# Tag Info

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I know how , in the physical sense - $$\frac {dv}{dt} = a$$ But, even after thinking a lot, I am not able to see the fault in this - $$\frac {dv}{dt} = \frac {d(st^{-1})}{dt} = \frac {sd(t^{-1})}{dt} = s*(-1)*t^{-2} = \frac {-s}{t^2} = \frac > {-v}{t} = -a$$ I know something is terribly wrong here but I'm just not able to ...

4

I think you're confusing acceleration with velocity. If a body experiences a constant force, the value of acceleration is also constant, which means the value of velocity increases/decreases in a linear fashion, depending on whether the body is accelerating ($a>0$) or decelerating ($a<0$). If the force changes, the acceleration also changes, assuming ...

4

It isn't clear from your question exactly what you are integrating and how, but this is the way to tackle problems like this. You know that: $$\frac{dv}{dt} = -kv$$ The way to solve equations like this one is to rearrange it by dividing both sides by $v$ and multiplying both sides by $dt$ to get: $$\frac{1}{v}dv = -k\,dt$$ Now we can integrate both ...

3

If the object is spinning close to the speed of light then it has significantly more energy than if it were at rest. This does contribute to an increase in gravitational pull and is significant in astrophysical phenomena like neutron stars! http://arxiv.org/pdf/1003.5015.pdf The Earth is also more massive because its spinning but we have no hope of ...

3

Keep in mind that the screw axis does not have to pass through the body. For your example place the axis of rotation parallel to the x axis straight above the cylinder, then rotate the cylinder 180° about it. The result will be equivalent to 180° rotation about the x axis followed by a translation along the z axis by twice the distance from the x axis to the ...

3

You are looking at a specific application of a more general formula $$v_q^2-v_{oq}^2=2a_q(q-q_o),$$ where $q$ is the coordinate direction, the $v$ terms are velocity components along the $q$ axis, $a_q$ is the constant acceleration component along the $q$ axis, and $q$ and $q_o$ are the positions along the $q$ axis, which match, respectively with $v_q$ ...

2

Physicists tend to be a bit casual about sign conventions when it seems to be obvious. So let's attempt to be completely rigourous. The key step is getting the flight time $t$ since the range is just $v\cos\theta\, t$. We do this using the SUVAT equation: $$v = u + at$$ We'll use the usual conventions that up and right are positive, so $v_y$ and $v_x$ ...

2

Fast things appear blurry to your eyes, and the distance from a point on the paper to your pencil is fixed. The distance from an ink dot on the paper, to the tip of your pencil where you press down, cannot change. Call this distance $d$. The only positions the point is allowed to be at when you spin the page around, are positions that satisfy the equation ...

2

Simplification. There are different formulas that are used to calculate drag at various speed regimes. The problem is, they are all of the form $$F\propto v^n$$ where $n=1$ or $2$. This means that $$a\propto v^n$$ which in turn means that $$\frac{\mathrm{d}^2x}{\mathrm{d}t^2}\propto \left(\frac{\mathrm{d}x}{\mathrm{d}t}\right)^n$$ You need calculus to solve ...

2

Yes. This approach is definitely viable. Coincidentally, equating the time-derivative of vertical position to 0 is the same thing as maximising the height mathematically (concept of maxima and minima). Another approach is to use the conservation of energy. Assume the vertical component of velocity to be $v_y$. Solve it by equating difference of potential ...

2

Let $\mathbb{X}^\prime$ be the image of body $\mathbb{X}$ (a set of points in Euclidean space) under a proper Euclidean isometry $E$. To work out how to find the rotation and displacement, I'll discuss the transformation generally, then reformulate the general discussion into a statement of Chasles' theorem. Thinking of a general proper Euclidean Isometry ...

2

You throw the ball upwards with velocity $v$ and it returns to your hand with velocity $-v$. Let's draw a graph showing the velocity as a function of time: Acceleration is defined as: $$a = \frac{dv}{dt}$$ so it is the gradient of the line in this graph. The velocity-time line is straight so the gradient is constant which means the acceleration is ...

1

Yes, the cart will move, due to the force applied by the string to the pulley. To solve, calculate the string tension while the weights are moving, and then note that the pulley has to provide an opposing force in order to change the string's direction. The reaction to that force acts upon the cart, accelerating it. Momentum is conserved, because the ...

1

Setting $$\theta := \theta_1 + \theta_2$$, the momentum of the Higgs boson (candidate) with respect to the lab $$\| \textbf p_{lab}[~H~] \| = \| \textbf p_{lab}[~\gamma_1~] \| ~\text{Cos}[~\theta_1~] + \| \textbf p_{lab}[~\gamma_2~] \| ~\text{Cos}[~\theta_2~] = (E_{lab}[~\gamma_1~] ~ \text{Cos}[~\theta_1~] + E_{lab}[~\gamma_2~] ~ ... 1 Tell me if I get something wrong... You are going to accelerate in the first part of the track (with constant acceleration, let's say, a_1) and then you are going to decelerate in the second part of the track (with constant acceleration a_2, negative), and ther's no speed limit, so you never travel at constant speed. Am I right? As I see, in this problem ... 1 The answer by ryanp16 gives a great derivation of the equation you're asking about using calculus, and in fact, that is the approach that I would have taken had I not seen his answer. However, if you're not familiar with calculus, there's a second, algebraic approach that you can use to arrive at the same conclusion. I like to take this approach with high ... 1 Let's approach this slightly differently; I am sure you would agree that:$$v a = a v$$This is the equivalent of your equation after being multiplied by v where v=\frac{dx}{dt} and a=\frac{dv}{dt}. Now integrate this with respect to time:$$\int_{t_1}^{t_2} v a dt = \int_{t_1}^{t_2} a v dt  From the definition of the derivatives we know that ...

1

Backing up what zeldredge said, what you asked about is known as "relativity without light". According to the intro of this paper (arXiv link) for instance, the original argument was given as early as 1910 by Ignatowski, and has been rediscovered several times. There is a modern version due to David Mermin, in "Relativity without light", Am. J. Phys. 52, ...

1

If the two cars start at the same time and place, they can only pass once: However, they can pass twice if B is given a head start in either time or distance. e.g. if they start at the same place but B leaves first: or if they start at the same time but B starts off some distance in front of A:

1

I will split this question up in to two parts. First it will be useful to figure out how fast the flower pot is going when it passed by the bottom of the window. Secondly from this velocity you can derive at which time/position its velocity should have been zero and thus when or from where it has been dropped. For the first part we know that after the pot ...

1

If the force on a body is constant, it will accelerate uniformly. The velocity will be a linear function of time, and the position will be a quadratic function. This is a very common scenario with falling objects. For objects near the surface of the Earth, the force of gravity is very nearly constant. Without friction, a body will move along a parabolic ...

1

Place the car's centre of gravity at the origin of a Cartesian coordinate system $x,y$, as shown above. Now determine the balance of forces along the $x$ axis. Assume the car will slide (up or down), as there's no friction. The equation of motion is: $ma_x=mg\sin\alpha-F_c\cos\alpha$ (Eq.1). $F_c=\frac{mv^2}{R}$ with $v$ the velocity of the car and $R$ ...

1

When you shoot the ball upwardly, gravity acts on it with a force $mg$ where $m$ is the mass of the ball and $g=9.81 ms^{-2}$ the Earth's gravitational acceleration. If the initial upward velocity was $v_0$ then the instantaneous velocity $v$ is given by: $v=v_0-gt$, so after some time $t=\frac{v_0}{g}$ the balls's velocity becomes $v=0$. However, we know ...

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