# Tag Info

2

I think you should take the sample velocities and divide them by the respective times after minusing the previous velocities to obtain the accelerations. If the time interval between calculating the discrete sample velocities are too small, then the above-got values may be taken as instantaneous acceleration. Now plot these over graph wrt time. Now here we ...

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A fairly simple way of treating the data is to present them as a histogram: Each data point (here 5 data points) is the quotient of the distance moved in that interval, say $\Delta y$, by the time interval $\Delta t$ and is the average velocity during that time interval: $$v_i=\frac{\Delta y_i}{\Delta t_i}$$ Where $i$ indicates interval number $i$. For ...

2

When deriving the wave equation we assume the horizontal component of the tension in the string is constant and equal to $T$ (the tension when the string is at rest). To calculate the tension in the string let's start with the wave then zoom in to a small segment of it. If we take a segment small enough that we can consider it as a straight line, then the ...

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Let two orthonormal systems $Oxyz$, $O'x'y'z'$ with a general motion (translational plus rotational) between each other and a point particle $\rm P$, see Figure. Symbol Conventions : 1.The vectors for position $\mathbf{R}$, velocity $\mathbf{U}$ and acceleration $\mathbf{A}$ of a particle with respect to $Oxyz$ expressed by coordinates of this same ...

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The tension of the string is a constant, if there is no vibration on the string. A wave is produced on the string when you give an unbalanced force on the string which varies the original tension of the string. The velocity of the wave now depends on the value of the tension. The given equation is valid only for small amplitude vibrations. The tension is ...

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Do u really think that velocity is constant?i think in this equation nothing is constant.if tension increases per unit mass decreases and it may make change in velocity or not if the ratio remains the same.further tension depends on some variables such as intermolecular force,elasticity etc

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The real answer is quite complex; I think we should break it into a couple of different pieces. First - the static case. If you submerge an open pipe into water, the pressure inside and outside will be the same at a given height, and the water level inside the tube will settle at the same height as outside. If you add the effect of surface tension, it is ...

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Before valve-A is opened, the pressure in the tube is atm. Pressure since the top of the tube is opened. When the valve-A is opened, assumed instantaneously, the water will rush into the tube at: $$V = \sqrt{2gh}$$ where: $g$ = gravitation const (32.2 ft per sec per sec.), $h = 100\ \mathrm m$ = Height from bottom of tube to tank ...

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Sorry, didn't complete my answer above. The area A of the 20 m exit hole you mentioned is 313 m^2. To determine that calculate the area of a circle with a 20 m diameter. Example: A = .785(20^2) = 313. If you want to calculate the vol. of water mult. The V by the A as follows: Vol(m^3) = 26(m/s) * 313(m^2) = 8,112 (m^3/s)

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So we have a tower (tube) 5000 m (16500 ft) tall. The bottom of the tower has a hole 20 m ( 66 ft ) in diameter. To determine the pressure the column of water (assuming specific gravity = 1.0) exerts on the bottom of the tower at point of water exit is: P = (H * SG)/2.31 where: H = height of the liquid in feet. SG = 1.0 ...

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In this case, the ansatz is to equate potential and kinetic energy. The kinetic energy at zero height will be converted into zero kinetic energy and maximal potential energy. This gives $$\frac12 m v^2 = m g h \,,$$ where $m$ is the mass, $v$ the initial velocity, $g$ the gravitational acceleration and $h$ the maximal height. You are interested in the ...

