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1

Balance torques around the corner of the step, so r x cos(38.7) x mg= F x 0.3. F = 114 N


0

THE MOMENTUM IS RELATED TO THE ENERGY AND POSITION OF ELECTRON IN THE SPACE , AND THE ENERGY IS RELATED TO THE WORK AS A FUNCTION STATE .thanks


1

The ping pong ball would lose a tiny amount of kinetic energy to the truck. The truck ends up with a momentum of just under twice what the ping pong ball had. However, energy is 1/2 m*v^2 = 1/2(m*v)^2/m. Since the truck is much more massive than the ping pong ball, it carries much less energy for a given momentum. The end result is that the small amount of ...


3

If $s_0$ and $u$ are zero, then the equaion for $s$ simplifies to: $$ s = \tfrac{1}{2}at^2 $$ so: $$ \frac{ms}{t^2} = \tfrac{1}{2}ma $$ and since $F=ma$ this becomes: $$ \frac{ms}{t^2} = \tfrac{1}{2}F $$ So, apart from the factor of $\tfrac{1}{2}$, the equation your teacher is using works for an object accelerating from rest. That factor of a half is a ...


-1

Assuming a small angle, the angular frequency of the oscillator can be found using this equation: $$ \omega = \sqrt{\frac{mgd_{cm}}{I}},$$ where $m$ is that mass of the oscillator, $g$ is the gravitational acceleration, $d_{cm}$ is the distance of the center of mass from the pivot of the oscillator, and $I$ is the moment of inertia about the pivot of the ...


0

Newton's third law states that if body 1 exerts a force on body 2 ($\vec{F}_{1,2}$), body 2 necessarily exerts a force $\vec{F}_{2,1}$ on body 1 that is opposite to $\vec{F}_{1,2}$. If body 1 doesn't move (or only in a direction perpendicular to $\vec{F}_{1,2}$), no work is done and no energy is transferred. Now, let's say body one does move in the same ...


0

There is actually not an energy exchange happening at all, but energy transformation. A common example may be 2 gravitationally massive bodies attracting. What you see is both of the bodies accelerating toward one another, or the transformation of gravitational potential energy into kinetic energy. The gravitational potential energy is a characteristic of ...


0

f=m*a force is mass times acceleration If one body is still (relative to the other body) then the force it enacts on the body that hits it is dependent on it's own mass. Meaning it is braking the body that hit's it or slowing it down. Or one body is speeding the other up thereby transferring it's acelleration. So the force opposite to the acting force is ...


0

The problem is that the equation should be $9m\cos 20= mg \sin 20+\mu mg$ since the friction force is going along the slope.


1

Assume this is on Earth, so you know $g$. Don't try to use any values for $M$ or $k$. Just use symbols. Draw a free-body diagram for the system at rest. From this you can get the relationship between $M$ and $k$ and $g$. You probably also have an expression telling you the relationship between the $T$, $M$, and $k$. Do the algebra and everything should work ...


1

Just a small conceptual hint will do No problem. A hint: Set up Newton's law, $\sum F=ma$. You will see that the sum of all the three forces must equal... yes, what should it equal? I'm confused with the condition at which the block will start sliding What is the difference in the equation mentioned above for a point just after it started moving, ...


1

However it makes sense that gravity can't travel faster than light because of the force-carrying photons Whilst it makes sense that gravity can't travel faster than light, we don't actually know this for sure. What we do however know is that the force of gravity is not conveyed by photons. Even electromagnetic force is not conveyed by photons - hydrogen ...


0

When you have a block with a mass of 10 kg, the normal force is approximately 98 N. If the block starts out stationary, it will continue to be stationary until you apply sufficient force to exceed the static friction. At that point it will start to slide, and will continue sliding until the force drops below the force of dynamic friction. For the ...


0

What about the car when there is no force applied to the block? The block will experience no frictional force, so when dealing with static friction, we have to say $f_s \geq \mu_s n$, since there are instances when the frictional force may not act at all or may at at a reduced capacity. So here, $f_s = 18N$, since that is the only force necessary to keep the ...


1

It is still accelerating, albeit with a lesser magnitude. The derivative of acceleration is called Jerk (See Physical intuition for higher order derivatives). So at point B it is accelerating with negative jerk.


