New answers tagged

1

Suppose given an equation of the form $$y = Cx^n.$$ If you plot $y$ vs $x$ then it would be hard to find the value of $C$ or $n$ by looking at it. But if we take log on both sides we get $$ log(y) = n ~log(x) + log(C).$$ Now plotting $log(y)$ vs $log(x)$ would give us a straight line with slope $n$ and intercept $log(C)$. In your question the relation is ...


2

You apparently expect the plot of force vs acceleration to be linear, but not everything you will plot (e.g. position versus time) will be linear. Another example: plot the mean orbital radius of the planets vs. the period of revolution the planets. It's not linear. But if you plot the log of the orbital radius versus the log of the period it will be ...


2

Solve for $F_b$ from the horizontal braking distance. Assume $F_b$ is constant, then during braking kinetic energy has been converted to friction work: $$F_b \Delta x = \frac12 mv^2$$ where $\Delta x=123\:\mathrm{ft}$ is the braking distance and $v=60.0\:\mathrm{miles/hour}$. I've not checked the rest of your work. You don't need to invoke friction ...


0

When a torque is applied to an object it begins to rotate with an acceleration inversely proportional to its moment of inertia. This relation can be thought of as Newton's Second Law for rotation. The moment of inertia is the rotational mass and the torque is rotational force. Angular motion obeys Newton's First Law. If no outside forces act on an object, ...


2

If we assume all other things being equal other than the downward force due to gravity, the vehicle on Earth would be capable of greater acceleration. The ability of the tires to grip the surface on which they are resting depends on the downward force keeping them in contact (Coefficient of friction). That will pretty much relate the gravitational ...


3

It is independent of the mass of the planet if you assume the bearings are frictionless. Also assume that the tyres do not make dents in the ground. However, in reality, the bearings have friction. Additionally, there is rolling resistance as the tyre makes small deformations in the ground as it is rolling. This is why you see a characteristic "W" shape in ...


1

No. If you want an answer beyond a simple yes or no have a look at my answer to How does "curved space" explain gravitational attraction?. The acceleration is caused by the curvature of spacetime. This curvature does also cause time to run more slowly as you approach a spherical mass, but the change in the rate taht time runs does not cause the ...


0

some context is missing here but it could be the difference between Eulerian and Lagragian quantities, i.e. the spatial derivatives vs the particles derivative. It is mostly used for continuous materials (e.g. fluids), but can extend to other cases (e.g. field or stream of objects). The spatial (i.e. Eulerian) velocity in a field is the one at a given ...


1

Newton's 2nd Law says that F=ma. This law says NOTHING about the physical properties of the object that you are accelerating. Thus, the answer is "yes", the two objects will accelerate at the same rate, so if they start at the same velocity, they will continue having matching velocities as long as they experience the same acceleration.


1

Let's imagine you walking down a tall mountain when the sun is setting. Let's say it is cold at the top of the mountain, so walking down the mountain tends to make you warmer. But also the sun is setting so you might get colder. How do you take both effects into account? Well the change in temperature you feel (LHS) is the change in temperature at the ...


2

This is how you do the calculation. The elapsed time on an observer's clock is called the proper time, $\tau$, and it is calculated by integrating the metric: $$ c^2d\tau^2 = \left(1-\frac{2GM}{c^2r}\right)c^2dt^2 - \frac{dr^2}{1-\frac{2GM}{c^2r}} - r^2d\theta^2 - r^2\sin^2\theta d\phi^2 $$ In this case we'll assume all motion is radial so $d\theta = ...


3

The equation for your curve is given by: $$ \frac{dv}{dt} = \frac{F(v)}{m} $$ where $F(v)$ is the net force on the car, which is a function of the velocity. we solve the equation by integrating to get: $$ \int \frac{dv}{F(v)} = \frac{t}{m} $$ The trouble is that the net force $F(v)$ is a complicated function that doesn't generally have a simple analytic ...


1

The Equivalence Principle of General Relativity holds that acceleration and gravity can be described identically. With an accelerometer, you can tell whether or not you are accelerating in empty space, regardless of whether another object is available to act as a reference point. Under acceleration, your weight will change just as though you were ...


0

Inertia is the term we use to say that an object can't change its velocity without a force acting on it. This does mean that a force always changes the velocity of an object (if net force is not zero). The wall scenario is a bit different because the wall will feel the force and will try to change its motion accordingly but in this case, inertia has less to ...


1

If you take your final expression $$ x(t) = \underbrace{\left(x_0 + \frac{b - a}{2}\tau^2\right)}_{x_0^*} + \underbrace{\left(v_0 + \tau(a - b)\right)}_{v_0^*} t + \frac{b}{2} t^2, \quad \text{with}\ t>\tau, $$ then $x_0^*$ and $v_0^*$ would be the position and velocity at $t=0$, however this is only meaningful if $\tau<0$ (however in that case $x_0$ ...


1

Firstly it's worth noting that such a discontinuity can never be 100 % real. To go from acceleration $a$ to $b$ instantaneously ($\Delta t = 0$) would require an instantaneous change in the net force responsible for the accelerations and that isn't possible in the material world. Secondly, I think you are over-thinking your problem. Just write the ...


1

This equation holds whenever there is constant acceleration. Here are 2 ways of deriving that equation, which I hope help you understand it. Energy conservation The change in kinetic energy must be equal to the work done on the particle. $$ \frac{1}{2}m v_A^2 - \frac{1}{2}mv_B^2 = \int F\cdot dx $$ For a constant force and mass $\int F\cdot dx = F (x_A - ...


