# Tag Info

4

Here are the steps you want to take. We need to find $v_0$. The equations are $$v_t = v_0 + g\cdot t\\ y_t = y_0 + v_0 t + \frac12 g t^2$$ Two equations, two unknowns. Eliminate $t$, then solve for $v_0$ (Note that I use a Y axis that increases as you go down - just saves thinking about the sign of $g$). Alternatively you can use conservation of energy. ...

4

The length along any segment of the Koch snowflake is infinite. It has finite area but infinite perimeter. So, for a particle to move from one place on the snowflake to another it would have to travel an infinite distance. This is why differentiability is important.

3

You have the momentum given by the vector: $\vec{p}=m\vec{v}$ $\vec{v}=$ being the velocity vector. Now if you integrate the momentum with respect to velocity then you have the integral: $\int\vec{p} \cdot d\vec{v}=m\int \vec{v} \cdot d\vec{v}$ Where $\vec{v} \cdot d\vec{v}$ is the dot(scalar) product between the two vectors $\vec{v}$ and $d\vec{v}$ ...

2

You can think of this question as someone a distance $2d$ ahead of you releasing a pulse of light (the man in the mirror). In that way, the problem simplifies to "how far can light travel in $0.80 \mu s$?" Solving this will give you the value of $2d$ ($2d = ct$). So divide by $2$ to get $d$, and the calculation is simply $d = ct/2 = 120m$. EDIT: I should ...

2

No, it's not possible, because one of the underlying assumptions of kinematics is that all paths are at least twice differentiable. Before you complain about this requirement, remember that physics is about building models that can be used to describe and predict measurements. Measurements always have some amount of uncertainty, and even if you suppose that ...

2

The "First Equation of Motion" you define is perhaps more accurately called the "First Equation of Motion with Constant Acceleration." One would need to use Calculus to calculate the change in velocity when Acceleration is not constant, but what you call a "varying variable." Your first equation which you arrive at by Algebra: $$V_f = V_o +a \Delta t$$ ...

2

However, I find that the net torque of the cylinder is 0 if I set the axis of rotation as the bottom of the cylinder (the contact point between the ground and the cylinder); my first question is, what accounts for this discrepancy? I suspect there should be a force about the center of mass, but I don't see what could be the source of such force. You're ...

2

you have to notice that this motion is accelerated, so if you define velocity as $d/t$ you will get the wrong result. For uniformly accelerated motion (which is your case, the acceleration is contant: g), you have to use the following relationship: $y(t)=y_0+v_0t+\frac{1}{2}at^2$ When you release the ball, $y_0=H$, $v_0=0$ and $a=-g$ so you get $... 1 The system is subject to a non-zero net force in the horizontal direction and no friction, so it will experience constant acceleration (of the center of mass). Superimposed on that motion with be the anti-symmetric oscillation of the two masses on the spring. If the masses are both$m$and the spring is characterized by constant$k$the angular frequency of ... 1 Jerk in a Car We should probably avoid talking about a car's gas pedal. Without going into too much detail, imagine when you floor it in a car. Once you've got the pedal all the way down, the car continues to accelerate, but eventually (assuming a long stretch of flat road) you will reach some steady state speed and stop accelerating. So with the pedal in a ... 1 I was able to determine the user status(static, slow walking, fast walking) by calculating the variance. The Va in the research was not velocity. It was my mistake to interpret it as such. It was the variance of the euclidian norm of the accelerometer data. I decreased the accelerometer update interval to 0.1s and every second I took 10 of the values and ... 1 You can decouple the horizontal and vertical motion of your rocket. In the vertical direction you have vertical thrust and gravity and horizontally you only have thrust (I ignore air resistance here). As you are interested in the altitude only, we only look at the vertical problem. All kinetic energy in the vertical direction is converted to potential energy ... 1 Kinetics: In physics and engineering, kinetics is a term for the branch of classical mechanics that is concerned with the relationship between the motion of bodies and its causes, namely forces and torques. Kinematics: Kinematics is the branch of classical mechanics which describes the motion of points, bodies (objects), and systems of bodies ... 1 This type of problem can be simplified if you use the frame of reference of the elevator. Now the bolt falls from rest, chasing the "starting point" which starts a distance$h$below and moves down at a constant$6 $m/s. The bolt catches up in$3$seconds. Same problem, but the equations are simpler... 1 So I came up with a graphical solution to this kind of problem. It might help you understand the process of collisions, without giving you a direct answer. Consider an Cartesian coordinate system xy for measuring momentum. Draw the initial momentum vectors$\vec{A} = m_A \vec{v}_A$and$\vec{B} = m_B \vec{v}_B$. Draw a circle with the two vectors as ... 1 Rotate the digram so the line connecting the circles is horizontal at the moment that they touch - since you know dx and dy, you just take the arc tangent. Now you move the frame of reference so the point where the two balls meet is stationary. The actual speed of the center of mass is just the vector mean of the velocities of the two balls (if they have ... 1 So if the path is described parametrically with$\vec{r}(q) = (x(q),y(q))$you need to define the tangent and normal vectors in order to describe the motion$\ddot{q}$and the reaction force$F$. The kinematic velocity vector is $$\vec{v}(q,\dot{q}) = (\dot{x},\dot{y}) = ( \frac{{\rm d}x}{{\rm d}q} \dot{q}, \frac{{\rm d}y}{{\rm d}q} \dot{q} ) = (x' ... 1 It is correct except for one sign, note that the work done by friction is negative (since you move the block in the opposite direction w.r.t. the friction force) and thus it is equal to$$ -\mu mg d \cos \theta $$with this solving for v gives you 11.49 m/s. 1 If you accelerate your car with constant acceleration AND we then assume that the friction decelerates the car with constant (negative) acceleration, then you simply consider each situation for itself:$$x_{acc}=x_{0,acc}+v_{0,acc}t_{acc}+\frac12a_{acc}t_{acc}^2=\frac12a_{acc}t_{acc}^2$\$ ...

1

I don't know the "formal" proof, but here is my proof: Time dilation and length contractions are given to us by the Lorentz transformations by: t’ = t/(1-v2/C2)1/2  and d’ = d/(1-v2/C2)1/2 (in other words “same” or proportional to each other) where: t = distance/length traveled through the T dimension in observers own frame of ...

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