1

This type of question keeps reccuring. I suppose either teachers don't teach this correctly, or students do not pay attention in class. $F=m a$ is really $$\sum F = m a_{cm}$$ These two distinctions (the sum, and the acceleration of the center of mass) make all the difference in the world (At least read on Newton's Laws of motion). You push on the rock and ...


0

Okay, I got the solution myself. The integral was causing problems, see my question on Math SE for the answer. The solution is $ \Delta \varphi = \frac{\pi R w_0}{v} \sqrt{\frac{2M}{2M+5m}}$


0

Because it is rotating with constant w0 you should not use conservation of angular momentum and for the same reason: The mass $m$ and $M$ are irrelevant to the problem. May be the author intended to say differently: INITIALLY a sphere of mass M is ... with constant w0 the problem is much simpler. Initially ... Yours response appears to indicate ...


2

For a particle in a gravitational field treated as a constant? Surely Newton's equations of motion in the fixed rectangular frame: $$\ddot{x}=0$$ $$\ddot{y}=-g$$ are as simple as it can get!


0

If the range is not too large, you can approximate the acceleration (due to gravity) vector to have the same direction over the trajectory of the projectile. It will probably be a good idea to use earth fixed 2D cartesian coordinates in this case, one axis horizontal and the other vertical to the Earth's surface. The acceleration acts only along the vertical ...


0

I want to give an answer for the case, that the mentioned particles are electrons. Let us consider that the magnetic dipole moments of this two electrons are aligned in a straight line through the points (0,a,0) and (a,a,0). Since both electrons are moving their magnetic dipole moments begin to turn when the electrons leave the mentioned points. Perhaps it ...


0

If your direction of motion is linear it does not necessary mean that the equation representing its position with respect to time is always linear.And if the velocity is constant it always means that acceleration is zero.


1

The main thing is that the total force you are applying on the body is not enough to move the rock.This means that the total force on the box is zero because the force of static friction is grater than that of your applied force,which cancels out the effect your force.As your forces increases the static friction also increases until a point comes when you ...


1

Force = mass * acceleration is the basic simplified version of the equation. There are more complected formulas available for this that take into account more complicated scenarios; like taking in the account of different forces coming from different directions like in your scenario. sum(forces) = mass * acceleration force = the derivative of its linear ...


14

The confusion comes from how you have written the equation. If you write it like this $$F_{net} = ma$$ it will be easier to see your error. You are exerting a force on the boulder, but net force is zero. This means that other forces such as friction are canceling your force out. In this case. Friction equals applied force.


1

With a strong grasp of Lie Algebra and Calculus of variations, "Invariante Variationsprobleme" should provide all the foundation one needs to build Newtonian Mechanics (and so much more). The deeper reason that we use either of these formalism is that they agree with experiment; that either formalism predicts the other is far less valuable than that they ...


4

No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be: \begin{align*} F_r &= m \ddot{r} - m r \dot{\phi}^2 \\ F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \end{align*} From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} ...


0

In a rotating frame of reference you usually consider the centrifugal force, one of the three possible fictitious forces. This is because you usually consider it as rotating at constant angular speed. There are other forces however: the Coriolis force that appear when the object moves in the rotating frame of reference. The Euler force when the angular ...


1

The rule is simple: regardless of which forces are acting, if the motion is accelerated then there is a net force, otherwise, there not net force. The only kind of non-accelerated motion is motion in a straight line at uniform speed. In particular you options: a)wrong, at it is accelerated in the curved part. We do not know in the straight part. b)wrong, ...


0

David Z, excellent answer - You comment that "... it does require that space defines some sort of absolute rotational reference frame". I believe that the existence of the reference frame is provable - In a nutshell - The proof is based on the relationship between angular velocity and the centripetal force it produces. It assumes that the same angular ...


1

Your question is very confusing, so I will first attempt to answer the spirit of your question with a cleaner scenario. It is possible to "move" from one place to another if there is minimal friction. You can do so with yourself, a large box, and a bag full of baseballs. But it isn't as cool as it sounds (that is why I had to put quotes around move). If ...


3

Your profile lists your age as 17, so I assume you're still at school. At this stage in your physics education you'll only have been exposed to differential equations that have relatively straightforward solutions. I assume the education system does this to avoid putting you off. If you continue studying physics you'll quickly learn that the vast majority of ...


