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0

I can see how you got that answer, but if you make a sketch of the projectile's trajectory (which is a parabola) you will see why it is wrong. You have not applied any equations which relate to projectile motion. Find equations for the height $H$ and range $R$ as functions of launch angle $\theta$ and launch velocity $v$, then eliminate $v$ and set $R/H=3$....


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.


0

The definition of uniform motion is that the object is supposed to cover equal distances in equal intervals of time. Well, it isn't defined from velocity. As you clearly write it here yourself. That is your own interpretation. You could just as well have said that "that means that the acceleration must be constant, so why do they define it from ...


-1

You can assume any arbitrary distance,say x ,and then the total distance will be twice of x..time taken in going will be x/60 and time in coming back will be x/50 so that the total time will be their sum..that would give your average speed that is 2*x divided by sum of x/60 and x/50 ...that nicely gives you the final answer and the x neatly cancels out.


-1

Assume the first trip takes a time of 1. Then the return trip takes 1.2 (see why?) and the total time is thus 2.2. From that you can calculate the average.


-2

You should use the average speed formula: $\bar v=\large{\frac{\textrm{distance}}{\textrm{time}}}$ For going: $t_1=\large{\frac d{v_1}}$ For coming back: $t_2=\large{\frac d{v_2}}$ Then, $\bar v=\large{\frac{2d}{t_1+t_2}}$


0

I question the physical possibility of the question as posed. If A is accelerating to the left, as I believe it would, then the pulley is accelerating to the left as well. Then B, hanging from the pulley, must accelerate to the left (as well as down, of course). Otherwise it would be "left behind" by the block A. The only possible source of a horizontal ...


0

I think you'd better take a clear force analysis first. This is not a hard problem, though it may take a few tricks. There's some hints for you: Take the pulley and A as a whole; Investigate carefully on the direction of every elastic force; A and B move simultaneously. The answer consists only of m(A), m(B) and g.


0

I am assuming that A is a free body on a frictionless plane and that the rope is fixed to the wall to the left. The key concept to understand here is that the tension in the rope as a consequence of the weight of B must be equal at all points. You already seem to have understood that concept. However, you state that the two forces are in the same direction. ...


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 ...


0

The question asks for the change in the velocity vector, which should be done by vector subtraction, not addition. This would give you pi*sqrt2/30, as you have, but directed at an angle of 225 deg - toward the center of the circle, as it should be. (The first vector is +i, the second is -j, so the subtraction is -j-i, which gives the correct angle.) Your ...


0

Your formula is wrong. The correct formula is: $$\bar {\vec v}=\large{\frac{\Delta \vec r}{\Delta t}}$$ $\bar {\vec v}$ is the average velocity.


0

You know where you start and you know where you go. Raised platforms or not. Then you only need something about the trow (like initial speed). Just mingle around the four fundamental equations of motion: $$x=x_0+v_0t+\frac{1}{2} at^2\\ v =v_0+at\\ v^2=v_0^2+2a(x-x_0)\\ x=x_0+\frac{1}{2}(v+v_0)t$$ Chose the first platform as the 0-point. So $x_0$ and $v_0$...


0

There is one golden rule in kinematics in two dimensions, i.e: you can treat the vertical motion and the horizontal motion individually. The problem you stated in your question can be split into two separate problems which are given below. The first step is to split the velocity into its constitute components (along x-axis and y-axis). This gives us the ...


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 ...


0

Most people believe that you need mass to transmit a force, or even to be able make one, but the more fundamental concept happens not the force but the momentum. The momentum is the capacity to interact with another entity and transfer it some change in speed. In the case of the photon, it is massless but however has momentum and energy. The energy and ...


0

This is newton second law: $$ \mathbf F = \frac{\Delta\mathbf p}{\Delta t} $$ As you can see, its variation of momentum that brings force. So, its the transfer of momentum. Granted, photons are massless, but they do have momentum. There are two ways of a photon transfer momentum to a solar sail: The photon is absorbed (also heats up the sail), or, the ...


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 ...


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 ...


0

I think where the issue lies is the question is asking about speed - not velocity. Speed is defined to be the absolute value of velocity, and is a scalar quantity. Thus, the first case could work if the velocity is negative and the acceleration is negative. This way, the velocity would decrease, becoming more negative, but the absolute value of velocity, ...


0

Peskin and Schroeder use natural units, where $c = 1$, or alternatively where factors of $c$ (and $\hbar$) are omitted. If you restore the necessary constants to make the units work out in an SI-like system, you would have $$\frac{k^z}{E} \to \frac{k^z c^2}{E} = v$$


0

The definition of acceleration is change in speed with time, and if the speed is increasing, by definition acceleration is positive. Think of continually adding the quantity of acceleration to speed. If you add a positive number it makes the result a bigger number, if you add a negative number it makes the result a smaller number. Over a tiny fraction of ...


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.


0

$\Delta x =v_{average}\times t$ In uniform acceleration $v_{average}$ becomes $\dfrac{v-v_{0}}{2}$ Hence; $\Delta x =\dfrac{v-v_{0}}{2}t=\dfrac{1}{2}\dfrac{v-v_{0}}{t}t^{2}=\dfrac{1}{2}at^{2}$


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 ...


