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0

Use the trigonometric table for finding the values ( if you want an easy way out) or try doing a bit of adjustments here or there to suit your needs.


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The hirizontal component of velocity is a constant $v_H=50 \cos(53^{\circ}) \approx 30.1 \mbox{m/s}$. The vertical component at time $t$ is: $v_v(t)=-9.81t+50 \sin(53^{\circ})\approx -9.81t+39.9\ \mbox{m/s} $ Now you want to find the time $t$ when: $$ \tan(33.7^{\circ})=2/3=\frac{v_v(t)}{v_h}=\frac{-9.81t+39.9}{30.1} $$ which you solve for $t$


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If you're a beginner to these type of problems, its recommended that you start by drawing a rough trajectory and a free body diagram for your projectile. The only force acting on the body is gravitation, so its acceleration will always be $g$ downwards. The horizontal component of velocity always remains the same. (because there is no horizontal force on ...


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Some dimensions I was able to dig up (mostly from Wikipedia). Draft of the Allure of the Seas: 31 ft (10 m) Length: 1181 ft (360 m) Beam at waterline: 47 m Height: 72 m above waterline Let's just draw the section based on these simple numbers: Now if the center of gravity were in the middle of the ship (31 m above the water line), it would indeed not be ...


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You have $x$ and $y$ (the displacements along those axes) represented as functions of some variable. Velocity is defined the instantaneous rate of change of the displacement, with respect to time. (i.e. its time derivative) The velocity in the $x$ and $y$ directions therefore, would be $\frac{dx}{dt}$ and $\frac{dy}{dt}$ respectively. (Because projection ...


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I doubt it is all fluid dynamics. The have to stay upright even with dead engines. If the integral of the lever below the water line is bigger than above then it should stay upright. Ballast at the bottom goes a long way as it has a long lever. Stuff like engines below deck tends to be heavy anyway. Weight is not a big deal as they are not going up ...


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Isn't a physical frame of reference useless for calculating speed? There aren't really any physical frames of reference. You can't step outside and point up to the clear night sky and say "Look, there's a reference frame". You can point to the Moon and the stars, but they are what they are. You can use them in your reference frame, but that reference ...


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Velocity and speed of a body are measured with respect to another body (see other answer). This is the only possible definition. The "actual" speed respect to the "empty space" is not well defined instead. And this is because we are not moving inside an "empty space" or an "aether" which fills the universe. To grasp the concept think about how you will ...


2

When physicists use the word velocity it has a precise definition that is meaningful and unambiguous. If I measure the displacement from me to you then the result is a vector i.e. it tells me how far apart we are and in what direction you are. The velocity tells me how this vector is changing in time. The point is that I can do this for any pair of objects: ...


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I find that sometimes intuition works better with more extreme examples. Let's change the problem up a bit. Instead of some itty-bitty difference in acceleration (a vs 2a), lets choose a big acceleration for a short period of time. In this modified example, both A and B are going to be fired out of a cannon. The firing is going to take just 0.1s, but ...


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I'm assuming you used the following kinematic equation in your solution: $$d = x_0 + v_{0}t + \frac{1}{2}a_{0}t^{2}$$ Remember that after the initial acceleration, the particles will have covered some distance and gained some velocity. Hence, in the second step where you have a new acceleration, you no longer have a zero initial position and velocity. I got ...


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Let's draw a graph of velocity against time for the two particles $A$ and $B$. For convenience I've made the total time $2t$: The red line shows the velocity for particle $A$ while the green line shows the velocity for particle $B$. When we draw a velocity:time graph the distance travelled is the area under the line. More precisely it is the integral of ...


0

I think your answer is incorrect. First the intuitive argument: Assume you had stopped both particles after half of the time. It is obvious that in this case they had covered the same distance. At first B is first, but in the second half A would catch up. If you don't stop the particles before they continue with a different acceleration also the velocity ...


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The same reasoning that goes for racing cars: Acceleration is important if you are slow, because then the relative change in velocity over time will be high edit: (also the velocity is the accumulated acceleration, meaning that the earlier you have high accelaration the better because you will "keep" that to the end)


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There are 3 ways for a molecule to store energy, excluding electronic energy. Those three ways are translation (the molecule just moving), rotation, and vibration. All molecules can translate in each of the three Cartesian planes. You can also imagine that a non-linear molecule, could rotate in all the planes (xy plane, xz plane, yz plane). A linear molecule ...


0

We probably don't really have enough information to answer this question. The missing radius is the least of your problems: the $a = v^2/r$ formula is only a radial acceleration, not a tangential one: so it will never slow you down. If your position describes a curve it has the form $\vec r = \left[\begin{array}{c}x\\y\end{array}\right] = ...


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With given vectors for acceleration and velocity, is there a way to determine if a body accelerates or decelerates at a certain time-interval? Given the velocity vectors of a (specific) body (with respect to specific members of a suitable reference system) throughout a trial, in particular the velocity vectors $\mathbf v_{\text{initial}}$ at the ...