1

You're correct that (B) is incorrect, but for the wrong reason. Kinetic energy doesn't have to be conserved since, in any closed system with only conservative forces, mechanical energy is conserved. There are many physical situations where kinetic energy isn't conserved (inelastic collisions for example), hence we do not know kinetic energy is conserved. ...


0

You can calculate the maximum height above the launch point with $$y_c = \frac{v_y^2}{2 g}$$ and draw a horizontal line at this height. Now you must know the initial direction of travel you can fit a parabola to this slope while being tangent to the height line at $$x_c=\frac{v_x v_y}{g}$$. The general shape of the curve is $$y = y_c - K (x-x_c)^2 $$ By ...


0

As a side note skidding is not a yes or no state with tires. See this answer for more details. The sum of the normal force and friction force that act on the car is the reaction force to the sum of the weight and centrifugal force of the car on the road. We can equate them in the coordinate frame parallel and perpendicular to the road. $$N=m\,g\, ...


0

No. W=o for internal forces is valid for all bodies irrespective of their rigidity. Because a non rigid body can b deformed only due to external forces. Not due to internal forces. Because internal forces acts along the same line. With equal and opposite direction causes net force to b 0


2

Answering your three questions: He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 ┬Ám is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size ...


0

With this answer I'll try to note the mistake committed in the original formulation "Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?" Well the problem is originated when you assert: "If one instead uses $F=\dot p$ so that $$f = ...


0

Can Lagrangian Mechanics be justified without referring back to Newtonian Mechanics? Sure; one can deduce Newtons Laws from it. The question is should one? By deducing Newtons Laws one is missing the crucial aspect of induction; the reverse procedure and in a sense more difficult; that is the discovery and invention of a theory that covers a wider ...


1

The answer to this question is very much analogous to the answer to how aeroplanes fly. See Physics SE Question "What Really Allows Airplanes to Fly?" and the best (IMO) answer is this one here. But basically the airfoils, sails or vanes - whatever they may be called - deflect the flow of air. They do this by pushing on the air and changing the latter's ...


8

Since the Lagrangian results in exactly the same equations of motion as Newton's laws, I'd say that based on their agreement with experiment both are on equal footing. Of course, to get the right equations of motion from Lagrange's equation you have to pick the right Lagrangian, so then you ask how we systematically pick the right Lagrangian. The recipe in ...


0

The parachute can not be opened instantaneously. The parachute is fully opened at the "vinicity" of the point $C$. The acceleration due to gravity is constant but there is another force one needs to take into account : the air resistance.


2

In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...


1

You are correct that centripetal force does not affect speed. So let us consider tangential speed, which is the speed the turning car would have if centripetal force were removed. Linear speed = tangential speed = distance / time = (2 * pi * radius) / time. (To simplify, assume the car goes through one full circle at uniform speed.) The linear speed of the ...


0

I would say that indeed the center of the car still move at 50km/h. The angular velocity is $\omega = v/r$, with v the tangantial velocity and r the radius. Therefore the wheels on the inside will go slower (smaller v) and the wheel on the outside will go faster (larger v) to keep the same angular velocity $\omega$. Best, Samuel


1

To answer your question directly, when the direction of the velocity of the front of the car does not match the front wheels' direction, the tire is deformed, acting like a spring which exerts a lateral force on the car. The deviation between these directions is called the slip angle, and to a first order approximation the lateral force on the tire is ...


1

There is an implicit assumption in this problem that the chain is perfectly flexible. In this limit, the part of the chain that just hit the table cannot transmit any force back up the chain. It just bends out of the way when it stops. If you consider the falling part of the chain as your "system", its surroundings include the Earth (acting via a distance ...


2

You've inserted the scale between yourself and the planet Earth. The scale's spring is compressed by the force exerted on you by Earth's gravitation. The compressed spring tries to uncompress itself and pushes back on you with linear restoring force. Linear restoring force is exactly equal to the force exerted on you by the Earth's gravitation. By ...


1

There is an omission in input data: glider's "fuel" is a sum of its potential and kinetic energies. So heavier glider (towed to same height and velocity) requires more work from the towing plane and starts with more energy than lighter glider.



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