# Tag Info

0

From what I remember $h=\frac{2r}{3}$ is indeed the corect answer. This the incorrect part of your reasoning. The component of the mass's weight along the centre disappears only when θ >becomes 90 degrees. At this point, it leaves the surface of the hemisphere. From what I understand your saying that at $θ = 90 degrees$ the radial component of the ...

1

Let me present a slightly different perspective to Alfred's answer, although I'm basically saying the same thing. I suspect you've got hung up on the idea that velocity causes the relativistic effects like time dilation, but the underlying cause is something different. All the weird effects in SR are caused by a fundamental symmetry of spacetime, which is ...

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We need to untangle this a bit but first: the cause of time dilation is the geometry of spacetime which is such that there is an invariant speed c. Now, remember that velocity or speed is not a property of an object; there is no absolute rest. Further, consider the case of three objects in uniform relative motion with respect to each other. If I choose ...

2

Consider a mass-spring system executing simple harmonic motion. If I draw the displacement, velocity and time graph, it would look something like this: You may see that when t=1 second, velocity is maximum and acceleration is zero. Another way to explain this is by using the definition of acceleration. Acceleration = change in velocity/time = gradient ...

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As taught in calculus, to find the local extrema of a function, one solves for the values of the argument where the derivative of the function is zero. So, at the maxima and minima of velocity, the (time) derivative of velocity is zero. But acceleration is the (time) derivative of the velocity. Do you see the answer now?

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At the point the bullet leaves the gun it is traveling 800m/s. You need gravity to run; I guess he could be running in a vacuum. This is with random values. You would have to construct a bullet decription for yourself in a Ballistic Calculator.

-1

because a bullet does not travel at a constant speed, it would most likely shoot ahead of you (more than 400 m/s) but than even out with you untill there is a change in velocity. (this is just a theory from me, and i have no education in the field of physics, im in 9th grade)

7

At low velocities like this you can ignore special relativity and simply add the two velocities. This is really easy to see if you imagine yourself standing still and the Earth moving under you. Relative to you the gun should fire just like you were standing still. This is called an inertial frame of reference. You see the bullet leave at $400\: ... 0 I don't think you can calculate the fraction for molecules with exactly 305 m/s speed. Statistical calculations have an inherent assumption about the 'precision'. Exact 305 m/s speed implies your precision for calculation of speeds has become infinite. You can only determine speeds upto some precision and hence there will be a 'dv' factor in your final ... 0 The impression I got form the equation is that you model your system as being composed of a collection of point particles. The index$i$indexes this set of point particles (as opposed to indexing the components of a vector). One could create a spatial velocity distribution by saying$\vec{v}(\vec{r})=\sum_i \vec{v}_i \delta(\vec{r} - \vec{r}_i)$, where ... 0 I'm just going to quote Wikipedia here: For the case of two colliding bodies in two dimensions, the overall velocity of each body must be split into two perpendicular velocities: one tangent to the common normal surfaces of the colliding bodies at the point of contact, the other along the line of collision. Since the collision only imparts force along ... 0 If I have a velocity which has some component$v_x$in the$x$-direction, then is there any reason for you to assume you know anything anything about the component of my velocity which might be in a perpendicular direction,$v_y$? No. So you can see that it is reasonable to assume that, if you know my$v_x$, my$v_y$is still unconstrained, i.e. you have ... 0 F=M x A= F 25 x 25=6.25 For the explanation is,the formula must be F=M x A=F For example 25 kg. is the mass and .25.00.00 is the acceleration of the speed limit.Now you must multiply the two 25 and the answer is 6.25 0 Note that the speed of the criminals relative to the river ($27\,\mathrm{mph}$) is larger than the speed of the stone relative to them ($11\,\mathrm{mph}$). Then, the resulting velocity would be expected to be in the negative direction 1 All correct. Your values for$ \dot{x}_P$and$ \dot{x}_C$are negative because you defined$ \dot{x}_R$to be positive (both boats are moving against the river). No problem there, this is completely up to you, but to satisfy the requirement of the question you are required to state that motion of the river has been designated by you as positive direction ... 1 I wonder if you're getting mixed up with propagation of waves in a physical medium like a string. If you have a wave travelling on a string then it has a velocity along the string, but the string is also oscillating normal to its length. So if you stretched the string along the$z$axis, as the wave travelled along the string (i.e. the$z$axis) the string ... 0 A photon is the quantized unit of the electromagnetic field. If you have en electromagnetic wave propagating in the x-direction, this must consist of a magnetic field and an electric field oscillating perpendicularly to the direction of travel, and to each other, i.e. in the y and z directions. If you have a wave with a frequency of, as an example, 50Hz, it ... 0 You're confusing the process of quantization with the wave-nature of propagating electromagnetic fields. When you look at a Electromagnetic waves as photons, this means you don't look at their wave-characteristics, and you consider them as particles travelling with the speed of light, and those particle could "hit" electrons and knock them our of the atom ... 0 Yes, for example, an object with a constant applied force and an oppositely directed speed dependent force has a terminal speed. The sum of forces is:$F_{NET} = F_A - kv$Clearly, when the speed is large enough that$kv = F_A\$, the net force is zero and the object stops accelerating - the object has reached terminal speed. See this hyperphysics article: ...

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