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Assuming you are asking undergraduate mechanics book. I found the one used by Caltech is unbelievably awesome. The Mechanical Universe: Mechanics and Heat, Advanced Edition by Steven C. Frautschi, Richard P. Olenick, Tom M. Apostol, David L. Goodstein (pick the advanced edition intead of introduction edition) The good parts of this book: It teaches ...


3

I am not too certain of what you are asking. If you are asking why do we not look at a molecule and then just identify the number of independent modes of vibration the answer is that it would be too hard. Real oscillations are a linear superposition of these normal modes, even undergrad spectra are often complicated. Only very simple diatomic give nice Raman ...


0

You asked two simple questions - I will give two simple answers. I don't know if I can use uniform circular motion equation since v is not constant At the very instant that the curve starts, the velocity is given by $\sqrt{2gh}$ - and for that first instance it is constant. So yes, you can use uniform circular motion Where is the g-force directed ...


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Let's say Bob is standing still while a one light second long photon formation flies past him. How long does the passing of the photon formation and Bob take according to us? Answer: It takes one second. 1 light seconds / c = 1 seconds. Let's say Jim is moving forwards at speed 0.1 while a one light second long photon formation moving to the opposite ...


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I don't know if I can use uniform circular motion equation since v is not constant The equation for centripetal force is independent of whether the motion is uniformly circular or not. However, irrespective of the radius of the track, the velocity at that point, and the weight of the roller-coaster, or whether the equation for centripetal force is ...


1

What happens if you first drive on a road at speed 0.99 c, and then accelerate to speed 0.999 c? You will pass milestones at much faster rate. That happens mostly because of length contraction, not because of larger speed, as the speed increase was only 0.009 c. Roughly the same contraction as above will happen to a line of photons approaching you, if you ...


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You're good. Yeah, you can pretty much assume that it's a constant velocity, as long as $h \gg r.$ As I recall, the expressions involved are extremely simple as long as you don't try to figure out exactly what's happening in time: actually solving the Euler-Lagrange equations gives you some sort of $\int d\theta / \sqrt{a + b \sin\theta} = t$ equation for ...


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There are two accelerations involved: The gravitational acceleration $g$ that points down, and the centripetal acceleration $a_r = \frac{v^2}{r}$ that points along the radius vector of the curve. The component of the gravitational acceleration that is tangential to the curve does not contribute to the g-force as it accelerates the cart and us in this ...


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This is one of those questions that can drive you crazy, since there is a great deal assumed and not stated. Let me try an alternate possibility. Conceivably, the problem wants you to assume that, when moving, the overall speed of the raindrop remains fixed at 5 m/sec, but it travels in a straight line at an angle due to horizontal wind forces, and this may ...


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If the raindrop's vertical velocity is constant as the train is both stationary and moving, the time taken for the raindrop to travel down the window would be: $$t = \frac{1\ \text{m}}{5\ \text{m}/\text{s}} = 0.2\ \text{s}$$ Remember, the time $t$ would not depend on the speed of the train. The exercise also specifically states that the raindrop's vertical ...


0

Another way of looking at this: The vertical component of the raindrop's velocity vector is 5 m/sec downward, and the horizontal component is 30 m/sec across. By the Pythagorean method, the resultant velocity vector is 30.41 m/sec diagonal. The vertical component of the raindrop's displacement vector is 1 meter, and as the raindrop is pushed horizontally ...


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You are right in that the speed of light doesn't change. It is a completely different effect to the rain drop analogy. If you had only light hitting you directly from the front and directly form the back, you would observe the same intensity in the moving frame (only blue/red shifted). But for light coming at you from an angle $\theta_s$ in the rest frame, ...


0

This time, you got the right answer using your intuitive reasoning because all the velocities are equal in magnitude. However, you will not be so lucky most of the time, in more complicated situations. Yes, definitely, concepts can be used here. Consider the four directions(north,south,east,west) as the ends of two dimensional co-ordinate axes: Now, I ...


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SECTION A : Free fall of roller coaster into circular motion (kinetics) Suppose that the roller coaster, called from now on "particle", is at rest at point A ($\:\upsilon_{A}=0\:$) and starts free falling till point B where it starts its circular motion. Well-known is that at B the speed is $\:\upsilon_{B}=\sqrt{2gh}\:$ under the assumption of no energy ...


2

The roller coaster falls a height $h$ before entering the loop (the path of the RC will look like a J. The straight bit of the J is of length $h$), and I assume it started with zero velocity. Within the loop it's total kinetic energy will equal the total potential energy lost, which is $$ E_\text{kin}=mg(h+r\sin\theta). $$ Here $\theta$ is defined such that ...


1

$v$ isn't referring to either $v_1$ or $v_2$, necessarily; $v_1$ is representing the vector before it moves, and $v_2$ is the vector after this movement. If we are working in polar coordinates (the reason he is using $v_\perp$ and $v_\parallel$), then let's suppose this small movement isn't changing the magnitude of the vector, it is just changing the ...


