New answers tagged

1

As the force $F$ is constant, and we know that the force of weight $mg$ is constant too; we can say the net force acting on the ball is constant whole the motion. But, from $t=0$ to $t=t_1$ the net force is $F-mg$ and after $t=t_1$ the net force is $-mg$. So, velocity graph will be linear but it has a breaking at $t=t_1$. For $0\le t\lt t_1$ the gradient of ...


0

You have a few questions here. They mostly concern the energy loss due to friction. In this problem there is no energy loss because of friction. Friction is present between the rope and the pulley. This is static friction. It makes the pulley rotate. Energy is lost due to friction (dissipated as heat) only when there is relative motion between the ...


3

The equations of motion for the position determine the accelerations: they are second-order differential equations in time: $$\vec F = m\vec a = m\ddot{\vec x }$$ So the acceleration, the second derivative of the location in time, has to be determined from the state of the physical system in some way. Typically, it is determined using the $F=ma$ formula ...


0

Imagine a body moving with velocity $\vec{v}$ in the $\hat{x}$ direction and now imagine having it losing speed (decelerating, so $\vec{a} \propto -\vec{v}$) the friction force should be opposite the $\vec{a}$ (acceleration) or the $\vec{v}$ (velocity)? if you think carefully about it, you'll convince yourself that it should oppose velocity and not ...


-1

You can assume any arbitrary distance,say x ,and then the total distance will be twice of x..time taken in going will be x/60 and time in coming back will be x/50 so that the total time will be their sum..that would give your average speed that is 2*x divided by sum of x/60 and x/50 ...that nicely gives you the final answer and the x neatly cancels out.


-1

Assume the first trip takes a time of 1. Then the return trip takes 1.2 (see why?) and the total time is thus 2.2. From that you can calculate the average.


-2

You should use the average speed formula: $\bar v=\large{\frac{\textrm{distance}}{\textrm{time}}}$ For going: $t_1=\large{\frac d{v_1}}$ For coming back: $t_2=\large{\frac d{v_2}}$ Then, $\bar v=\large{\frac{2d}{t_1+t_2}}$


2

It is like water in a hose. If the hose is full of water, water flows out the end immediately when you turn on the faucet. A drop of water at the faucet pushes a drop next to it, which pushes the next drop. Water doesn't flow that fast. If the hose is empty, it takes a while to reach the end.


2

A bit of 1, a bit of 3... The technical name is flow velocity, as correctly stated in the Wikipedia article about NS equations. But one could ask what "flow velocity" means. From the Wikipedia article: flow velocity [...] is a vector field which is used to mathematically describe the motion of a continuum. Although correct, this definition is ...


2

You are correct, it is the velocity of a small volume of fluid centered at the point, that is a macroscopic motion, but it is also the result of the average velocity of the particles in that volume.


1

This is a simple variation on the so called "Twin Paradox", which is not a paradox in the logical sense (i.e. not a logical contradiction). Each cycle of the oscillator's motion is like the journey of the spacefaring twin. One possible cycle on a spacetime diagram is drawn below (source: Wikipedia "Twin Paradox" Article with my own additions). The ...


0

Lorentz transformations are transformations between inertial frames of reference, i.e. frames of reference that move with constant velocity relative to each other. An oscillation frame as you described it is not an inertial frame of reference, so Lorentz transformations do not apply here.


1

First of all, recall that one may vary the velocity $v$ independently of the position $q$ in the Lagrangian $L(q,v,t)$. In fact, the (Lagrangian) canonical momentum is defined as $$\tag{A} p(q,v,t)~:=~\frac{\partial L(q,v,t)}{\partial v}. $$ This is explained further in e.g. this, this, and this Phys.SE posts. Let us define for later convenience $$\tag{B} F(...


2

According to definition of 4-force: $$F^{\mu}=m_0 \frac{d V^{\mu}}{d \tau}$$ $$\to F_{\mu}V^{\mu}=\eta_{\nu \mu}F^{\nu}V^{\mu}=m_0 \eta_{\nu \mu} \frac{d V^{\nu}}{d \tau} V^{\mu}=m_0\frac{d}{d\tau}(V^{\mu}V_{\mu})-m_0 \eta_{\nu \mu} \frac{d V^{\mu}}{d \tau} V^{\nu} $$ But we know that: $$V^{\mu}V_{\mu}=c^2$$ and $\eta_{\nu \mu}$ is symmetric so: $$m_0 \eta_{\...


2

Recall that the four force $F$ on a particle of constant invariant mass $m$ is $$F = m\, \mathrm{d}_\tau V\tag{1}$$ where $V$ is the four-velocity. Moreover, recall that the Minkowski norm of a four-velocity is unity (or $c$, more generally, in nonnatural units). So therefore we have $$\mathrm{d}_\tau (V_\mu\,V^\mu) = 0 \tag{2}$$ as a universal identity....


1

The position vector of a particle in polar coordinates is given by $$\mathbf r ~=~ r\mathbf e_\mathrm r$$ Velocity $\bf v$ is \begin{align}\mathbf v&=\frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\mathbf e_\mathrm r}{\mathrm dt}\\ &= \frac{\mathrm dr}{\mathrm dt}~\mathbf e_\mathrm r+ r\frac{\mathrm d\theta}{\mathrm dt}~\mathbf ...


-1

I would start by measuring the pebble and the weighing it to dtermine it's size. Then I would ask "am I simply dropping it" like Galileo or am I trying to make an inference based upon throwing it then estimating the effects on landing. If it's the former then you should just get a number as no matter the rock size they all fall at the same rate.


0

So I think I've got an answer, but I'm not completely convinced it's the right one. A friend (thanks Ricky!) pointed me to several places online where people had solved this problem: Falling Body with Air Resistance Velocity-time graphs for falling objects Calculating time to reach certain velocity with drag force Each graph of $v(t)$ vs. $t$ looks ...


1

This is a simple question of conceptual understanding. The only force acting on the ball after it is released from the hand is that of the ball's weight due to gravity. Since it is the only force, the consequent acceleration is also the only acceleration. Gravity is being taken to be $10 m/s^2$ in a downward direction. For the answer to be correct, our ...


0

This is a trivial kinematic deduction. \begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align} Integrating this, we would get $$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6}...


3

Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $ Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it) Where you are wrong is here: According to your question v is the final velocity since $(v=v_{0}+gt)$ So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ ...


1

Let's take the first equation of motion which is : \begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the ...


0

I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong: $ \Delta x = v_o t + gt $. A unit analysis will show you why: $$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds$$ $$ meters \not= meters + \frac{meters}{seconds}$$ And we can see that the assumption is simply not true ...


0

When the ball is travelling in your hand before you let go, it is being accelerated by the amount of force you exert on it, this increases its velocity before you let go. (The initial velocity) When you let go of the ball, it immediately begins to slow down due to the acceleration of gravity acting downwards. There is no upwards acceleration, as this ...


2

While I don't know the terminal velocity of a cat (which may or may not be 100 kph). I know a lot of people say 100 kph, but I don't know if that is verified or just a calculated estimation someone came up with. I have experimented quite a lot with falling objects, including small animals and bugs. I have documented the falling speeds of many small ...


0

you are correct in starting with $mgh=1/2mv^2$. You just need a value for h, the height the ball falls. In this case $h=2tan 10^{\circ}$. Substitute that value of h in your equation and find v.



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