0

This is a trivial kinematic deduction. \begin{align}s(t_2) &=\int_{t_1}^{t_2}~v(t)~\mathrm dt +s(t_1)\\ &= \int_{t_1}^{t_2}~\left\{v(t_1)+\int_{t_1}^{t}~a(t')~\mathrm dt'\right\}\mathrm dt+ s(t_1) \;.\end{align} Integrating this, we would get $$s(t_2)~=~ s(t_1) + v(t_1)\{t_2-t_1\} + a(t_1)\frac{\{t_2-t_1\}^2}{2} + \dot a(t_1)\frac{\{t_2-t_1\}^3}{6}...


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 $v_{average}$ ...


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 ...


0

I don't have a conceptual answer as to why it is that way. But mathematically, your first suggestion is wrong: $ \Delta x = v_o t + gt $. A unit analysis will show you why: $$ meters = \frac{meters}{seconds} seconds + \frac{meters}{seconds^2}seconds$$ $$ meters \not= meters + \frac{meters}{seconds}$$ And we can see that the assumption is simply not true ...


0

When the ball is travelling in your hand before you let go, it is being accelerated by the amount of force you exert on it, this increases its velocity before you let go. (The initial velocity) When you let go of the ball, it immediately begins to slow down due to the acceleration of gravity acting downwards. There is no upwards acceleration, as this ...


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 ...


0

You say: the motor's velocity is always $>=0$ but this is not true. The magntude of the velocity is indeed always greater than or equal to zero, but velocity is a vector so the direction matters as well. The motor will oscillate alternately up and down as the masses rotate around it. The combined centre of mass of the motor and the two masses will ...


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 ...


0

I'm confused that area is a 2 dimensional concept and it may indicate distance or displacement , which are 1 dimensional quantities. You are right. Area is a 2 dimensional quantity but what you have missed out is that you didn't use dimensional analysis properly. The dimensions of velocity is $[LT^{-1}]$. So if you multiply velocity with time, you get the ...


0

If we consider the velocity-time graph area under the whole line is the distance. For example, Now look at the second part of the diagram (rectangular, dark blue) $$v=\frac{dx}{dt}\rightarrow\int{vdt}=\int{dx}\rightarrow x=vt$$ But the first part (triangle), we should consider acceleration $$a=\frac{d^2x}{d^2t}=\frac{dv}{dt}\rightarrow\int{adt}=\int{dv}\...


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. ...


0

It is sometimes easier, particularly when first using the constant acceleration kinematic equations, to draw velocity vs time graphs for the motions. The area under a velocity against time graphs gives you the displacement, so use the graphs and equate the area. When you do this you should see the kinematic equations appearing within your solution. There ...


0

As this is a bit too long for a comment I put is as an answer. If you take $x_0$ and $t$ you extracted and put it in some equation of motion, you will get something, but not what you want. I will try to answer your conceptual problems: You have to figure out the equation of motion for the motorist (constant $V$) and the cop (accelerated motion) $x_0$ is ...


0

The crux of the problem is to recognise that the trooper will catch the motorist when their respective distances are equal. So, you need to express their distances in terms of the known quantities, and equate them to find the unknowns. Note that you have forgotten one of your known quantities (the velocity of the motorist), and you have made an erroneous ...


0

Your explanation is almost exactly the same as the one which you say that you do not understand. So what exactly is confusing you? If $v_p-v_q$ is due south it does not necessarily follow that $r_p-r_q$ is also due south. For example, Q could be stationary while P moves due south, but the initial position of P could still be east or west or even south of Q ...


0

Relative to Q, Q is not moving, and P is north of Q. $\vec v_p - \vec v_q$ is the velocity of P relative to Q. So basing on that, and knowing that P is exactly north of Q initially, we only need to know if P is going south relative to Q. Also, this means that $r_p-r_q$, the position of P relative to Q, is due north. Then, we should expect the velocity of P ...


0

I also would like to give you a hint. Forget about back and forth trip of bird, let it fly straight, until boy reach the finish line and see how far the bird is. First Calculate the time boy takes to reach the finish line and then solve for the bird and see how far can bird go in that time. Even, we can get answer without calculation. Bird's speed twice, ...


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 ...


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 ...


0

If this is a projectile problem, and assuming no air resistance or buoyancy, the only force is the weight mg acting constantly downwards, and the only acceleration is g. Neither of these is affected by initial velocity.


0

Initial velocity has absolutely no effect on force. Any inertial frame must obey the same laws of physics as any other frame. You asked what you need to know to calculate force and then said F=ma. I'm not entirely sure what you are getting at but you clearly need to know the mass and the acceleration. Now acceleration is the change in velocity over time (a=...


0

The vertical speed is $$v=\frac{dh}{dt}=\frac{2u}{x}e^{-gt/ux}$$ and the acceleration reads $$a=\frac{dv}{dt}=-\frac{g}{xu}\frac{2u}{x}e^{-gt/ux}=-\frac{g}{xu}v.$$ This would be a particle going upwards under action of an effective force $$F=ma=-\frac{mg}{xu}v.$$ This is drag type force. Since there is neither a term corresponding to the weight nor a term ...


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. ...



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