Velocity Definition and 1000 Threads

solution is here; I just need to understand this part ##14.7 = -14.7 =9.8T##... why initial velocity upwards is a negative value? or i am interpreting it wrongly. ........... In my reasoning, ##v=u+at## ##0=14.7 + (-9.8)t## ## t_1=1.5## in reverse direction, from top to start point ##T##...

I need to find ##v_1## and I know what are the initial conditions: ##\theta(0)=\pi## and ##\dot{\theta}(0)=0##. Then what is ##v_1## and how to find it? Thanks!

part i) i did 1/2 * 1700 * v^2 i dont know what v is... so how do i solve it? part ii) i calculated it correctly by 440*25 please explain in detail why i used 440? and part d) i did 1.7*10^4 = 48000/t my t= 2.82 s but correct answer is 3.5s

Several different sites, including PhysicsForums, mention/discuss a formula for computing a bullet's Ballistic Coefficient based on measuring two downrange velocities where the measurements are made a distance L between them. This can be done with traditional skyscreen (optical) chronographs or...

Can someone explain to me Bernoulli's principle as to why the pressure across a restriction varies as the square of the velocity? I'm looking for an understanding (conceptual) as to why this is without a gazillion math examples please. Thanks.

Hello I'm thinking of how to attack this problem. Outside a channel-inlet, wind is passing perpendicular to the opening with a wind speed of 9 m/s and a fan sucks in a certain amount of air into this channel so that the velocity in the channel is 3 m/s. When the outside air is 0 m/s I have no...

Hi everyone, I have created a question which I thought would have a single simple solution, but have noticed there are two possible answers. This makes me think that the question's scenario is impossible with the numbers I made up. I think we all can agree that the horizontal component to...

So in my textbook there's a basic problem where you solve for the final velocities of two hockey pucks, which happen to have different colors which are red and blue, using conservation of momentum. The notation that the textbook uses to express the final velocities of the pucks is ##v_{1,f}##...

My answer is d = (e+3)/e x(t) = ∫01 3t2 dt (0 ≤ t ≤1) = t3 |01 = 13 - 03 = 1m x(t) = ∫1∞ 3e-t dt (t > 1) = -3e-t |1∞ = lim(t->∞)[-3e-t] - [-3e-1] = 0 + 3e-1 = 3/e m Therefore total distance = 1m + 3/e m = (e+3)/e m However, the textbook answer gives...

This is the problem: I am a little confused at the solution below. Since ##\vec r_{P/C} = (R/2)*(5 + 4cos(\theta))^{0.5}\ \vec e_r##, I am wondering why there is no vector e_r component for the velocity. I know that d(R)/dt = 0, but you can still get a derivative in the ##\vec e_r## direction...

this is the formula v is velocity g= gravity h= height ro= outer radius of cylinder ri = inner radius of cylinder please help

In my lines i have, ##h(t)=ut+\dfrac{1}{2}at^2## where ##u=0##, with ##a=10##, ##h = \dfrac{1}{2}×10t^2## ....................1 ...also, ##h-14.7 = \dfrac{1}{2}×10 (1.4-t)^2## ..................2 1- 2 gives, ##14.7 = 14t -9.8## ##24.5=14t## ##t=1.75## Therefore, ##v = 10×...

I have been trying to solve this problem for hours using the mentioned equations but no matter what I do I cannot get the correct answer, that is v = 22.4 m/s. I thought that maybe if I could get an expression where v is a function of time I could solve the problem but I don't know how to do...

To boil down the question, if you have a body at rest and apply a constant force, it will accelerate and the work done on it will be F*s (or the integral version of that statement). However, as the body accelerates due to the force, does that mean, per a given time unit, more and more work will...

Statement of the problem : I copy and paste the statement of the problem to the right as it appeared on the website. Given below is the graph of the ball as its distance from a fixed point with time. Attempt : Where does this fixed point, say ##\text{P}## lie? Imagine the fixed point lied...

