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I will quickly post an answer here and will edit/expand on the answer with time. I used the model for nitrogen gas viscosity based on the PhD thesis by Kegang Ling from Texas A&M university, 2010 The model requires temperature and density. Density was calculated using the real gas law. The real gas law requires the compressibility factor of gas at ...

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Initially a small force needs to be applied to get the desired velocity but this small force is neglected and we start our observation after the body acquires the velocity and the force is removed .

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Note that when you apply chain rule , you assume dx not to be zero . This will clear it up for you .

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You can apply chain rule if $v$ is differentiable wrt $x$ and $x$ is differentiable wrt $t$. I think there are no other conditions,as this post on MathSE seems to say, http://math.stackexchange.com/questions/688152/necessary-conditions-for-the-chain-rule-of-differentiation-to-be-valid#= and this condition is not always available. When $v=0$,make sure ...

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No, it doesn't imply that $a = 0$. If, at some value $t = t_0$, the acceleration is non-zero while the velocity is zero, the position function is either a minimum or maximum. That is, $x(t)$ is stationary there: $$x(t_0 + dt) = x(t_0)$$ which means that at $t = t_0$ $$\frac{dx}{d\dot x} = \frac{dx}{dv} = 0$$ thus $\frac{dv}{dx}$ is undefined at $t = ... 16 The correct thing to say would be that "if v=0 and dv/dx is finite then a=0". A simple example, to help illustrate what's going on, is the well known case of constant acceleration "-g" near the earth's surface. In this example, we consider "x" to be the height above the ground, and assume the initial x is zero. In this case $$x=-\frac{gt^2}{2}+v_0t$$ $$... 1 The simplest way to look at this is to consider separately the horizontal and vertical velocity/position. For a projectile launched at angle \theta and velocity v, the components are: Horizontal velocity$$v_h = v\cos\theta$$Vertical velocity$$v_v = v\sin\theta$$The position at time t is then given by$$(x, y) = (v_h\cdot t, v_v \cdot t - ... 0 The motorcycle is moving at$10\frac{m}{s}$, so his distance from a starting point of 0 each second can be modeled as: $$y = 10x$$ The car is moving at a constant acceleration, so that distance from an origin of 0 can be modeled as: $$y = 2(x^2 + x)$$ Therefore, when they are equal to each other, they are at equal distances from the starting point. ... 2 Let's say car and bike be at rest at$1pm$so,$v_c=0$and$v_b=0$. Calculations for motion of car: Since car is moving with constant acceleration, At 1:00:00pm,$v_c=0m/s$,$S_c=0m$At 1:00:01pm,$v_c=4m/s$,$S_c=4m$At 1:00:02pm,$v_c=8m/s$,$S_c=12m$At 1:00:03pm,$v_c=12m/s$,$S_c=24m$At 1:00:04pm,$v_c=16m/s$,$S_c=40m$Calculations for motion ... 1 First, you can use your velocity expression to determine the time that it takes to reach a certain final velocity,$v_f$. $$v_f=v_0e^{-k\Delta t}$$ Given that$\Delta t$is the amount of time spent decelerating. If you solve this for$\Delta t$you'll find: $$\Delta t=-{1 \over k}\ln\left({{v_f \over v_0}}\right)$$ Next, you can determine an expression ... 1 You need to solve system like following:$\begin{cases} d_0=\int_{0}^{\tau}v_0 \exp(-kt)dt=\frac{v_0}{k} (1-\exp(-k\tau)) \\ v_f=v_0\exp(-k\tau)\end{cases}$You need to solve for$\tau$which is time till stop, and$k$which is essentially rate of acceleration. 1 Your calculations are correct. They differ from your model (which uses ABS braking) however, because they don't take into account the duty cycle of the braking. If this is added to your calculations, then the two results should be similar. 0 When you're using the equation F=ma, the F is ALWAYS the total/resultant/net/unbalanced force, NOT one of the individual forces. It describes the effect (the acceleration) that happens due to the cause (the total force on an object). Here you happen to be right because (at least horizontally) there is only one force, the frictional force, so you should get a ... 0 So I'm assuming you're saying that the work done on the car in distance$d$has to be equal to its kinetic energy$\frac{1}{2} m v^2$. Then, using$W = F d$: $$F d = \frac{1}{2} m v^2 \\ m a d = \frac{1}{2} m v^2 \\ d = \frac{v^2}{2 a}$$ So, yes, this equation is correct. Your relation between the two forces is also correct. Since mass drops out, the ... 0 Assuming that the acceleration is$a$m/s$^2 \neq 0$, then the distance travelled after reaching a velocity of$\sqrt{a}$m/s is always the same, namely$x =0.5$m. Therefore, as the acceleration tends to zero, the time taken to reach this distance obviously tends to infinity, so your answer makes perfect sense. However, the same result cannot be applied ... 1 Velocity has units of m/s and acceleration m/s² so the root of acceleration would be √m/s. This unit does not make any sense. If you just want the numerical value, then v=a·t → √a=a·t then √a/a=t. 0 I think you should clarify the question. If it has an acceleration of a=0 it cannot attain a velocity of 1/sqrt(a). 