# Tag Info

1

I agree with kevinsa5 that the variation in $a$ is due to rounding errors, but I'd like to suggest a better way to analyse the data. Generally speaking, the best way to analyse data is to find a way to convert it to a straight line, then you can graph it and do a linear regression. In this case the way to procede is to note that if the acceleration is ...

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The best thing to do is draw a graph of velocity versus time. Knowing the acceleration is constant, you draw the best straight line. The inclination will give you the acceleration. You can even draw the maximum inclined and least inclined line to determine your uncertainty.

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The object will continue accelerating to just below light speed (speed limit of the universe), but only in a vacuum. Unfortunately I only know a little about air resistance, but apparently the faster an object travels the more the atmosphere tries to resist the object. This is where aerodynamics comes in. I'm sure you're familiar with the classic image of ...

2

You are right in that gravity did not change during data collection. You are a victim of uncertainty, which is a very important part of experimental physics. I'm sorry in advance for the "wall of text", and I hope that this clears up some confusion. The problem is that $1.50$ may not be exactly $1.500000000...$. Because the numbers are provided rounded, ...

0

I used to think the answer to this question might be something like: c^2 (reasoning that if c is the universal speed limit for anything (energy, matter) that moves, it would probably be unreasonable to expect that changing velocity faster than, say, from c to zero or the reverse in a time interval that is less than the Planck time interval, would be a pretty ...

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Assuming there are black holes with singularities, then the acceleration can be as a high as you want inside a black hole, the larger the closer you get to the singularity.