The correct answer is u=vcos\theta. I have understood so far to be able to conclude that \text{displacement of string} = PA - PC \approx AB Also, \overline{AB}=\overline{AC}cos\theta or, more generally, \vec{S}_{along\ the\ string}=(\vec{S}_{along\ the\ horizontal})cos\theta Now, I had hoped...

position and velocity expressions for the initial conditions 1) x(0)=Acosφ = xi 2) v(0)=-ωAsinφ= vi Dividing 2nd equation by equation results in: -ωAsinφ/Acosφ = xi/vi tanφ = xi/(-ωvi) tanφ = 0.4 And here they (the textbook) got φ = 0.127π, This is where I...

Here is my attempt. At ymax the velocity turn to zero so we get time t*=v0/g and ymax=1/2 (v0^2/g). At the height y max, since the velocity at this point is 0, i get another equation y= 1/2(v0^2/g)-(g/2)t^2, this equation could be considered as continuation of first equation. Set ymax/2=1/4...

Since space is curved within the Earth's gravitational field, every body that moves there will follow the curvature of space no matter what speed it has, so what will its trajectory be, how will it be straight, only if the launch is made absolutely vertically towards sea level? Only then can it...

Here is the exercise: Which one seems logical and correct ? this one: [ Normally when we increase distances, the velocity of sound decreases?] Or this one? You might wonder why. Well, my friend in class told me that the second table could be correct because the experiment was conducted at...

the answer to this question uses the above formula with the tangent function and solves for the initial velocity, i used the equation (v.sinθ^2) = (v.sinθ)^2 - 2gΔy, setting final velocity equal to zero and solving for initial velocity. this kinematic equation gives a different answer. can...

Hello everyone, I'm reading Morins book which I like, and I feel I kind of understand the part on time dilation, however I'm a little confused by the geometry of the Pythagorian theorem when applied to velocities. On the moving clock he shows the velocity of light on the diagonal it traces...

What is velocity V in mass flow rate formula, velocity of inlet ,outlet, velocity of rocket speed in relation to freestream?

I understand that through process of elimination the only plausible solution is (E), but a question that rises up: When the ball bounces, does the velocity change from negative to positive instantly (as shown by the dotted lines) or gradually (a very small time period, but still solid line)?

Problem statement : I copy and paste the problem as it appears in the text down below. I have only changed the symbol of the given acceleration from ##a\rightarrow a_0##, owing to its constancy. Attempt : I must admit that I could proceed very little. Given...

1. The first equation between velocity ##v## and time ##t## can be derived using the graph I have drawn for the purpose as shown on the right. Since acceleration ##a_0## is a constant, the graph of ##v-t## is a straight line. The slope of the line is ##\dfrac{v-v_0}{t} = a_0\Rightarrow \boxed{v...

Hi, I am having problems with task b I then defined the velocity vector and the acceleration vector as follows ##dot{\textbf{r}}'(t) = \frac{1}{||\dot{\textbf{r}}(t)||} \left(\begin{array}{c} \dot{r_1}(t) \\ \dot{r_2}(t) \end{array}\right)## and ##ddot{\textbf{r}}'(t) =...

My solution was that the final velocity of the fly is equal to the mass of the elephant divided by the mass of the fly, and then multiplies by the delta in the elephant's velocity. My teacher said it was the wrong answer and that the calculations are presumably pretty long

$$ \ ω = \frac{Δα}{Δt} \ $$

dont know where to start. Other than it will take 5 seconds for v = 0m/s

so then when we model the velocity of the motion of a rocket (e.g., in a mathematics report), is it reasonable to assume that the initial velocity equals zero? I also apologise for my lack of information if I made a huge mistake in my question 🙏🏻

Here is only my solution: ##A_1 \frac{\mathrm d h}{\mathrm d t}=-A_2\sqrt{2hg}##, so by integrating we get ##h(t)=\left(\sqrt{h_0}-\frac{A_2}{2A_1}\sqrt{2g} t\right)^2.## Setting ##h(T)=0## we get ##T=\frac{A_1}{A_2}\sqrt{\frac{2h_0}{g}}.## By doing the first time derivative of ##h## we...

