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14

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 ...


11

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, ...


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 ...


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 ...


3

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 ...


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: ...


2

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 ...


2

$|\frac{d\vec{v}}{dt}|$ = magnitude of rate of change of $\vec{v}$ = |$\vec{a}$| while $\frac{d|\vec{v}|}{dt}$= rate of change of |$\vec{v}$| = rate of change of speed. If you consider any situation where speed of object is not changing and object is not moving in straight line , i.e. , moving in any curved path , both these conditions will be true. This ...


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 ...


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 ...


2

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 ...


2

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 ...


2

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 ...


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 ...


2

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 ...


1

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 ...


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 ...


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 ...


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

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 ...


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

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.


1

When neglecting air drag, the vertical and horizontal motions decouple and we have the equations \begin{align} \ddot{x} &= 0,\\ \ddot{y} &=-g. \end{align} The initial conditions (at $t=0$) are $x=x_0$, $y=y_0$, $\dot{x}=v_0\cos\alpha$, and $\dot{y}=v_0\sin\alpha$ with $v_0$ the initial speed and $\alpha$ the angle of projection. This gives the ...


1

Friction comes into play whenever there is relative motion between the surfaces in contact or a tendency of motion between the same.There need not necessarily be a force on either of the bodies,that is,there need not necessarily be a relative acceleration,merely relative motion or a tendency for the same.


1

Sorry, but the currently accepted answer is not precise enough. In the situation you discussed, there is no applied force acting on the object, but there is still the normal force $N$, which prevents the object from "falling through" the surface due to gravity. The most common model of kinetic friction is $F_k = \mu N$, where $\mu$ is the coefficient of ...


1

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 ...


1

There are 4 standard kinematic equations from Newtonian mechanics, and you need what is usually considered to be the fourth equation. $\mathrm{(final\ velocity)^2 = (initial\ velocity)^2 + 2 \times acceleration \times distance}$ You know the initial velocity, final velocity, and distance. Solve for acceleration.


1

A particle moving along a curved path has its motion decomposed as $$ \vec{v} = v \hat{e} \\ \vec{a} = \dot{v} \hat{e} + \frac{v^2}{\rho} \hat{n} $$ where $\hat{e}$ is the unit tangent vector, and $\hat{n}$ the unit normal vector pointing towards the inside of the curve. $v$ is the scalar speed and $\rho$ is the radius of curvature at that moment. 1) You ...



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