Kinematics Definition and 1000 Threads

Here I'm not worried about the solution as I got it. There are two ways I could come up with: Either finding the time it takes the cat to leap through the window, use that time to find what distance does the dog cross through the room Or: finding both times, for the dog and the cat to cross the...

I know That the velocity of each particle of the thread along the thread must be ##v_{0}## since the thread is inextensible. Now let's say the bead moves with some velocity ##v## For the particle attached to the bead to move with a velocity ##v_{0}## along the thread: ##v##cos##θ##= ##v_{0}##...

This is my solution below.

no clues

The time scale on which the change (such as a change in external parameters or a external parameters or an addition of heat) takes place is referred to as τ_exp. The relaxation time τ_relax, on the other hand, is the time that the system needs to return to a state of equilibrium after a sudden...

x=x0=vt v=v0+at x=x0+v0t+(1/2)at2 That (1/2) there does not make sense to me. It makes sense mathematically, if you integrate the velocity formula you get the (1/2) as a result of the integration rules and if you differentiate its needed to cancel out the 2 and arrive at the velocity...

Ive done this problem two different ways (sorry it’s messy) and keep getting hB = 1/9 hA, but my homework says it’s wrong. I’m guessing it’s because I assume that v0 is 0, but I’m not sure what other formulas or steps I need to take to either find or omit the variable. Are there any other...

Carlton writes on page 89: "The thrust of a marine propeller... may be expected to depend upon the following parameters: (a) The diameter (D) (b) the speed of advance (Va) (c) The rotational speed (n) (d) The density of the fluid (ρ) (e) The viscosity of the fluid (μ) (f) The static pressure of...

Attempt is attached, thanks for the help :)

For the first question I thought of using an energy balance, there is friction ##\Rightarrow \Delta E_m = -W_f##. Both at the start and at the end, the block has no velocity. Therefore ##E_{\text{initial}}= \frac 1 2 m_s v_{s,i}^2## and ##E_{\text{final}}= \frac 1 2 m_s v_{s,f}^2##. This means...

I'm trying to make a very basic physics engine. So far I've got a variety of small things worked out but I've been driving myself crazy trying to work out collisions. From one sense I get I can use momentum and impulse to determine the velocity of an object after a fully elastic collision (no...

For this problem, My solution to (a) is, We have constraint ##x + y = L##. There are many places we could define our (x,y) Cartesian coordinate system. However, the most easiest I think for the problem would be to attach a ##x^*## and ##y^*## coordinate system at the COM of ##m_1##. We define...

I managed to solve this by tilting the axes along the hill, and calculating the range, and then differentiating wrt ##\theta## (angle launched from hill) to get the answer. However, I recently came across the alternative solution below: The parabola it refers to represents the parabolic...

I get that: ##x(t) = A\cos(\omega t + \phi)## ##y(t) = A\sin(\omega t + \phi)## (from the above relevant equations). This agrees with the solution for part (a). However, the solution manual claims in part (b) that: In the case where ϕ1 = ϕ2 = 0 and A = B, the mass moves in a circle centered...

My guess was simply that as acceleration changes from the north to east direction, the total magnitude change of v is ##v \sqrt 2##. Acceleration is ##\mu g##, so time would be ##\frac {v \sqrt 2} {\mu g}##. This agrees with the textbook solution. What I do not understand is the trajectory...

This is the problem set. I am stuck from this point... If anyone could give me a hand I would really appreciate it. I know this is probably really simple, but I don't know any of this and have been trying my best with youtube, and other peoples posts. PS this is for high school

The distance covered by the first box is :s1max=v²/2|a|=v²/2μg where a=-μg by second newtons law Similarly S2max=(2v)²/2|a|=4v²/2μg It gas to be s1max+s2max≥S => v²/2a +4v²/2a ≥s => 5v²≥2aS =>v²≥ 2μgS/5=> v≥√(2μgs/5) But this is in the possible solution, am I wrong somewhere? I appreciate your help

My initial approach to this question was breaking the components of acceleration in the x and y axes and applying the three equations of motion to find the final velocity as well as the final position. As we were expected to find the net final velocity of the particle, I found the resultant of...

I'm using rigid body dynamics/kinematics in robotics stuff but I don't have a background in mechanics, I'm interested in understanding the kinematics of frame transformations for rigid bodies. Suppose we have two reference frames fixed on a rigid body, F_1 and F_2 and a transformation T which...

