Velocity Definition and 1000 Threads

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

edit: I don't know why my latex isn't rendering, any help would be appreciated. Edit 2: The question was due to a misunderstanding I had, I thought integrating instantaneous velocity would give me average velocity. I have attached what I have tried so far. I had a doubt. Can you calculate the...

TL;DR Summary: Velocity and the human experience As I was walking into work this morning I looked up and noticed an airplane flying overhead. Airplanes travel at speeds of up to 550 miles per hour while in flight. This question is really less about physics and more about our everyday...

On many websites etc the velocity of a freefalling object in a vacuum is shown as follows. After 1,2,3 and 4 seconds respectively; 9.8m/s; 19.6m/s; 29.4m/s; 39.2m/s I worked out the distance travelled by a freefalling object in a vacuum using d = at^2/2, or, d = 0.5gt^2 and got, for after...

For this, Does someone please know whether they assume for the equation highlighted that ##\frac{v}{f} ≥ \frac{v_S}{f}## since otherwise the wavelength would be negative (which I assume is impossible)? Many thanks!

Hello Physicsforum! This is my attempt: First I realised: ##a_s=a_n## Secondly I used since previus known formulas: ##a_n=\frac {v^2} {R}## ##v=v_0+a_s*t## Although now I do not know how to continue, any suggestions would be appriciated! Thanks for your help on beforehand :smile:

im not sure where to start

I did try to solve the problem by forming the derivative and my result was: v=(rcos(phi), -rsin(phi),0). My solution is wrong, the tutor corrected the task but he didn’t give us the results. My question is what the solution is. Thanks in advance.

The question was this: My calculations show that the answer should be equal to work done on crate to make it reach the same velocity which is equal to 216 J but the answer given is 432 J It is believed that extra energy is needed to overcome friction but friction is an internal force and...

If an object is going close to the speed of light, will outside frames of references perceive the moving object as going slow or fast?

Hello, I'm new to the forum, I want to ask help on this problem here, above is a press mechanism and I'm tasked to draw a velocity polygon based on this mechanism. The point of contact between the two gears is A and consider it as a swivel hinge (rotating but stay still), so I guess it's O2...

If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both...

I don't understand the question. how am I supposed to find the magnitudes and directions of the velocity from the figure?

This is the problem: And this the answer provided by the examiner: And this is my own answer: So what did I get wrong??? Also I want to know if the Velocity is the same for both masses.

Assume there is a force (vector field) on the space .....does the effect of this field on the particle(the change of momentum) at some position depend on the speed at that position? And is it related to the time interval dt the particle experiences this force ? Can i say dt=dx/v? And is that...

Hi I have come across something confusing in rolling motion. If an object moves with a positive V_cm meaning to the right its angular velocity will be clockwise or negative. The formula is V_cm=wR but for a positive V_cm you get a negative w as it moves clockwise if V_cm is to the right...

For this, The solution to (a)(i) is 0 ft/s. However I got -4 ft/s. The formula I used was, ##v_{avg} = \frac{s_f - s_i}{t_f - t_i}## ##v_{avg} = \frac{\frac{1}{2}t^2_f - 6t_f + 23 - \frac{1}{2}t^2_i - 6t_i + 23}{t_f - t_i}## ##v_{avg} = \frac{ \frac{1}{2}(t^2_f - t^2_i) - 6(t_f - t_i)}{t_f -...

For this problem, The answer is ##6.28 m/s##, however, I got ##-314 m/s## and I am not sure what I have done wrong. My working is, ##H'(t) = \lim_{t \rightarrow 1} \frac{H(t) - H(1)}{t - 1}## ##H'(t) = \lim_{t \rightarrow 1} \frac{-1.86t^2 + 10t -8.14}{t - 1}## ##H'(t) = \lim_{t \rightarrow 1}...

What is the velocity of a photon through the four dimensions? What is the velocity of a photon through the three spatial dimensions x1, x2, x3? What is the velocity of a photon through the fourth dimension x4?

This is a UK A-Level question that I'm really struggling with, and can't seem to find any resources online that explain it well. I've been given the following details: mass of gokart + driver = 520kg radius of track = 42m Maximum frictional force between tyres and road on flat track F = 20%...

The question is: A uniform rod of length ##L## stands vertically upright on a smooth floor in a position of unstable equilibrium. The rod is then given a small displacement at the top and tips over. What is the rod's angular velocity when it makes an angle of 30 degrees with the floor, assuming...

