# Tag Info

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 ...

0

This is an excellent question. It is pedantic to whine about non differentiability ,however there is in fact a point to be made on that topic . We seem to be conflating the derivative dy/dx with the time derivative dx/dt(x is the position here). One may have a differentiable path but still the instantaneous velocity can remain undefined. Whether the converse ...

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 ...

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.

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 ...

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: $$x_{acc}=x_{0,acc}+v_{0,acc}t_{acc}+\frac12a_{acc}t_{acc}^2=\frac12a_{acc}t_{acc}^2$$ ...

0

There are a continuum of answers. For any velocity, there is a distance that the car will travel to get to zero velocity with a frictional force of f. Let's say you pick halfway point (1/2 kilometer) as that distancee, there is a certain velocity v at the halfway point that the car will coast to a stop under the friction f when it reaches 1 kilometer. ...

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As you've already figured out, roll the sphere down a frictionless hill with zero slipping and use $$\frac{1}{2}I\omega^2 + \frac{1}{2}m v^2 = mgh.$$ To use this equation, you will need to understand the relation between $v$ and $\omega$ and then you will need some way to measure either $v$ or $\omega$. How you measure the velocity will depend on the ...

0

Experimentally measuring a sphere's moment of inertia See: Measurement of the Moment of Inertia of the Rotating Platform and Attached Cylinder. Once you build the platform and determine its moment of inertia you put a sphere on the platform and determine the moment of inertia for the system sphere - platform and then by subtraction the moment of the ...

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 ...

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 ...

0

Yes it will. The spring behaves like friction here. When the first block is pushed, it transfers the energy to the spring which converts the kinetic energy to its potential energy. Once the second block overcomes its inertia, it will also start to move. Think of two blocks on a surface with friction without a spring in between. Pushing the first block ...

