$ Then you'd get an equation of the form $$v\frac{dv}{ds} = \frac{a}{v}-bv^2-c$$ $$s = \int {\frac{v}{\frac{a}{v}-bv^2-c}}dv$$ probably best solved with a graphical calculator such as Desmos. You would then have a graph of $s$ against $v$ with $v$ on the $x$ axis. Then do ...

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Actually the formula which you are using will be applicable for constant acceleration also. Let initial velocity will be $u$ and constant acceleration be $a$ then distance travelled in $t$ seconds will be: $$S=ut+\frac{1}{2}at^2$$ At root level average velocity means total displacement (not distance) divided by total time taken. So $$V_{avg} = \frac{X_f-X_i}... 0 Assuming the acceleration a is uniform, then:$$v(t)=v_0+a\Delta t$$where v_0 is the velocity at t=0. If by average velocity \mathbb{v} we mean average over time then strictly speaking:$$\mathbb{v}=\frac{1}{\Delta t}\int_0^{\Delta t}v(t)\mathrm{d}t=\frac{1}{\Delta t}\int_0^{\Delta t}(v_0+at)\mathrm{d}t=\frac{1}{\Delta t}\left(v_0 \Delta t+...

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Assuming that your acceleration is constant, then you can simply use $\bar v = \dfrac{v_1+v_2}2$, and you can calculate both $v_1$ and $v_2$ by using the equation $v=at+v_0$. Do note that if the acceleration is not constant, then the average acceleration is $\bar v = \dfrac{\Delta x}{\Delta t}$ where $\Delta x$ is the change in position.

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Acceleration is the change in velocity, to understand acceleration intuitively start with the constant acceleration constant acceleration if you have a moving particle star moving from rest with Constant acceleration, say $a=5 m/s^2$,so its speed will be changing as following time $(S)$ speed $(m/s)$ acceleration $(m/s^2)$ 0 0 5 1 5 5 2 10 5 3 15 5 4 ...

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In a situation where the applied force acting on a body increases to a point where it exceeds the maximum force of static friction the two surfaces in contact will start to slide relative to each other. When this happens there will be a switch from static friction to kinetic friction. For two given surfaces the coefficient of kinetic friction is typically ...

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Well, with a>g, the ceiling is the new floor, so it still drops to the floor, in a manner of speaking. With a=g, it does not drop at all. So the answer is "impossible scenario" ;-).

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When answering these sort of questions you do not just have to understand the physics, you also have to make a model of the examiner's mind. Yes, if your accelerate downwards a rate greater than $g$ the briefcase will not hit the floor. At $a>g$ (downwards positive) it will hit the ceiling. Quite possibly the question creator did not think of this ...

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If the briefcase remains in equilibrium with the elevator, then the elevator is accelerating downwards. This means that $a=-g$, where the minus sign means the motion is downwards. If the briefcase rises to the elevator's ceiling, then $a \geq -g$.

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As the air cannon works quickly, you can reasonably assume that no heat is exchanged with the surroundings --- in which case the expansion adiabatic so $PV^\gamma$ is constant.

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You are modeling the expansion process as being both isothermal and reversible. For the process to be reversible the expansion needs to occur very slowly . That means the acceleration would have to approach zero. Hope this helps

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See StephenG's answer for why eccentric orbit doesn't let you gain any more energy leaving the mass than you had when you started approaching the mass. You will probably get vaporized by other high-energy particles particles in your path, but let's ignore those since you asked about gravity problems only. There won't be any any felt gravitational ...

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This won't work unless the black hole is orbiting something else. It doesn't make a lot of difference to this whether we use a black hole or any other type of body - it's just gravity and conservation fo momentum and energy at work. If you just had an isolated body and approached it, a slingshot won't happen. All that happens is that your trajectory will ...

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When you have an expression of the form $(a + b)^n$, where $|a| \gg |b|$, you can rewrite it as $$(a + b)^n = \left(a\left(1+\frac ba\right)\right)^n = a^n \left(1+\frac ba\right)^n$$ and then, given that $\left|\frac{nb}a\right| \ll 1$ (and $a + b > 0$), apply the binomial approximation to obtain $$a^n \left(1+\frac ba\right)^n ≈ a^n \left(1 + n\frac ba\... 4 There is nothing mischievous about massless pulleys in this exercise. Here's why: Let me rewrite the system of equations but this time assuming massive pulleys$$ \begin{cases} m_1 a_1 &= T_1 - m_1 g \\ m_2 a_2 &= T_2 - m_2 g \\ m_A a_A &= -T_1 \\ m_C a_C &= 2 T_1 - T_2 \\ 0 &= (a_1 - a_A) + (a_C - a_A) + (a_C - a_B) + (a_A - a_B) \\ 0 &...