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If you neglect air resistance the mass of the golf ball does not influence the range with same initial velocity , only with air resistance it does. But if you hit the ball the kinetic energy you transfer gets divided through a smaller mass, so the velocity of the lighter ball should be higher if hit both balls the same way: $$e_{kin}=m\cdot v^2/2 \to ... 0 In this case, it is the momentum that must be considered. Impulse is defined as the change in momentum of an object. The golf ball will always start with 0 momentum. If we assume that all of the clubs momentum is imparted on the ball (unlikely, but simplifies the math), then they both receive the same impulse. The impulse is also equal to force multiplied by ... 1 Harmonic motion is sinusoidal: x(t)=x_0\sin(ωt)=x_0\sin(2πt/τ). The argument of the sine function is called phase, here an increasing funtion of time: x(t)=x_0\sin(φ(t)) with φ(t)=ωt=2πt/τ. Hence, a phase difference δφ corresponds to a time difference δt=δφ/ω=δφ\,τ/2π. With δφ=π/2 you find δt=τ/4. Remember: 1\ \text{period} ↔ 2π ↔ τ 2 If you draw similar triangles, then you'll find that r_A/r_B = v_y/v_x, and so the product r_A v_x is equal to r_B v_y. Try drawing a line from the tip of your lower \vec{v} vector to the tip of your lower v_y component to see this. 1 Hint: Since the car begins from rest and keeps accelerating, the equation d=vt is no longer valid, because velocity is not constant. Note that you can, however, say that d=v_0 t + {}^1/_2 a t^2 = {}^1/_2 a t^2 and the acceleration will be the same for both time intervals. Hope that gives you a nudge in the right direction! Also, these equations may be ... 0 A particle following a prescribed path has its velocity vector parameterized as$$ \vec{v} = \vec{e} \,v $$where \vec{e} is the tangent vector and v is the speed at that instant. This is kind of obvious. But you use the above to find the tangent vector if you know that radial vector \vec{r}. Use \vec{v} = \frac{{\rm d}}{{\rm d}t} \vec{r} = \vec{e} ... 1 As per the comments, I wasn't taking into account the relativistic addition of velocities, which is becomes relevant when designing scenarios with such high velocities. So for a observer in the point specified in my argument, the fastest objects (object #1 million, object #999.999, ...) would appear to have velocities close to light speed, but they would ... 1 To add to Dilithium Matrix's answer, to simplify, you can look at v_1 and v_2 as any vectors, not necessarily velocity vectors. We know that if a = b and b = c, then a = c. Hence, to say that \theta = \beta is to say that \alpha = \beta. You can then use geometry to show that there is only one other case where, for O' \neq O, \alpha will ... 2 No. Trivial counterexample would be if the origin, o, is just below the tangent point for v1, directly below it. In that case, beta clearly can be 90 degrees or more, even if v1 and v2 are actually in the same direction! 0 If acceleration is linear, use the 'kinematic equations'. If not, use differentiation/integration (in this case, you will probably be given a formula which expresses acceleration). 1 All you have shown is that the instantaneous velocity of the particle at point D is can be equated to the translation of an arbitrary axis together with a rotation about that axis. The statement "while \vec \omega is independent from the chosen rotation axis, v depends on it" refers to all the particles in the body. So the all have the same ... 0 The velocity of any point is always same. 0 The initial answer is exactly right, and your response shows you understand that the dot product of the force and velocity will give you a scaler value that leads to the right answer. However when I read the question, I was impressed at how hard it is to visualize what is really going on, and the incurable teacher in me thought you might be interested in ... 0 Joules per seconds is Power. Power is work done per unit time. (J/s) Work is Force x Distance x Cos(angle) Dividing by time to get power yields Force x Distance / Time x Cos(angle) Velocity is Distance / Time and therefore Power = Force x Velocity x Cos(angle) Since the direction of the application of the force doesn't matter in this problem you can ... 0 It does indeed have something to do with time dilation. You can use the formula$$s=\frac{v+u}{1+(uv/c^{^{2}})} where s is the speed of one spaceship relative to the other while u and v are their speeds relative to the Earth. I think you will find whatever values of u and v you use s will always be smaller than c.

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The position-time graph you've sketched looks roughly right (though be careful to make the scale on the vertical axis consistent). The data don't look at all linear and don't justify a straight line. The second graph has a couple problems. Is it a speed vs. time graph or a velocity vs. time graph? In what you've written you refer to velocity so, if ...

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If you plot the time versus position of 'the high point on a see-saw', there is an abrupt change when end A goes from high to low (while end B goes from low to high). This is not entirely trickery, there are lots of useful items that exploit some kind of discontinuity (a toggle switch or an electronic astable 'flip-flop'). What happens when light ...

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As you said, the next derivative of the velocity with respect to time is the acceleration. And the acceleration could in principle have a step somewhere due to a force starting to act on the object.

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In your solution you have found an acceleration and then used a constant acceleration kinematic equation. In terms of energy think about what $\int F_x\;dx$ might be equal to. PS Multiply your constant acceleration kinematic equation right through by $\frac 1 2 m$ and you might notice something interesting?

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This is a "famous" problem for showing what different things "average" can mean. If "the first half" means half of the time, then your distance will be $(v_1+v_2)*t/2$, and therefore the average velocity just the arithmetical mean $(v_1+v_2)/2$. If "the first half" means half of the distance, then your time will be $\frac{s/2}{v_1} + \frac{s/2}{v_2}$, ...

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