0

Inertia is what we simply called 'quantity of material'. The word material has been used here to specify the matter of body. For example, a plastic chair, a wood chair and an iron chair. Among them, a plastic chair will have less inertia because it will apply less reaction force, so it is easy to lift it. And the word quantitative is used to define the ...


2

Define the origin of your coordinate system at the top of the hill with the vehicle travelling in the $+x$ direction with initial velocity $v$ and initial position $s=(0,0)$. The road then "falls away" from the vehicle at a rate of $v \tan(\theta)$, so the altitude $h$ of the vehicle above the downward sloping road can then be written as a function of time ...


-1

It just oblique projectile motion U= speed of truck h= height of hill If you want to calculate how far from hill the truck land after flight use U multiply by the whole root of 2H/g g is gravitational acceleration =10m/s2


2

First I will make a few assumptions: the hill is flat at the its top (so its surface is perpendicular to gravity); the side of the hill is flat and makes an angle $\theta$ relative to the top; the end of the hill is sufficiently far away such that it will never be reached during the "jump" sideways of the top of the hill; the tradition from the top to the ...


3

The car becomes "weightless" when the curvature of the road is sufficient that the car does not stay connected to the road. Angles don't really matter - what matters is a change in the direction of the road. As you know, an object going around in a circle needs a force $F = \frac{mv^2}{r}$ to stay in the orbit at radius $r$. Normally, when you drive over a ...


1

For a right angle triangle you can use SOH CAH TOA to find the relations between the angles and the lengths of the sides. If you want to find the relation between the horizontal and vertical components of the velocity you need to use the tangens and $\tan(45^\circ)=1$, thus the ratio between the horizontal and vertical components of the velocity is one.


0

Is this true for all horizontally-thrown, free falling objects then i.e. that at the 45 degree point the x and y components are equal?! It's true for all vectors, whether they represent velocity, position, acceleration, or whatever. If a vector has a direction at 45 degrees to the x-axis, then the x and y components of the vector will have equal ...


1

The introductory quote defines $\tau$ and $\nu$ to be unit vectors. That means that their magnitude is 1. $\mathbb{i}$ and $\mathbb{j}$ are unit vectors along $x$ and $y$ axes. That's a standard notation. Some of the comments are stressing the fact that the basic understanding of vectors is missing. I have a feeling that you've jumped into a 'complicated' ...


1

So you can use the uniform circular motion equations, but you must use different values for $v$ since the v is not constant due to acceleration from gravity. I will show the g-force experienced at the beginning of the loop, the side of the loop (so a height $r$ above the ground), and the top of the loop (a height $2r$ above the ground). For the cart ...


0

I feel like you are asking more about language rather than about physics. An important concept is velocity. Velocity is a vector that points in the direction of travel of the object and its magnitude is the speed of the object. In the following I assume you are talking about velocity whenever you talk about vectors in your question. In that case, ...


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First be aware that, so far, you have been dealing with projectile velocity as a 2-dimensional phenomenon. This is going to have to change. In order to conform to standard notation, you'll need to refer to altitude (vertical motion) as the z-coordinate, while horizontal will be handled by x and y coordinates. For ease of calculation, you can assume that you ...


1

You conclude that constant acceleration is appropriate for one of a few reasons: You read in the question text that you should treat it that way. If (1) does not apply, you read in the question text that some physical situation obtains and you know or suspect that this situation is usually well represented by a constant acceleration. If neither (1) nor (2) ...


1

Events can never univerally only be said to happen in a particular order if they are space-like events, i.e. $(\Delta r)^2 > c^2(\Delta t)^2$. If this is true, the there is no universal agreement about the order in which the events happen. If, however, $(\Delta r)^2 ≤ c^2(\Delta t)^2$ holds true, then every observer will agree on the order.


1

Let me describe a geometrical way to approach your question. We start by drawing a spacetime diagram with distance along the horizontal axis and time on the vertical axis. We'll also take the speed of light to be $1$, so on our diagram a light ray travels along the line $x = t$ i.e. at 45º. If we have some other observer moving at a speed $v$ relative to ...


0

We can draw a tangent (Call it 't') at any point (call it 'p')on a curve. We can also draw a circle to which 't' is a tangent. The radius of this circle is the radius of curvature to the given curve at the point 'p'. An analogy from motion of a body along a curved path may help easier understanding. When a body moves along a curved path, its velocity keeps ...


0

So the x part of the drag is $$\begin{align*} f_x &= f \cos\theta \\ &= (k v^2) \cos\theta \\ &= k v (v \cos\theta) \\ &= k v v_x \,, \end{align*}$$ and $v$ is dependent on the $y$ component of motion as well as the x-component. Similar consideration, of course, apply to the y-component of drag. So the independence is explicitly lost and ...


1

I am going to assume that you have not yet studied linear algebra, sorry if it seems as if I am talking down to you at any point. You are correct in that we can split a vector into two components in the plane. This is because any two linearly independent(not parallel or anti-parallel) vectors form a basis(a set of vectors from which you can "build" other ...



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