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1

Since drag is the resistance force of an object moving through a fluid (a fluid friction term), then you can make the physical argument for the value of $b$. This frictional force exists only at the boundary between the object and the fluid itself (a surface force), this means that the drag force must be independent of the mass of the object, thus $b=0$. ...


1

The way I like to phrase pretty much all of dimensional analysis is, "you can only take an arbitrary mathematical function of dimensionless parameters: mathematics doesn't directly deal in any other sorts of functions." When you see $[[R]] = \text m, ~~[[M]] = \text {kg},~~[[v]] = \text{m/s},~~[[\rho]] = \text{kg}/{\text m^3}$ your first question needs to ...


1

You'll find a one parameter family of solutions, because you have 4 independent quantities while in this problem you have the 3 independent units for mass, length and time. In your solution, you can see that the freedom to choose the parameter b comes from the fact that the density of air divided by the density if the object is dimensionless. To fix b ...


1

Well, it's not energy, its power $P$. $~~~~~~~~~~~~P = \int F \cdot dv$ And since power is the derivative of energy $P = \dot E$, your world makes sense again ;). Regarding your problem I agree with Yanping Cai, the kinetic energy of the car $E_{kin}$ must be converted into potential energy of the spring $E_{spring}$. $~~~~~~~~~~~~\frac{1}{2} m_{max} ...


2

This answer may not look like a typical answer, but I am attempting to instill a key concept, so please bear with me. For this type of problem, where you are investigating a possible solution, UNITS are EXTREMELY important. What are the units of momentum? What are the units of force? Note that if units do not match across an equal sign, the answer is ...


1

I'm sure you can consider the force and integrate it over displacement to calculate the total work has been done. But why don't you step back and consider the conservation of energy? In short, $$\dfrac{1}{2}mv^{2} =\dfrac{1}{2}kx^{2}$$ $k$ is the minimum spring constant it requires.


1

Solution based on wind energy and cost aspects: Typical design range 10-20 m/s: Wind Turbines are designed to produce maximum power under somewhat above normal mean wind speeds such that overall energy output per total cost of ownership is maximised. This typically results in optimum operating velocities in the 10-20 m/s range. Wind Turbines already ...


1

Problems that depict situations where the tensions are same on ropes on both sides of the pulley are ideal situations.It is stated so in order to minimize any complexities that may arise if the pulley was to rotate.Now, if the tensions were not equal on both sides, the pulley would experience a net non-zero torque and hence a net angular acceleration and ...


1

Because the pulley possesses mass, you need to apply a non-zero net torque to it to increase its angular acceleration (assuming that is the goal here). If the tensions were the same on both sides of the contact point between the string and the pulley, there would be no angular acceleration.


1

While I'm not willing to spend the money to get access the paper, one issue jumps out at a casual reading of the abstract - turbine design. Honeste_vivere's answer mentions the possibility of destroying a farm, and the abstract includes "The reduction in wind speed due to large arrays increases the probability of survival of even present turbine designs." ...


1

I think you need to be careful here. The total power contained within a hurricane ranges from $1 \times 10^{12}$ to $6 \times 10^{14}$ Watts or 1 to 600 TW. The world energy consumption in 2008 was 20,279 TWh. There are 8760 hours/year, thus our consumption rate was ~2.31 TW (1 TW = $10^{12}$ W). The point being, the energy you would need to dissipate ...


0

This text is excerpted from Richard Fitzpatrick's Newtonian Motion: "It should be noted that Newton's third law implies action at a distance. In other words, if the force that object $i$ exerts on object $j$ suddenly changes then Newton's third law demands that there must be an immediate change in the force that object $j$ exerts on object $i$. Moreover, ...


1

Typically, the Friction force will be proportional to the velocity of the object it affects (or at least it is usually assumed to be proportional). The Force you describe is constant and pointing in negative x-direction. The situation you are describing resembles an object on a Hookean spring in a gravity field. At first, F1 pulls it upward (positive x) far ...


1

Why is it that two carbon atoms fired at each other will bounce off and not stick together? It is because as the atoms move close together their orbital electrons begin to repel more than their nuclei attract each other's electrons. The result is greater potential energy as they approach and this leads to the tendency to move apart much like compressing a ...


0

Mathematically, the friction is described by a couple of Heaviside functions $\Theta$, asuming your body is of length $d$ and the transition from non-friction to friction occurs at $x_0$ with $x_1 = x_0 - d/2$ and $x_2 = x_0 + d/2$. $~~~~~~~~~F_\mu (x,v) = -\mu \cdot v \cdot \left[ \Theta(x-x_1) \cdot \Theta(x_2-x) \cdot \cfrac{x-x_1}{x_2-x_1} + ...


0

If the person is moving in a straight line and then takes a right turn then the answer simply is because of inertia. The person initially moving in a st line would continue to move in that direcn and hence when he takes a right turn it would appear that he turns left but actually he is simply trying to maintain his earlier motion.. If he is moving,in a ...