3 She asked us if the body was accelerating or slowing down Acceleration is defined as the time rate of change of velocity and, in this example, the acceleration is constant and positive. So, the full answer is: the velocity of the body is always increasing while the speed is decreasing for$t<1$and increasing for$t>1$. In this plot, the ... 2 If you start with the bag stationary at 300m then drop it the bag is going to fall straight down, and its maximum height would indeed just the 300m point it started from. However you're not starting with the bag stationary. You're starting with the bag moving upwards at 13 m/s. So the bag is going to start at 300m then move up, come to a halt, then start ... 0 in your answer you ignored the initial velocity the bag will rise a little until its velocity is zero then it will fall again , it is like he found the total distance 0 I understand what your teacher is saying, but I think she's wrong. In my physics classes we were told never to use the word "de"celeration, only acceleration. Why? For elegance reasons mostly. Acceleration is a vector quantity, therefore its magnitude is always positive or zero with a direction. The sign (or direction) of the acceleration depends on your ... 3 The initial velocity and acceleration here are in opposite directions. The magnitude of velocity (represented by$S=|\vec v|$) decreases upto a certain instant. (i.e. where$\vec v=0$). Edit: Also, consider these graphs. ($t^.$being the time where$v=0$) Note how the velocity increases but the magnitude of it (in the$S$,$t$graph) decreases till$t^.$. ... 0 1.So you have a position vector given by $$\vec{r}=\left(8t-4.5t^3\right)\hat{i}+\left(10-2t^2\right)\hat{j}$$ If you use the definition of velocity as $$\vec{v}=\frac{d\vec{r}}{dt}$$ then we differentiate each component of the vector to get the velocity vector. $$\vec{v}=\left(8-13.5t^2\right)\hat{i}+\left(-4t^2\right)\hat{j}$$ Acceleration$\vec{a}$is ... 0 From what I understand, the question is pretty straight forward. For acceleration: $$L=\frac{a_1 \cdot t_1^2}{2} \implies t_1 = \sqrt\frac{2L}{a_1} \implies v=\sqrt{2a_1\cdot L}$$ For deceleration: $$a_2\cdot t_2+ v=0 \implies t_2=-\frac{\sqrt{2a_1\cdot L}}{a_2}$$ The minimum time is thus$t_{min}=t_1+t_2$. 0 (Classical Physics only) Any massive body has a property known as inertia, thus even a body floating in outer space would require some kind of force to be accelerated. Using Newtons second law, you would find $$\tag{NII} \sum \vec{F} = \frac{\mathrm{d}}{\mathrm{d}t}\vec{p},$$ which for constant mass and one-dimensional motion simplifies to $$\tag{NII'} ... 1 The answer is negative. The gravitational acceleration is the same for all the atoms. Thus, the internal structure of the body is not affected. Using the terminology in your question, the average kinetic energy of atoms inside this object doesn't change. This is the reason for which given a body left free in the gravitational field, we usually consider the ... 0 Yes. The kinetic energy from translatoric motion of the object rises as a whole. Any small piece of it contributes with its' own fractional mass, since any small piece also achieves a net translatory motion. Each piece is a fraction of the whole object:$$K_{trans}=½m_{total}v^2=\sum K_{trans,atom}$$1 I suspect your teacher glanced at it very quickly, and didn't realize you were using timesteps. When you work with timesteps you are doing a discrete approximation of the differential equations which describe the relationships, namely: \frac{dv}{dt} = a and \frac{dx}{dt} = v which in the discrete form, and arranged to match your code, become: \Delta ... 0 Your solution is correct at the level of your course. We don't see what else is in your code, so there might be another problem. I can only guess what he means by "acceleration / 2". Perhaps he didn't read your program carefully enough; he might have been expecting a solution involving 1/2 a t^2, and when he didn't see it, moved on. Your solution ... 0 s = v \cdot t is only true if v stays constant. If v changes with time, the proper relation is$$ s(t) = \int_{t_0}^{t} v(t') \mathrm dt'$$For v = a \cdot t with a constant acceleration a this becomes$$ s = \frac{1}{2} a t^2 $$So instead of calculating the position from the velocity you could calculate it from acceleration directly. 0 There is no problem to get the time T_1 to reach the peak, and the time T_2 to fall on the ground if you know v_{y,0}, the projection on the vertical of the total velocity under which you threw the ball. For this calculus you can consider only the center-of-mass movement of the ball. About the spinning movement, I don't understand in football but I ... 0 Assuming that you are talking about an uniform gravitational field \vec{F_G}={0,-mg} than it's possible to write y(t)=y_0+v_{0y}-\frac{1}{2}gt^2 and x(t)=x_0+v_{0x}t if you impose the initial conditions on x you shall find that \forall{t} \rightarrow x(t)=0 so if b \ne 0 the particle will never reach it beacuse it will just fall along a ... 