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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 ...


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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} ...


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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 ...


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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 ...


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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 $b=0$. ...


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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 ...


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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 ...


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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 ...


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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 ...


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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 ...


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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." ...


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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.


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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 ...


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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 ...


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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.



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