2

1) why $a_c$ has to have opposite sign. $a_c$ is the centrifugal acceleration in the rotating frame experienced by the tennis ball. $$\vec{a_c} = (\vec{\omega} \times \vec{r}) \times \vec{\omega}$$ You’re answer for $\vec{a_c}$ gives the correct sign and magnitude. If you draw the vectors for $\vec{\omega}$ and $\vec{r}$ and apply the right hand rule ...


0

Think of a position axis $x$ starting at zero, positive to the right and negative to the left. It is like a number line. A positive velocity means that the value of $x$ is increasing eg going from $x=+3$ to $x=+5$ or $x=-7$ to $x=-4$ or $x = -3$ to $x=+4$. A negative velocity means that the value of $x$ is decreasing eg going from $x=-3$ to $x=-5$ or $x=+7$ ...


0

well you have given answer in your own question! Velocity and acceleration are both vector quantities, meaning they have magnitude and 'direction'. The sign (+/-) will depend on the direction. To simplify, let me give you an easy example.. Case 1: An object is moving down from the top of a mountain. The acceleration (in this case 'g') will act in the ...


0

The punch can be landed on the table and the ball is pushed through the liquid column and naturally the height the ball rises the force can be calibrated, Now you want to build the machine that can be put to a wall. so you can make adjustments to the design or even get a different idea from this. You drop a known weight at a measured height and ...


1

The acceleration at the point of reflection is actually quite complicated. It is caused by the elastic forces of the surface and the ball and has a complicated time dependence. However, the timespan in which the ball touches the ground is very short (especially if the ground and the ball are very rigid), therefore we can simplify the actual acceleration ...


0

In addition to the other answers, you can forget about the board altogether and simply use a punching bag. You can measure the angle that is displaced by your hand on the punching bag using some kind of video tracking device or a gyroscope attached to the punching bag. Thus you will be able to detect angle displacement, potential energy and/or torque which ...


0

(1) Measure out and draw a fine grid on the surface of an air hockey table. (2) Set up a camera of known frame rate above the table facing down. (3) Place a large, flat bottomed, relatively light object of known mass on the table, make sure it hovers freely. (4) Punch the object horizontally, then analize footage to find impulse. (Note) If you want to ...


4

So for starters you need to make sure the center of gravity is in the middle of the tray (so in you scenario of lighter/heavier cups, balance it out). This is assuming that you make as many right turns as left turns. If this is not the case, say 70% of your turns are right turns, then you would want to slightly offset the orientation so that it's not quite ...


1

If you have no means to change the inclination of the tray while you are driving, then the answer is: flat. You'll make as many left turns as right turns, so you can't favor left tilt vs right tilt. And you will speed up as well as slow down, so you can't favor forward tilt vs backward tilt. The distribution of cups ... which cup goes where ... ...


2

It depends on various things: the energy $E$ of each neutrino, how likely each of them is to interact with a particle in your body (expressed as a cross-section $\sigma$, which depends on $E$), and how large an acceleration $a$ you consider "significant". Further let $m$ be your mass, $75\;\rm kg$. I've pilfered some rough numbers from here, namely ...


0

The formula for acceleration at a given height above the Earth would be: \begin{equation} a = -gR^2/r^2 \end{equation} where R is the radius of the earth and r is the radius you are at. This is because gravitation varies with the square of the distance between two objects. I'm afraid you no need calculus to be able to derrive escape velocity due to the fact ...


3

You don't need acceleration. Just calculate change in potential energy with altitude and make sure your initial Kinetic energy is sufficient that your final velocity is greater than zero. You cannot use your formula without integration. As for actual formula it would be $$g= GM/r^2$$ where M is mass of earth and r is varying distance.


0

Are talking about Newtonian (Classical) mechanics - Physics I and II? if so, in that case all you need are Conservation of Energy and Conservation of Momentum. You ignore external forces.


4

You do not need the kinetic energy. Working with the total energy $\gamma m c^2$ produces the same result. Assuming both the total initial energy $\bar E_0 = \gamma_0 m c^2$ and the additional energy $E_i$ are known, write $\gamma_1 mc^2 = \frac{mc^2}{\sqrt{1-\beta_1^2}} = \bar E_0 +E_i$ for $\beta_1 = \frac{v_1}{c}$, then $$ \sqrt{1-\beta_1^2} = ...


0

In short, the answer to your question is no. Time dilation can't ever run backwards. Consider a simpler case: Let there be a person A on a massive planet, and a person B floating in space far away. If both had huge clocks with them, then B would see A's clock running slower than his own. And although it seems contradictory, A will also see B's clock ...


0

Well the body will accelerate as long as you are applying a force on it. So if u apply a force for forever it will accelerate forever but there is an interesting thing that happens. Force applied for ever You would notice that if u apply a force for an infinite amount of time then by F=ma there should be acceleration for an infinite amount of time leading ...


2

The average velocity is given by $$ \bar v=\frac{1}{T}\int_0^T v(t)\mathrm dt=\frac{1}{T}(v_1t_1+v_2t_2) $$ where $t_1$ is the time spend on the first interval, $t_2$ is the time spend on the second one, and $T=t_1+t_2$. Using $$ v_1t_1=v_2t_2=d $$ you get $$ \bar v=2\frac{v_1v_2}{v_1+v_2} $$ I believe you can take it from here.


0

The point where the log touches the ground is the point where the log has zero velocity. If you now draw a small displacement, you see (from equal triangles) that if the center of the log moves $x$ the top moves $2x$ .



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