The answer should be 10 m/s^2 but I don't know how to solve it

The equation that connects final velocity with distance traveled is ##v_f^2 = v_i^2 + 2a \Delta y## Since the system starts from rest ##v_i = 0## and the above equation becomes. ##v_f^2 = 2a \Delta y## Since there is rotation in this system we need to connect ##a## to the rotation of the...

my attempt: i solved it all correct but i dont understand a few things mentioned above... 82.04 * v = 56 so i got v as 0.68 m/s which is correct but i dont understand the concept...

I think I have completed the exercise but since I have seldom used polar coordinates I would be grateful if someone would check out my work and tell me if I have done everything correctly. Thanks. My solution follows. Since ##\left(\frac{x}{a}\right)^2+\left(\frac{y}{b}\right)^2=1## it follows...

ay = -9.8. Vy = -9.8 + v0*Sin0. y = -4.9*t^2 + v0*t*Sin0. By using formula of y, my solution was -4.9(10s)^2 + (10m/s)(10s)Sin0. I assumed that y was equal to 0. Since -4.9 is m/s^2 the answer would be -490m + 100m*Sin0. Therefore I assumeed Sin0 = 490/100.

I tried using the formulas x=xi+vit and y=yi+voyt-1/2g(t^2) I assumed voy would be 0 and I almost arrive to the answer but idk how to get rid of the negative

this is how far i have come with my model, i am trying to first the most simple model, meaning no friction involved and then testing that against an actual stick falling by using tracking software. I am currently stuck as my model still has an acceleration in the y direction that i cannot seem...

A ball is thrown with an initial speed vi at an angle 𝜃i with the horizontal. The horizontal range of the ball is R, and the ball reaches a maximum height R/8. In terms of R and g, find the following. (a) the time interval during which the ball is in motion: Sroot(R/g) Correct (b) the ball's...

Why is it 10.92 seconds and not 10 Cars displacement = 30*t +1/2(0)t^2 police displacement = 0*t + 1/2(6)(t)^2 30t = 3t^2 t = 10 seconds ???

I propose to you a kinematics problem described by classical physics. Three space beacons A, B and C are 300,000,000 m (approximately one light second) apart. Beacon A emits a bright flash every three seconds. Beacons B and C respond instantly to the flash of Beacon A by emitting...

idk

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...

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...

$$ v_1= 115km/h $$ $$ v_2= 60km/h $$ solution: $$ \frac {v_1 + v_2}{2} $$

$$t_A= 3s$$ $$ s= 100m$$ $$v_A= 5m/s$$ $$v_B=7m/s$$ $$s(t)=7(t-3)+0$$

Hi, I am learning. Rotational Kinematics and I was given this formula in class: a=mgsin(theta)/(m+I/R^2); however, I couldn't understand the professor's explanation of where it comes from. Could someone provide some insights about it? I appreciate any help you can provide.

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

When two remote events are observed/measured to be simultaneous in one inertial frame, the same events will not be simultaneous when observed from a second frame in uniform motion relative to the first. Why is this distinction in the kinematics of light not considered an operational distinction...

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...

what i tried to do is to write y=v_0tsin alpha - 1/2gt^2 and x=v_0 cos alpha tand that t=x/v_0 cos alphai plug t in the formula for y and get that y= x tan alpha - gx^2/v_0^2 (tan^2 alpha -1)since jaan klada said there should be a quadratic equation (because its a parabola) i thought that...

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 realize I can solve the other way too. But I want to solve using the equations ##v = u +at## ##S = ut +0.5(at^2)## and I don't know why I didn't get the right answer. Thank you for your help!

Statement from the text : I copy and paste the portion of the text that am struggling to understand and underline in red the claim the author makes which I can't believe to be true. Doubt : As you can read in the first line of the paragraph and in the one I underlined, the author believes that...

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I'm struggling with section a. This is my calculation: The expression remains depend on the variable t, while in the answer is a concrete number:

My approach is as follows: a = dv/dt = (dv/dx) * (dx/dt) = (dv/dx) * v Putting v = 0: a = (dv/dx) * 0 = 0 m s^(-2) But, what I don't understand is this: If v=0, then dx/dt must also be 0. Consequently, dx must also be 0 at that particular instant. But, we are writing acceleration as (dv/dx) *...

Even though I have all the formulas needed to do this problem, I cant figure out how to get rA, rB and rA' aswell as rB'

So I tried the following: Getting the velocities for x and y V_xi = 5.2cos(30) = 4.5 V_yi = 5.2sin(30) = 2.6 Then I use v^2 = u^2 +2as to get the final velocities before she leaves the ramp: for V_x the final is the same as the initial since the equation becomes V_xf = V_xi for V_y the final is...

all i could accomplish was calculating the distance between P and the ceiling in a horizontal line(6)

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...