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9

The zig-zag strategy seems trivially obvious - but it might not be the better strategy in a particular situation. I suggest rather than asking under what conditions this strategy is preferable, you ask under what conditions the counter-intuitive straight-line strategy is preferable. The advantage of zig-zagging is that it presents a smaller "collision ...


3

Consider the graph only from the moment when the brakes are applied, that's the time $t=0$ here. The uniform deceleration means that the distance travelled after that is $$ s = v_0 t - \frac 12 a t^2 $$ while the velocity is $$ v = v_0 - at $$ We want to draw $v(s)$ as a function of $s$ but the second formula depends on $t$. So we have to find $t$ from the ...


3

Indeed your first suggestion is wrong :$ \Delta x = v_o t + gt $ Instead it should be $ \Delta x = v_o t + gt^{2} $(You can recheck it) Where you are wrong is here: According to your question v is the final velocity since $(v=v_{0}+gt)$ So $\Delta x\neq vt$ but instead it should be $\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ ...


3

First, if you only have one line of atoms in the three-dimensional space, it will almost certainly bend if you push it. So it's better to make the sequence thicker. But let's assume that you have guaranteed that the thing doesn't bend. It looks like this: The boxes represent the individual atoms. And the springs represent some forces between the atoms. ...


3

What should be the minimum value of $v_0$ in order to hit the monkey while it's in air? Minimum value for $v_0$ is when arrow hits to the monkey just before it (monkey) reaches to the ground. Or, minimum value for $v_0$ is when arrow's range is equal to horizontal distance between hunter and monkey. So you need to find the time of monkey's fall. Or, you ...


2

The drone flies by its propellers exercising force against the air inside the airplane. It flies with respect to the air inside the airplane. Since the air is being carried by the airplane, the drone will fly with respect to the airplane. It'll fly forward or in whichever direction you point it, with respect to the X. Assuming the airplane is airtight.


2

$$ a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = \frac{dv}{dx} v $$ The physics notation makes the truth of the derivation above clear because $dx/dx$ cancels. In mathematicians-style presentation of the rule, it's the chain rule for the derivative of a composite function, $v'(t(x)) =\dots$


2

There are 4 vibrational modes in a molecule like CO2 like you are considering. These would be a symmetric stretch, an antisymmetric stretch, and 2 bends (just like your animations). In case of non linear molecules there is only one vibrational bend: Lets consider H2O. You might be tempted to think that the same argument of two bends should apply here, but on ...


2

While I don't know the terminal velocity of a cat (which may or may not be 100 kph). I know a lot of people say 100 kph, but I don't know if that is verified or just a calculated estimation someone came up with. I have experimented quite a lot with falling objects, including small animals and bugs. I have documented the falling speeds of many small ...


2

The symbol $e^Y$ is almost certainly not a misprint. This function (the exponential function) is omnipresent in mathematics of functions, physics, and any discipline of sciences or social sciences that use mathematics. It is an extremely important function – basically the most important operation after addition, subtraction, multiplication, and division. ...


2

The displacement is equal to the area under a velocity time graph so in your example the change in displacement $dx$ in a time $dt$ is equal to $v\; dt$ with $v=\frac 3 x$ So $dx = v\; dt = \frac 3 x \; dt \Rightarrow x\; dx= 3\; dt$ and then you do the integration.


1

You can consider an object to be a point particle, when you only think of its translatory motion, means it is moving in 3-D like a point would, like, if a point particle moves around, it is so small that you don't care about its orientation, whether it is rotating. So, when you are studying objects like a block sliding down an inclined plane or something, ...


1

An extended object can be considered a point object or simply a particle whenever its dimensions are negligible compared to the characteristic dimensions involved in the problem and it does not have an internal structure (or it may be also neglected). For example, in studying the motion of the Earth in the solar system we can treat the planet as a point mass....


1

We define the average value of a quantity by : $$ \langle v \rangle=\frac{\displaystyle\int_{t_0}^{t} v ~\mathrm dt}{\displaystyle\int_{t _0}^{t} ~\mathrm dt} \ .$$ Now using the first equation of motion we get: $$v=u+at$$ Putting this in the integral we get: $$\langle v \rangle =\frac{\displaystyle\int_{t_0}^{t}(u+at)~\mathrm dt}{t-t_0}$$ which ...


1

This is a simple question of conceptual understanding. The only force acting on the ball after it is released from the hand is that of the ball's weight due to gravity. Since it is the only force, the consequent acceleration is also the only acceleration. Gravity is being taken to be $10 m/s^2$ in a downward direction. For the answer to be correct, our ...


1

Let's take the first equation of motion which is : \begin{equation} v=u+at \end{equation} Integrate this equation to get: \begin{equation} \int\frac{dx}{dt}dt=\int{u}dt+\int{at} dt \end{equation} this gives: \begin{equation} x=ut+\frac{1}{2}at^2+x_0 \end{equation} The integration constant can be done away by putting the proper limits on $x$.(Assuming the ...


1

Whilst the motion you intend is not altogether clear, the motor will indeed move as you say. But the attached rigid bodies also move, such that the center of mass of the whole system is stationary (or, more precisely, its state of motion unperturbed by the system's internal motions). Work out the path of the center of mass in your system to check this ...


1

There is no relation between the area being two-dimensional in your graph and what it means. For example, consider you make the $x$-axis as an indicator of the temperature, so what is the difference between $x= 5 K$ and $x = 8 K$? of course it's $ \Delta x = 3K $. Now, isn't $x$ a one-dimensional quantity? So the graph gains its meaning from you not from ...


1

The interpretation of the area under a curve, depending upon the curve, will vary. If it is a Velocity v. Time Graph, the area from a given time to another time, will be the distance traveled between those times. If it is an Acceleration v. Time Graph, the area from one time to another, will be the change in velocity of the object between those two times. ...



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