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Landau's hint goes as follows: Four-velocity is $$u^{i} = \frac{dx^{i}}{ds} = \gamma_{v}(1,\frac{1}{c}\mathbf{v}),$$ where $$\gamma_{v} = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}.$$ We have used $$ds = \frac{cdt}{\gamma_{v}}.$$ To proceed, we note that $$\dot{\gamma}_{v} = \frac{\gamma^{3}_{v}}{c^2}\mathbf{v}\cdot\dot{\mathbf{v}},$$ therefore $$w^{i} = ... 0 Considering forces on each mass separately :$$F_1-T=m_1aT-F_2=m_2a$$adding these 2 equations :$$F_1-F_2=(m_1+m_2)a\Rightarrow a=\frac{(m_1-m_2)g}{m_1+m_2}\Rightarrow >a=9.8/3=3.27$$1 Your m should be m1+m2....... 1 Hint: The floor goes down at 13 m/s^2, the person goes down at g \approx 9.81 m/s^2, so the net acceleration is? 1 What is angular speed? Clearly it is \frac {v_\perp} {r} where symbols have their usual meanings. Rod rotates about its, say, rightmost point, say O. We will consider left side as positive x-axis. Now consider a point A at distance r_1 from it. Let the rod have instantaneous angular speed \omega. All points on the rod will have this \omega ... 1 This is correct, assuming constant acceleration, we have for this problem$$a = -\frac{v^2_i}{2d} = -\frac{(6.26)^2}{0.05} = -784\frac{m}{s^2}$$First, I applaud you for asking the question. Too often, I have graded homework and tests where numbers were submitted for answers without any thought as to whether they were reasonable. However, this is a ... 1 Well, let's think about this: An object is traveling at 6.26 m/s during impact and travels only 0.025 meters before stopping. The force causing the object to decelerate needs to be extremely high. It's just like having a force being applied for a very short period of time, such as a bat hitting a baseball and the time of contact is extremely small, you ... 0 The problem as is stated is somehow ambiguous, but using some simplifications we can manage to get something: if we assume that forces don't change depending of the angle (i.e. there is no "correcting trayectory rocket" that acts depending of its orientation), and that the center of mass is fixed, then you can express net force F' and net torque T' with ... 5 One must distinguish between instantaneous velocity and average velocity. 1 Let's take a look just from the point of view of someone reading the problem. First of all, we can say "our car has motion", because it's changing its position each second. Ok. So, how is its motion? Well, it is moving in 1 dimention, it is a linear movement. Then, we can say "our car has linear motion". Also, we can see our car's velocity is changing ... 0 in v-t graph we see how velocity changes as time passes. if the graph is a straight line making acute angle with time-axis (x-axis), it means velocity is increasing at a constant rate. Now the problem here is, if i understood you correctly, is not thinking about it without using mathematics. In your question, the acceleration is constant and it is 10 meters ... 2 For some vector function \vec {v}(t) that gives the velocity: \vec{a}(t) = \dot{\vec{v}}(t) If velocity is not given as a vector function, then you need additional information to find the direction of the acceleration. 1 Take the velocity vector a two points separated by an infinitesimally small time interval. Subtract the later vector from the earlier, and divide by the infinitesimally small time interval. The resulting vector is the instantaneous acceleration (in the limit that the time interval goes to zero). The direction of that vector is the direction of the ... 1 The car probably experiences a constant acceleration of 10{m/s^2}. You can see from the chart that the velocity follows this as after every second the car is going 10{m/s^2} faster. However, this is clearly not the whole picture. We do not know the acceleration at 1.5 seconds, or 1.55 or 3.14. We can get confirmation that our acceleration model works ... 0 You must consider the acceleration that is acting on the object. Let me give you an example, a ball falls from rest to the floor. This ball is 5m above the ground and the acceleration on the ball due to gravity is 10m/s^2. When the ball is let go, clearly the ball is not moving at 0m/s which is what it would be if there were no acceleration, it would just ... 9 The velocity is constantly increasing due to a constant acceleration. Exactly at 1 s the velocity is 10 m/s, but this does not mean that velocity was at 10 m/s in preceding second. In fact, given the distance 5 m moved in this second, the average velocity in this second was 5 m/s. And this should make sense to you, because in this first second the velocity ... 0 Velocity is a vector, meaning it's direction must be taken into account. Acceleration is the change of velocity over time. On the crest the direction of the velocity changes continously, so there's an acceleration. In your hand drawing the sled passes over a horizontal terrain. Velocity does not change direction so there is no acceleration on the flat. 0 You are right that the difference between a crest and a trough is qualitatively just a mirroring. However, there is another subtle difference between the two cases. The skier is only on "level" ground for a single point along the path, while the sled on the hill is on level ground for an extended region. That is, your paths are not truly mirror images of one ... 1 This has a simple closed-form solution. Denoting m_0,m_1 as the initial and final person's mass, v_r as the rice speed and \delta=m_0/m_1, if the bag is thrown in one single parcel, we have$$\Delta v_1=(\delta-1)v_r$$By the rocket equation, if the rice is thrown continuously, we have$$\Delta v_2=v_r\text{Log}(\delta).$$But$$\text{Log}(\delta)\leq ...

1

You need a model for how you throw the rice. The obvious one is that you can expel any mass at the same velocity $v$ relative to you. Letting $M$ be your mass (without the rice), $V$ your velocity in the CM frame, if you throw it as one lump we have momentum conservation. You start with no momentum in the CM frame, so $10v=MV, V=\frac {10v}M$. If you ...

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Given a bag of rice of mass $m_b$ that you can throw with a maximum acceleration $\vec{a}_b$, by Newton's second law, the most force $\vec{F}_b$ you could exert on the rice is given by $$\vec{F}_b = m_b \vec{a}_b$$ By Newton's third law, the reaction force (acting on you) $\vec{F}_{you}$ is given by $$\vec{F}_{you} = - \vec{F}_b = - m_b \vec{a}_b$$ Again ...

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I believe that the reason why specific impulse has the units seconds is to prevent confusion between people who use the metric or imperial system. Because the "efficiency" of a rocket is actually indicated by the effective exhaust velocity: $$v_e=\frac{F}{\dot{m}}$$ However this has an unit of length, which is different between the metric and imperial ...

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The Wikipedia article you link states: For all vehicles specific impulse (impulse per unit weight-on-Earth of propellant) in seconds can be defined by the following equation ... The quantity $\dot{m}g_0$ is the weight flow rate of the propellant when the local gravitational acceleration is $g_0$, so the weight-on-Earth bit of the definition implies ...

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