1

We often treat small objects as dimensionless points but that in that case we take the assumption that the object's mass and size does not have a meaningful impact on the behaviour of the system we're measuring. The force between a planet and the object is $F = G\frac{m_1 m_2 }{r^2}$ It's true that the object can be "torn apart", however in a real-world ...


0

This is a subtle problem. Assuming that the normal force that is involved in friction is strictly dependent on how much of the block is over that surface, it is seen that the friction force increases steadily as more and more of the block moves over that surface. This means that the amount of friction force and the acceleration are dependent on the ...


0

Assuming the body comes to rest before it completely enters the region, the body will trace out a partial sine curve in the space of $(t,x)$. Suppose the body is of length $L$ has uniform mass $M$. We let $x$ correspond to the positive horizontal distance from the interface. Then the force of kinetic friction (with coefficient $\mu_k$) is \begin{cases} ...


0

NO Acceleration does not change at all because it is proportional to force applied and the force does not change. The effect of force does increase (sometimes decrease) something and that is velocity (speed if we are talking one dimensional). Also velocity increases continuously. It is not like it changes after every one second. If acceleration is 1 ...


0

This is a mechanical advantage problem. For fixed pulleys, only the direction of motion is changed, and there is no mechanical advantage. A 1 N force directed downward on one side of the fixed pulleys (the small ones) produces a 1 N force directed upward on the rope on the other side of the fixed pulleys. For a movable pulley (the large one), there are ...


0

For a constant acceleration $a$ the speed is given by $v(t) = at + v_0$, where $v_0$ is the initial velocity, as you correctly used. But to find the distance you would have to integrate the last question to find that: $x(t) = \frac{1}{2}at^2 + v_0 t+ x_0$, where $x_0$ is the initial position. You can then check your results and you will find that they are ...


0

There are three main types of van der Waals forces: Keesom forces: Dipole to dipole attraction between oppositely charged ends of permanent dipole molecules. Hydrogen bonding is an especially strong form of Keesom force. It's responsible for water condensing into liquid and solid form at temperatures prevailing on our planet, and for loosely maintaining ...


1

Observe that $v\dot v=\frac12\frac{\text d}{\text dt}v^2 = \frac12\frac{\text d}{\text dt}\mathbf v\cdot\mathbf v = \mathbf v\cdot\dot{\mathbf v}$ and therefore the integral of the OP becomes (assuming a constant mass $m$, e.g. a point particle) $$\Delta\left(\frac12mv^2\right)=\int_a^b mv\ \text dv = \int_a^b \frac{\text d}{\text dt}(m\mathbf v)\cdot\mathbf ...


0

You are talking about two derivatives, one derivative to time and the other to velocity. If you would derive the kinetic energy two times to the velocity, you would end up with m, not F.


1

Both can happen simultaneously. This happens in binary star systems all the time. Work done on object A as it moves from position 1 to position 2 is calculated by $$\int_{\vec{r}_{1A}}^{\vec{r}_{2A}} \vec{F}_{total\,on\,A}\cdot\,d\vec{r}$$. That changes the kinetic energy of A. It could be negative which means the kinetic energy decreases. Some might say ...


0

I hope that you don't seriously intend to try this project in the real world. There are several issues that WILL lead to failure and/or serious bodily harm: 1) The water will expand as it is heated. If you start with a container that is too full of water, it will burst the sides of the container before you get to 300 C. 2) Pressure vessels lose some ...


0

Fundamentally you just need to know which direction energy flows. If energy flows from object (or system) A to object (system) B, object A is doing work on object B which is the same as saying object B is being worked on by object A.


1

Assuming that the angle theta is measured relative to the vertical (e.g., the position of the string when the pendulum is at rest), a careful free body analysis indicates that the acceleration of the pendulum is g * sin(theta). This means that the acceleration of the pendulum continuously varies as it swings. This is relevant because the kinematic ...


4

You usually cannot push your hand through the table, because it's a single solid. The atoms are held together by covalent bonds, which are electromagnetic in nature. Sand on the other hand is grainy - the $SiO_2$ grains do not interact with each other and are only held "in place" because of gravity. You can run your hand through sand similar to driving a ...


0

Both objects are responsible for the field, and the total gravitational (or electrostatic) force field is the linear superposition (sum) of the force fields arising from each object on its own. So, for example, if you have two point masses of mass $m_1$ and $m_2$ at positions $\vec{r}_1$ and $\vec{r}_2$, the field at point with position $\vec{r}$ will be: ...


-3

The following explanations are for electrons, but the are true in some kind for protons, positrons and antiprotons too. Every electron has the same electric field. It rotational symmetric. Additional every electron has a magnetic dipole moment (a tiny magnet) and an intrinsic spin. This spin is really a rotation of the electron around the axis of the ...


1

Based on the title of your question, there are only two objects. I'm going to infer that you are analyzing the motion behavior of one of those objects based on the influence of the other. I'm also going to assume that the influence is gravitational and not electrical. That's the context of your question. You will only show the gravitational force vector ...


