🔑:The argument that Newton's third law is absurd because equal and opposite forces should cancel each other out, resulting in no acceleration, is a common misconception. However, this argument is incorrect because it fails to consider the fact that the forces act on different objects.Key point: Forces act on different objects, not on the same object.When two objects interact, they exert forces on each other. According to Newton's third law, these forces are equal in magnitude and opposite in direction. However, these forces do not act on the same object, so they do not cancel each other out.Example: A person pushing a wallImagine a person pushing a wall with a force of 100 N. The wall exerts an equal and opposite force of 100 N on the person, according to Newton's third law. Now, the person is exerting a force on the wall, and the wall is exerting a force on the person. These forces do not cancel each other out because they act on different objects: the person and the wall.As a result, the person will experience a force of 100 N in one direction (from the wall), and the wall will experience a force of 100 N in the opposite direction (from the person). Since the person has a smaller mass than the wall, the person will accelerate backward, while the wall will not move significantly (assuming it is fixed to the ground).Example: A car acceleratingConsider a car accelerating forward. The car's engine exerts a force on the ground, and the ground exerts an equal and opposite force on the car, according to Newton's third law. Again, these forces act on different objects: the car and the ground.The force exerted by the ground on the car is what propels the car forward, causing it to accelerate. The force exerted by the car on the ground is what causes the ground to exert a force on the car. The two forces do not cancel each other out, and the car accelerates forward as a result.Physical principle: Net force and accelerationThe key to understanding Newton's third law is to recognize that the net force acting on an object determines its acceleration, not the individual forces acting on it. According to Newton's second law, the net force acting on an object is equal to its mass times its acceleration (F = ma).In the examples above, the net force acting on the person and the car is not zero, even though the forces acting on them are equal and opposite. The net force is what determines the acceleration of each object, and the equal and opposite forces do not cancel each other out because they act on different objects.In conclusion, the argument that Newton's third law is absurd because equal and opposite forces should cancel each other out is incorrect. The forces act on different objects, not on the same object, and the net force acting on each object determines its acceleration. By considering the physical principles of net force and acceleration, we can see that Newton's third law is a fundamental and correct description of the natural world.

🔑:## Step 1: Understanding the ProblemTo calculate the mean pressure on a wall due to molecular collisions, we must consider the forces exerted by individual molecules over very small time intervals. However, because the forces and time intervals are variable and extremely small, a direct calculation of pressure from these forces and time intervals is impractical.## Step 2: Momentum Change and Its Relation to Force and TimeThe force exerted by a molecule on the wall during a collision, multiplied by the duration of the collision, equals the change in momentum of the molecule. This relationship is based on the impulse-momentum theorem, which states that the impulse (force applied over a time interval) is equal to the change in momentum.## Step 3: Importance of Momentum TransferThe product of the force and the duration of the time interval (which equals the momentum transferred) is crucial because it represents the total effect of the collision on the molecule's momentum. This is more relevant than the force or time interval alone because it accounts for both the magnitude of the force and the duration over which it is applied.## Step 4: Calculating Mean PressureThe mean pressure on the wall can be calculated by considering the total momentum change of all molecules colliding with the wall over a given time period, divided by the area of the wall and the time period. This approach averages out the varying forces and time intervals of individual collisions, providing a macroscopic measure of the pressure exerted by the molecules.## Step 5: Relation to Kinetic Molecular TheoryThe kinetic molecular theory of gases provides a framework for understanding how the momentum changes of individual molecules contribute to the overall pressure of a gas. According to this theory, the pressure exerted by a gas is due to the collective collisions of its molecules with the walls of their container. By summing the momentum changes of these collisions, we can derive an expression for the pressure in terms of the density of the gas, the average velocity of the molecules, and the area of the wall.The final answer is: boxed{P = frac{2 times text{density of gas} times text{average velocity of molecules}^2}{3}}