First, I have always consider that the angular momentum equals to inertia times angular velocity, but that’s not the case from the options perpective, is my memory wrong, or is there something wrong with the options? Another, I think I need to figure out the angle it went through, I think it has...

so far what i have gotten to is that 300/0.75 = 400km/h but I dont know how to draw the diagram for this

Given that the ions are initially at rest my initial velocity is 0. Therefore my Vavg is equal to vf/2 Using the formula Vavg = Change in positon/time, I can solve vf to be equal to 2r/t. Using the momentum principle, I get an equation of 2r/t = FnetT/12m -> Given that the mass of the ion is...

I think it's not possible. In 3D for constant acceleration we have : ##\vec v = \vec v_0 + \vec a t## It's a line in 3 dimension so velocity's magnitude(speed) is changing with time. I appreciate any better idea.

TL;DR Summary: Almost as difficult as the Physics Olympics semi-finals or finals The coefficient of friction of the two arcs is the same.The radius of the ball is not taken into account.The roll of the ball is ignored.Can it be calculated or estimated quantitatively?The difficulty is estimated...

1) a=0 and v=0 2) 0.7 m/s

EDIT: I've finally found the solution, so here's what I did. First, calculate Work using the equation: (F-mgsin(theta))*displacement, where F=the force being applied by the push, theta is the angle of the ramp, and displacement is the length of the ramp. Now that you have the value for work...

800 - (32 x 9.8) = 32v/0.18 where v = velocity this gives me v = 2.736 m/s The answer given, however, is 800 = 32v/0.18, i.e. v = 4.5 m/s The difference, of course, is the weight of the child. I don't understand why this is not allowed for in the net force acting on the child. Can someone put me...

initial velocity y component is (cos30) * 20. t = 25m / ((cos30) * 20)m/s = 1.45 seconds d = vt + .5at^2 v= 20sin30 v= 10 , d= 10(1.45s) + .5(-9.8m/s^s)(1.45s)^2 d=4.2m 4.2-2.5 = +1.7m, so the ball will not hit the fence I need confirmation please

is it possible? I've given it quite some thought, and my conclusion was that it would be possible, cuz right after I jump, I'll still have the same velocity in the same direction as the running train?

I am going through https://itp.uni-frankfurt.de/~hees/pf-faq/srt.pdf, here I don't understand how we wrote eq. 1.5.3 with the help of three-vectors to get eq. 1.5.4.

Not sure what r would be in this scenario. I tried adding the radius of the earth to the altitude but that wasn't correct either.

From the bus driver's point of view, who is at rest, the ball's initial velocity is ##u+v##. After the collision, its velocity has to have the same value, but an opposite direction, so ##-(u+v)##. So that means that relative to me standing on the ground at rest, the ball's new velocity is...

1) Using "The person catches the ball at exactly the same height it was thrown from.", we can isolate t by solving yb(t) = v0*sin(theta)*t - (1/2)*g*(t^2) = 0: yb(t) = v0*sin(theta)*t - (1/2)*g*(t^2) = 0 v0*sin(theta)*t = (1/2)*g*(t^2) 2*v0*sin(theta) = g*t t = 2*v0*sin(theta) / g 2) At the...

Is it possible to trace the trajectory of an object using only its velocity and position, both of which are given as components. My method of doing so involves using the time until max height is reached, and using that time value to calculate the max height itself (h,k), then plugging in the...

u is the speed of the message relative to the station v is the speed of the spaceship relative to the station u’ is the speed of the message relative to the spaceship u=(-0.5c+0.7c)/(1+((-0.5c)(0.7c))/c^2 ) =0.2c/0.65=0.308c This just seems way too high, and I'm not sure if I'm doing it...

This is a famous book in India. I was wondering if one could say if the answer should include velocity or speed. I mean, I don't think there are any details which hint at velocity. We are gives speed in the question and we are asked to find out the distance traveled, this hints we are asked to...

I'm not sure. A source said New Velocity = Current Velocity - TerminalVelocity every second until New Velocity = TerminalVelocity But it doesn't seem right