1

Gravitational field vectors conventionally are represented pointing toward the gravitating body whose gravitational field is being analyzed. If two gravitating bodies are mutually attracted to each other, and one is in orbit around the other, the center of mass of their system is called the barycenter, and both orbit around the barycenter. This results in ...


-1

you are right, it is your formula a), as there is no thrust of the flowing water against the bucket, although the official NASA formula for the rocket chamber thrust implies that there is. Imagine that you instantly open the bucket floor, would the bucket jump to equalize he momentum of the flushed water? If I'm not mistaken, this problem is related to the ...


3

Actually, you have it backwards. The magnetic field isn't made of photons. Photons are made of magnetic (rather, electromagnetic) fields. To be specific, photons are ripples in the electromagnetic field. So, a magnet is surrounded by a magnetic field. If the magnet is not moving, then the field is stationary, and there are no photons. Wiggle the magnet, and ...


0

So indeed, the torque $\vec \tau = \vec r \times \vec F$ slows this thing down, and assuming that the force is never radial but only tangential, this is just $\tau = r ~ F$ where r is the radius of the circle and $F$ is the magnitude of the braking force. Newton's law $\frac{d\vec p}{dt} = \sum_i \vec F_i$ becomes for angular problems $\frac{d\vec L}{dt} = ...


-1

If a force f (Net force) acts on the mass(Spring Balance), it will accelerate with a = f/ m. Spring balance will only show the reading if there is net force causing deflection of spring (F= K*x). Here f = Net Force on Spring causing acceleration F= Force that causes deflection of spring(Which always acts in pair - equal in magnitude and opposite in ...


0

In terms of representing the answer: F would be for the reading because that is the basic notation for total force, and as there is no need to differentiate between the two forces Left or Right due to them being equal, then only F should be used. The actual answer as you probably will know is F = 0. When L is not equal to R, the reading should be along the ...


1

Other people have answered your questions about the signs; I want to address your comment, "why does this sound like the normal force?" By Newton's third law, pairs of forces always come in equal and opposite pairs. If the ground exerts an upward normal force on you, you exert a downward normal force on the ground. If sliding on the ground exerts a ...


1

What everyone said, but also: the way the question is phrased, the model it uses, is a 1-dimensional model. Imagine a point travelling along the real number line/the x-axis on a cartesian plane. The model in the problem assumes there is no vertical movement, or 'side to side', but only 'pure' 1-D movement along a straight horizontal line. There is only that, ...


0

The wheels use regenerative braking drive a propeller/turbine to provide thrust not the other way around. In a typical sailboat or other wind powered craft the wind provides thrust and the resistance given by the conveying mechanism (water or wheels) is typically minimized. In this scenario however, the wheels have a little bit of regenerative braking ...


2

No, Newton's Second Law of Motion does not require the acceleration to be constant. However, for any given acceleration (and mass) there can only be one value that the force can yield, namely $m \cdot a$. In other words, if the force and mass stays the same, then the acceleration stays the same as well. Contrary, if the acceleration and mass stays the same, ...


1

It's all a matter of how you've chosen your coordinate system. Think of how the velocities were calculated. They essentially looked at two points in time and the two corresponding points in space. If your positive direction is to the right, and the guy has advanced to the right, then its a positive number for velocity. Force carries the same logic here. The ...


0

The force/ acceleration of the skater comes out as negative because: the force or acceleration is in the opposite direction of motion. You can also think of it like this, frictional force always tries to oppose relative motion. In this case the skater is moving forward w.r.t the ice and the ice is moving backwards w.r.t the skater; so the ice exerts a ...


14

It's very common to get mixed up about signs. The only recommendation I can give is to establish a clear sign convention and stick carefully to it. To show what I mean let's consider your skater: I'm going to use the convention that positive is to the right and negative is to the left. remember that quantities like velocity and acceleration are vectors, ...


2

When the skater is on the ice, friction stops him/her in 3.52 seconds as you said. The molecules in the skates rub against the molecules in the ice, and the ice molecules absorb some of the skate molecules's energy, slowing the skater down. The reason the force is negative is because the friction is acting in the opposite direction of the skater's motion. ...


0

So within a cord/string there is a property called tension which is a measure of the force exerted along the string. If the string stretches homogeneously (the same at all parts of the string) then it turns out that this tension is the same at all parts of the string: you pull with force $m$ Newtons, then everywhere you see the string you need to think of ...


5

Energy kinematics I have this question, because typically problems that can be solve using conservation of energy or just energy-related principles, can usually be solved sing kinematic equations. Yep. In fact, there are two profound pieces of math, Hamiltonian and Lagrangian dynamics, which say that you can use energies to derive the actual kinematic ...


2

Work is transfer of energy from one system to another OR transformation of energy from one form to another. Either way, work does not create energy. When I lift an object, I am transferring energy from my body/muscles to the object-earth system. The energy goes into potential energy of the object-earth system because the separation between the object and ...



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