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 $... 0 Rather than starting with the object-surface interaction and adding a lift force, start with the object-lift interaction and add the surface later. If you have an object that weighs 10 Newtons and pull up on it with a force of 8 Newtons, then there's an imbalance: there's a net downward 2 Newton force and thus a net downward acceleration. Your object ... 0 What would happen here is that the object would not be lifted off of the surface. Instead, you would be relieving some of the normal force that is exerted by the table. Instead of$\vec{R} = -M\vec{g}$, you'd have$\vec{R} = \vec{F}-M\vec{g}$, where$|\vec{F}| < Mg$. The net force would still be zero. Imagine the following: If you put a block on ... 0 Change of velocity is definitely -8m/s, according to vector summation. Magnitude of change of velocity is 8m/s. '-' sign indicates that the change is taking in the other direction,i.e. in the direction opposite to the initial direction. 4-4=0 is the change in speed , not velocity 0 If you define your axes in such a way that the runner has a speed of 4m/s in the y direction in point A, the the velocity in point A is:$V_A = 4 \, \text{m/s} \, \hat{y} + 0 \, \text{m/s} \, \hat{x}$Your runner must have a speed of -4m/s in the y direction in point B, you get the velocity:$V_B = -4 \, \text{m/s} \, \hat{y} + 0 \, \text{m/s} \, \hat{x}$... 0 If no energy is wasted, it will take no energy to move the object. You started by saying "at constant speed". This must mean that the object has speed at the start of your time interval - otherwise it would have to accelerate and speed is not constant. Ditto at the other end - it can't decelerate at the end. And in between there is no dissipation ... 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 ... 0 A simple way of explaining jerk to a layman is to have them imagine how their head and neck feel when riding a roller coaster. When traveling in a straight line with no slope there is constant velocity but no acceleration. This results in no neck stress. When in the middle of a long constant-radius turn there is a constant acceleration but no jerk. The ... 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 ... 0 The claim that the jerk is proportional to throttle position rests on (at least) three assumptions: Engine power output responds instantly to throttle position. If there is any delay (think turbo lag, but other effects enter as well) it doesn't fit. Aside from turbo lag (caused by inertia of the turbine) it is probably too fast for people to notice, but ... 0 Imagine a race car starting to accelerate. You would immediately be pulled back to your seat, meaning that there would be immense positive jerk within the first fraction of a second. After the initial acceleration, determined mostly by the tires and the maximum torque of the engine, we are mostly limited by the specs of cars engine to deliver power. When ... 0 There is no particular answer to this problem. The force which will act on the chopstick upon collision is dependent on how quickly the paper comes to rest upon collision by decelerating (specifically, it's rate of change of momentum will determine the force exerted). If we assume the chopstick to be brittle in nature (assuming it is built in a way which ... 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 ... 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: $$V_f = V_o +a \Delta t$$ ... 0 You need initial velocity. Use these two kinematics equations. So first find time using the second formula, then use the first formula to solve for initial velocity. 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. ... -1 You have three knowns, and one unknown. Pick the newtonian kinematic equation that just has the 4 variables. You don't need to use two equations. 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$d\vec{v}$... 0 Alright, I think i figured this out by myself: If you set $$y(t) = v_y * t$$ then you get $$v_p(t) = c* v_y*t*(b-v*t)$$ after that we can now take the integral: $$s_x = \int_{0}^{s/v} v_p(t) dt = -c *s_y^2/v * (2s_y-3b) / 6$$ with$s_y = b$we get $$c*b^2/v*b/6 = 0.2*10^{-3}*b^3/v$$ 1 In going from this step:$\displaystyle 1=\frac{h}{λp}$(divided by$c$) to this step:$\displaystyle 1=\frac{hλ}{λ\color{red}{p}h}$(substituted$p=\frac hλ$) shouldn't the$p$on the RHS vanish after the substitution? 1 Your attempt is right since$1=\frac{h}{\lambda{P}}$. Then you say that you substitute for$P=\frac{h}{\lambda}$then how you get next relation with$P$included? You must replace$P$by$\frac{h}{\lambda}$and your equation become$1=\frac{h\lambda}{h\lambda}$and this gives$1=1$. So, this relation have no physical significance. But mathematicaly this ... 1 There is a mistake in equation (2). Its denominator should include the total mass of the system that you're considering, so the denominator should be '2m+m'. You correctly used this value for equation (1), but apparently incorrectly believed that since the position (and velocity?) of the lighter mass 'm' is zero that the value of 'm' shouldn't be included in ... 0 In the second case, you made a mistake in the denominator. You always put the overall mass. The lighter mass is part of the system that you are trying to find the center of mass of, even when it has a zero value of acceleration. So, the denominator should be 3m. In general, we put in the denominator the mass of every body that is in the system of bodies for ... 0 The velocity of center mass is as follows(for two particle system in straight line) as per the question, $$v_c=\frac{m(0)+m(v)}{m+m}$$ Now find the kinetic energy, 0 yes it is possible,if the total of those five forces is equal to the force of friction or gravity.then it is possible . eg. suppose if the sum of five forces in one same direction is 1N and the friction applied on that object is also 1N then the net force on that object will be zero. as they will cancel out .... 0 The most important thing that is constant is momentum in an inelastic collision. The force is constant for a specific wall and a moving bullet(considered to be appoint mass) irrespective of time. But, since in this case you take the wall to be immovable then its only the force which is constant. As you may want to say that the momentum that Earth gets due ... 1 Newton developed a formula for penetration depth of projectiles traveling at high speed. $$D\approx l_\text{bullet}\frac{\rho_\text{bullet}}{\rho_\text{wall}}$$ To a good approximation, the depth of penetration is constant. 1 Let me try and answer your question by applying it to the integration of: $$ads=vdv.$$ $$f(s)ds=vdv.$$ Multiplying both sides with$m$(mass of particle), we get: $$mf(s)ds=mvdv.$$ Here$mf(s)=ma$, so it's the force acting on the particle. So far this means that if we take the force$mf(s)$applied to effectuate an infinitesimal displacement$ds\$ and ...

0

Throwing a ball in the air and having it come back down is a basic concept of why there is non zero acceleration at a zero velocity. You throw a pencil up in the air and at the apex of the projectile, the velocity will always be zero but because of acceleration due to gravity (9.81m/s^2), the acceleration will never be zero.

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Here are some hints: 3c,d,e) The projectile will be at its highest point when the vertical component of its velocity is zero. (Otherwise, it would be going up towards or coming down from a higher point.) f,g) You should be able to see that f) is the same as e). Try to work it out with the formulas before plugging numbers in, if you can't. g) exhibits a ...

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