Is it generally the case even with light like paths that ##\dot t>0##?

Here is the problem statement along with the figure. Here, I take the right-ward and anti-clockwise directions to be positive. After the ball collides with the wall, its angular velocity remains the same and its velocity changes direction while remaining the same in magnitude. Using the...

So at first I thought that the time would be 10 years, and that I’d have to consider the outbound motion as v = .87c and inbound motion as v = -.87c but I’m struggling with addition of the velocities and whether or not this is even the right approach?

The car covers half of the road with an average velocity of v, so the elapsed time is equal to: ##t_1=\frac {d/2} {v}=\frac {d} {2v}## And it covers 1/4 of the road with an average velocity of 2v, so the elapsed time is equal to: ##t_2=\frac {d/4} {2v}=\frac {d} {8v}## Then it covers 1/8 of the...

a = 9.8*sin(25*pi/180)=>a=4.1417 m/s^2 vf^2=vi^2+2*a*s=>vf=sqrt(0^2+2*4.1417*3)=>vf=4.9850 m/s Meanwhile the correct answer is: (vf+vi)/2=>(vf+0)/2=2=>vf=4 m/s Why is my answer wrong? It seems that the acceleration is what is wrong, but I don't understand why.

A) I just did what it said to do: $$\sin\left(4x_{1}\right)=1\implies x_{1}=\frac{\arcsin\left(1\right)}{4}\ m=\frac{\pi}{8}\ m\approx 0.392699081699\ m$$ B) I modified the method from an example from the lecture the other week: $$U\left(x\right)=-\int...

Diagram : I draw a picture of the problem situation and paste it to the right. Attempt : Let me assume that the position of the smaller mass ##m## at a given instant of time ##t## is ##x(t)##. (1) Gravitational potential energy ##\frac{GMm}{x} = \frac{1}{2}mv^2(x)##, where ##v(x)## is the...

It didn't work and I don't know how to do it.

A problem I was thinking about when talking about natural circulation in a closed loop in another thread: What is the velocity of the buoyant plug? The loop is filled with water. There are no viscous forces acting on the plug or in the flow. The flow is assumed incompressible. 1) Is the...

The integral is this one: ##\int (\dot x)^2 \, dt,## With ##x=x(t). ## I don't know how to solve that integral and I haven't find nothing to read about on how to proceed with this kind of (implicit function?) integrals without having the initial function.

Vix=Vcosθ Vix=0.76604*V Xf=Vixt Xf=0.77604V*2 V=Xf/1.55208

Let's say we have some observer in some curved spacetime, and we have another observer moving relative to them with some velocity ##v## that is a significant fraction of ##c##. How would coordinates in this curved spacetime change between the two reference frames? For example, imagine a...

Ok, i have some rocket data from a 20,000 foot launch. I have the times, altitudes, and velocity at said time. Is it possible to find the terminal velocity from this long list of numbers? I tried graphing the velocities to see where the curve flattens out, since usually that is where terminal...

The instantaneous velocity at time a is defined as derivative of motion function f(t). It is not similar to average velocity in an interval of time. From the Newton law. If an object is at rest, we must exert a force to make it move, assume that there is no friction. Depend on the weight of...

Question : For uniformly accelerated motion ##a(t)=a_0\;\; \forall \text{times}\;t##, we can say that the average velocity for the entire motion ##\bar v = \frac{v_0+v}{2}##, where ##v(t)## is the final velocity at some time ##t## and ##v_0## is the initial velocity. How do we show that? Issue...

For calculating the total impulse, I'm taking the individual impact force times the elapsed time and adding them all together to get the total impulse. Given F = 25N, ## \Delta t_1 =3 secs##, ##\Delta t_2 = 4secs##, ## \Delta t_3 = 5secs##, ## \Delta t_4 = 6secs##. $$ Impulse 1 = F \Delta t_1 $$...

I've understood that between time t=0 to t=1 sec (moving backward), the object is moving with increasing velocity in the negative direction, slows down, and comes to rest at t = 1 sec. At t = 1 sec, the object returns to its starting position, briefly rests, and then begins to accelerate (moving...

i used t=d/v for the x direction using the formula t=250/vcos55 and subbed that into the formula for the y direction d = (v) (t) + (0.5) (a) (t)^2 35 = (vsin55) (250/vcos55) - (4.9) (250/vcos55)^2 canceled out the first two v 35 = sin55 (250/cos55) - (306250/v^2cos3025) 35 = 357 -...