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Just do $$\frac{1}{(r_A + 22)^2}=\frac{1}{ (22)^2 (\frac{r_A}{22} + 1)^2}$$ Now you have a term $$(1+\frac{r_A}{22})^{-2}\approx 1-\frac{r_A}{11}$$

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For Binomial approximation, you need take out the $22m$. So: $$(r_A + 22m)^{-2} = \frac{1}{(r_A + 22m)^2} = \frac{1}{(22m)^2(\frac{r_A}{22m}+1)^2}$$ $$\therefore \ \ \ \text{binomial expansion:} \ \ \ (1+x)^n \approx 1 + nx$$ Also: $$(1 + a x)^n \approx 1 + n a x \ \ \ \ \ \text{where} \ \ \text{a} \ \ \text{is} \ \ \text{constant}$$ We have: $$\text{... 1 No. A rocket in space would need less fuel since the gravitational force acting on it is less than on Earth. :) 4 The system you have shown is known as a fool's tackle (see: https://www.lockhaven.edu/~dsimanek/TTT-fool/fool.htm) The funny thing about it is that it is impossible for the pulley to remain in a stable, static configuration for any choice of m_1 and m_2. The reasoning is quite simple. The tension along the string, assuming it is ideal, is constant ... 1 This is the free body diagram for each mass you can write the Newton equation ~m_i\,a_i=\sum F_k$$m_2\,a_2=-m_2\,g+T_4\\ m_c\,a_c=-T_4+T_1+T_2\\ m_a\,a_a=T_3-T_2+T_5\\ m_1\,a_1=-m_1\,g-T_5$$additional equation are$$T_2=T_1\\ T_2=T_5\\ T_1=T_3$$you obtained 7 equations for 9 unknows$$a_2~,a_c~,a_a~,a_1~,T_i\quad i=1..5$$to solve the problem you need ... 2 \newcommand{\bl}{\boldsymbol{#1}} \newcommand{\e}{\bl=} \newcommand{\p}{\bl+} \newcommand{\m}{\bl-} \newcommand{\gr}{\bl>} \newcommand{\les}{\bl<} \newcommand{\greq}{\bl\ge} \newcommand{\leseq}{\bl\le} \newcommand{\plr}{\left(#1\right)} \newcommand{\blr}{\left[#1\right]} \newcommand{\lara}{\langle#1\rangle} \newcommand{\lav}{\langle#1|}... 0 Hints: To find \frac{d^2y}{dt^2} in the case where the speed along the cycloid is constant, using \omega for \frac{d\theta}{dt}, the formulae in the question and using the chain rule \frac{dy}{dt} = \frac{dy}{d\theta}\times \frac{d\theta}{dt} the speed is constant v_x^2 +v_y^2 = 2\omega^2R^2(1+\cos\theta) = c^2 so$$\omega^2 = \frac{c^2}{2R^2(1+\...

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Massless, frictionless pulleys are an idealization. You never encounter them in real life. But they do simplify this kind of physics problem. And that is a good thing. The problems can be hard enough without further complications. In this case, the idealization isn't helping. You have $3$ moving masses: $m_1$, $m_2+m_C$, and $m_A$. Note that $m_2+m_C$ count ...

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My question is, why is it okay to apply Newton's second law to pulley A, if we assumed its mass is 0? Shouldn't that lead to a contradiction? Yes it is OK to apply Newton's second law to pulley A. It is OK to apply the law to any object, even if its mass is zero. If the lengths of the strings are constant (i.e., the strings are inextensible) then it is a ...

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There is nothing magic about accelerometers that changes the way a Fourier transform works. If you’re murky on FFTs, I suggest you create some “fake data” that is the sum of one or more pure sine waves, feed it through your analyzer to see what its power spectrum looks like, and run the power spectrum through an inverse FFT (which is secretly the same as an ...

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how the acceleration can become 0? The acceleration can only become zero when there is no net external horizontal force acting on the mass, per Newton's second law. Given a frictionless surface and no air resistance, that means it will only become zero when the 10 N force is removed. It will then move with constant velocity of $v=at=0.5$m/s$^2$t per Newton'...

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Will a car consume the same amount of fuel when accelerating from 0−100km/h and when accelerating from 100−200km/h? No. It is easier to understand supposing a constant acceleration. In this case: $V_2^2 - V_1^2 = 2ad$ where $d$ is the traveled distance. $d$ is greater when going from $100km/h$ to $200 km/h$. As Work is $w = Fd$, and $F$ is constant, $w$ is ...

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Also, physically, what happens to the applied force? How does it vanish? How long the force is being applied? Or how long the force 10 N will be in picture? Nothing happens to it as long as you apply it. When you stop applying it, then it obviously vanishes. So the $10\,\mathrm{N}$ force will be 'in the picture' as long as you dictate it to be so.