1 When you take the average of two quantities, you need to consider the "weighting". In this case, the time spent at each of the velocities matters, and becomes this "weight". In general, when you have a weighted average you multiply each value by its weight, and divide by the sum of the weights. When all the weights are 1, that reduces to the familiar ... 0 In the moment the bottle falls to the water, it travels downstream at v_{stream}. It travels a distance X=v_{stream}∗1h in the first hour and a distance Y=v_{stream}∗t in the unknown time t. Furthermore, 3km=X+Y=v_{stream}(1+t). About the canoe, it traveled 2km in the first hour, and in the following time t it traveled back 5km. So it traveled ... 4 If you bring an object from infinity to a distance R then the potential energy change is:$$ \Delta U = -\frac{GMm}{R} $$Assuming your object starts at rest, the potential energy change is equal to the change in kinetic energy, so we have:$$ \frac{GMm}{R} = \tfrac{1}{2}mv^2 $$so:$$ v^2 = \frac{2GM}{R} $$You want v \ge c, so:$$ \frac{2GM}{R} ... 1 You don't have to use the standard equations of motion, you can just deduce them from Newton's laws: You have a force$\mathbf{F}=m\times -9.8 \: \mathbf{\hat{e}_y}$acting on the arrow. Newton's second law is$\mathbf{F}=m\frac{d\mathbf{v}}{dt}$. Integrating for the$x$and$y$coordinates yield: $$\frac{dv_y}{dt} = -9.8$$ $$\int_{v_{yi}}^{v_{y}}dv_y = ... 1 The symbol s_{NN} is in OP's context of RHIC the Mandelstam s-variable in a Nucleus+Nucleus collision. The s-variable is also known as the square of the center-of-mass energy. 0 With respect to ground: After push-off, man moves upwards at 0.8\text{ft}/\text{sec}, pillow downwards at 8\text{ft}/\text{sec}. It takes the pillow 1\text{s} to hit the ground, by which time the man has moved to 8.8\text{ft} up. Pillow (being weirdly elastic) returns at exactly the same speed it departed with, has to close the 8.8\text{ft} gap at ... 10 No they aren't. Suppose we have some velocity v(t). The differential with respect to time is just the acceleration:$$ \frac{d}{dt}v(t) = a(t) $$Now differentiate it with respect to distance s:$$ \frac{d}{ds} v(t) = \frac{dt}{ds}\frac{d}{dt} v(t) = \frac{dt}{ds} a(t) = \frac{a(t)}{v(t)} $$0 You already have t_{on\ the\ ground} expressed through initial velocity, angle and landing distance. Plug this time into equation for vertical displacement, and you will find the height. 0 You just solve for the t (time) corresponding to your x (the landing distance) and evaluate your y (height) by that t: If x_{(t)}=t\cdot v_0\cdot \cos (\beta ), then t_{(x)} = x/v_0\cdot \sec (\beta ) So you plug that in for t and evaluate the height y by distance x: y_{(x)}=t_{(x)}\cdot v_0\cdot \sin (\beta )-g\cdot t_{(x)}^2/2 Also ... 0 In the case you described the function a(v) gives you the differential equation$$ \frac{d v}{d t}=a(v).$$You just have to solve this differential equation for$v(t)$with some initial condition$v(t=0)=v_0$. After solving the equation you get the acceleration by differentiating the solution$a(t)=\dot{v}(t)$. For the example you mentioned: The drag ... 12 Just imagine an expanding shell. Then all the parts are moving at the same speed. If the mass is$M$and the radius is$R$, then the kinetic energy is$\frac{1}{2} M \dot{R}^2$. A sphere is just a series of shells. A shell at radius$r$has mass$\mathrm{d}m = \frac{M}{4/3 \pi R^3} 4\pi r^2 \mathrm{d}r$and is expanding at a rate proportional to the ... 0 The angular momentum of the system is the same before and after the collision. Since one object is stationary before the collision, the angular momentum is just the momentum ($mv$of the moving puck multiplied by the perpendicular distance between them (which is$2r\sin(30)$). The moment of inertia of the two pucks stuck together is a little bit tricky, ... 0 A key idea is that the path of the center of mass is unaffected whether the two pucks collide or not. So take the snapshot of the situation at any moment of time before collision. Find the angular momenta of the two pucks with respect to the com at that instant of time. Since angular momentum is conserved for an isolated system (i.e$\tau_{net}=0\$), the ...

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s, u, v and a are all vector quantities. When the vector points to a direction opposite to the "default" direction, then its sign is negative. The "default" direction can be any direction that you want it to be, but note that all these variables share the same "default" direction, so switching the sign of one variable will require you to switch the sign of ...

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It all boils down to choosing the direction in which the axis increases in the vertical direction. In your case that direction is the upward direction. Distances upward from origin are positive and negative downward. velocities in which the position increases with time are positive and those in which the position decreases with time is negative.

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