🔑:## Step 1: Understand the Problem and AssumptionsWe are dealing with two objects in a circular orbit, where one object has a mass M and the other has a mass m, with M being significantly larger than m. The central mass M is time-dependent, meaning it changes over time. We need to derive an expression for the distance between the two objects in terms of the angular momentum, gravitational parameter, and mass of the central body.## Step 2: Recall Relevant EquationsFor a circular orbit, the gravitational force (F) equals the centripetal force. The gravitational force is given by (F = G frac{Mm}{r^2}), where G is the gravitational constant, M is the mass of the central body, m is the mass of the smaller body, and r is the radius of the orbit. The centripetal force is given by (F = frac{mv^2}{r}), where v is the velocity of the smaller body.## Step 3: Equate Gravitational and Centripetal ForcesSetting the gravitational force equal to the centripetal force gives us (G frac{Mm}{r^2} = frac{mv^2}{r}). Simplifying, we get (G frac{M}{r} = v^2).## Step 4: Introduce Angular MomentumThe angular momentum (L) of the smaller body is given by (L = mvr). For a circular orbit, (v = frac{L}{mr}). Substituting (v^2 = frac{L^2}{m^2r^2}) into the equation from Step 3 gives us (G frac{M}{r} = frac{L^2}{m^2r^2}).## Step 5: Solve for rRearranging the equation to solve for r, we get (r^3 = frac{L^2}{Gm^2M}). Taking the cube root of both sides gives us (r = left(frac{L^2}{Gm^2M}right)^{frac{1}{3}}).## Step 6: Consider Time-Dependent Central MassIf M is time-dependent (M(t)), then the radius of the orbit will also be time-dependent. Substituting M(t) into our equation for r gives us (r(t) = left(frac{L^2}{Gm^2M(t)}right)^{frac{1}{3}}).## Step 7: Implications of Changing Central MassA change in the central mass M affects the orbit's radius. An increase in M would decrease r, potentially increasing the effects of tidal friction, which could further alter the orbit. Experimental errors in measuring M, L, or other parameters could lead to inaccuracies in predicting r(t).## Step 8: Tidal Friction EffectsTidal friction would cause the smaller body's orbit to decay over time, especially if the body is close to the central mass. This effect would be more pronounced with a larger central mass, as the tidal forces increase with the mass of the central body.The final answer is: boxed{r(t) = left(frac{L^2}{Gm^2M(t)}right)^{frac{1}{3}}}

🔑:Begin with a labeled diagram.Use the Law Cons. of Energy to solve for the final height ((h)). In the initial state ((i)), the watermelon is at rest ((v_{i}=0)) at a height ((h_{i})) of 5.00 m. In the final state ((f)), the watermelon has a kinetic energy ((KE_{f})) equal to 4.61 kJ and is at a height ((h_{f})) that we are trying to find. Note that (4.61text{ kJ}=4610text{ J}).[PE_{i}+KE_{i}=PE_{f}+KE_{f}]Substitute the equations for potential and kinetic energy into the conservation of energy equation.[mgh_{i}+0=mgh_{f}+frac{1}{2}mv_{f}^{2}]Note that (v_{i}=0) (because the watermelon begins from rest) and that we are solving for (h_{f}) (not (v_{f})). To solve for (h_{f}), first subtract (mgh_{f}) from both sides.[mgh_{i}=mgh_{f}+frac{1}{2}mv_{f}^{2}][mgh_{i}-mgh_{f}=frac{1}{2}mv_{f}^{2}]Factor out the (mg) from the terms on the left-hand side.[mg(h_{i}-h_{f})=frac{1}{2}mv_{f}^{2}]Divide both sides of the equation by (mg).[h_{i}-h_{f}=frac{v_{f}^{2}}{2g}]Add (h_{f}) to both sides of the equation.[h_{i}=h_{f}+frac{v_{f}^{2}}{2g}]Subtract (frac{v_{f}^{2}}{2g}) from both sides of the equation.[h_{f}=h_{i}-frac{v_{f}^{2}}{2g}]Before we can plug in numbers, we must find (v_{f}) from the equation (KE_{f}=frac{1}{2}mv_{f}^{2}).[4610=frac{1}{2}(118)v_{f}^{2}]Multiply both sides of the equation by 2.[9220=118v_{f}^{2}]Divide both sides of the equation by 118.[v_{f}^{2}=frac{9220}{118}=78.1][v_{f}=sqrt{78.1}=8.84;mathrm{m/s}]Now we can use the equation that we derived for (h_{f}).[h_{f}=h_{i}-frac{v_{f}^{2}}{2g}=5-frac{(8.84)^{2}}{2(9.81)}=5-frac{78.1}{19.6}=5-4.0=1.0;mathrm{m}]The watermelon's height above the ground is (h_{f}=1.0;mathrm{m}) when its kinetic energy equals 4.61 kJ.