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The force cannot just be applied at the instant $t=0$. It must be applied over some period, from $t=0$ to $t=t_1$, say. While the force is applied, the block accelerates with acceleration $a=0.5$ m/s/s. Once the force is removed the block stops accelerating and continues to move with a constant velocity. For example, if the force is applied for $10$ seconds, ...

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The frame you are choosing must be the one that is instantaneously co-moving with the object. If instead you are looking at some frame for which the object has $u^\mu$ at all times, then either the coordinate frame is non-inertial (in which case $\nabla_0 u^\mu \neq \partial_0 u^\mu$) or the coordinate frame is inertial (in which case the object is moving ...

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I think you need to work on defining your problem a bit more or you have defined things incorrectly. If your metabolic equation only accepts acceleration as a single positive or negative number, it must be assuming some axis to work along, and that axis need not be x, y, or z. Or you are working with an inappropriate single axis form of the equation and a 3D ...

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If you know the $x$-, $y$-, and $z$-components of the velocity of the object, then the quantity which will tell you whether the object is "speeding up" (positive acceleration) or "slowing down" (negative acceleration) is $$a_v = \frac{\vec{a} \cdot \vec{v}}{|\vec{v}|} = \frac{a_x v_x + a_y v_y + a_z v_z}{\sqrt{v_x^2 + v_y^2 + v_z^2}}$$ ...

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Remember kinetic energy represents how much energy I would have to do on a body to get it up to that speed OR the amount of energy a body has Thought experiment, thinking about doing work via an intermediatary electron can give good insight into why more work is needed to change the velocities between different reference frames, regardless if jts the SAME ...

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The Unruh temperature depends on the magnitude of the acceleration, which is a scalar.

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In the context of special relativity, the description of "harder to accelerate" is from a fixed inertial frame [as @m4r35n357 said in the comments] and it's as @Marco says that it's due to the addition of velocities in relativity. This visualization I made in Desmos might help: https://www.desmos.com/calculator/tjngj63cat Constant proper ...

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It is a consequence of relativity. You might know that the simple addition of velocities is an approximation that holds almost exactly true at very low speeds but becomes increasingly invalid as speeds approach the speed of light. For example, if you are cycling along a road at ten miles an hour, and you throw a ball ahead of you at ten miles an hour, the ...

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Burn more fuel is not the same as more power, power is the rate at which energy is being used up/ rate at which work is done on an object ( explanation about why fuel is needed to maintain speed at all)The rate at which work is being done and therefore more fuel burning) on an object given resistance would be higher for an object traveling at higher speeds, ...

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As per Newton's second law and in the absence of any loss functions (as you specified) the car power requirement will be independent of driving speed and zero, despite the fact that at higher speed the car has more kinetic energy, acc. $K=\frac12 mv^2$. Power is only needed when the car accelerates because then energy has to be added to increase the kinetic ...

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Since KE depends on velocity squared, each additional km/h requires more energy than the last. Suppose you have a car that rolls down a hill, converting potential energy into kinetic energy, and achieves some speed X at the bottom. If you double the height of the hill and double the kinetic energy gained, the speed of the car at the bottom is less than 2X. ...

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Let $u$ denote a time-dependent vector. Working at second order in $\mathrm dt$ throughout, we'll use the identities\begin{align}\sqrt{1+A\mathrm dt+B\mathrm dt^2}&=1+\frac12A\mathrm dt+\left(\frac12B-\frac18A^2\right)\mathrm dt^2,\\\frac{1+A\mathrm dt+B\mathrm dt^2}{1+C\mathrm dt+D\mathrm dt^2}&=1+(A-C)\mathrm dt+(B-D+C(C-A))\mathrm dt^2.\end{... 3 The high school kinematics problems normally assume one or two dimensional travel on a Euclidean plane. The earth is NOT a Euclidean plane, leading to the zero displacement issue. I suggest you reformulate this problem for your high school student such that you restrict the automobile to a flat two dimensional surface and have the automobile travel in a ... 1 In general, the acceleration relates to the position through a second order timer deivative.\mathbf a=\frac{\text d^2\mathbf x}{\text dt^2}$$I am not sure what you mean by "assuming a velocity at rest" though. If you have the position as a function of time then the velocity is also determined by a derivative:$$\mathbf v=\frac{\text d\mathbf x}{...

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I’m really looking for...more of an intuitive answer. How about this? Velocity is the rate of change in position. If we measure position in meters ($m$), then we measure velocity in meters per second ($m/s$). (I.e., how much the position ($m$) changed in each second.) Acceleration is the rate of change in velocity. If we measure velocity in meters per ...

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When an object initially at rest has uniform acceleration $a$ for time $t$, it reaches speed $at$. Its average speed up to time $t$ is $(0+at)/2=\frac12at$. Therefore, the distance travelled is $\frac12at\times t=\frac12at^2$. If the object has constant mass $m$, a force $ma$ was responsible for this acceleration. (We'll neglect any wasting of energy due to ...

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