Ask the Physicist! I have asked readers to suggest answers but nobody ever has. ADDEDANSWER. I have found the answer. The center of mass. Then, as shown. above, if the beam is off horizontal for the empty scale or equal.

Nomenclature. Khat goes by various traditional names, such as kat, qat, qaad, ghat, chat, Abyssinian Tea, Somali Tea, Miraa, Arabian Tea, and Kafta in its endemic.

QUESTION. I made a statement to somebody that a plane hitting a building was the same as if the building hit the plane at exactly the same speed,the plane now stationary. The results would be the same. In other words, if a man with large hands slapped my hand at 5. Let me try to set up a simple example to. Imagine we have a 2 lb ball of putty moving.

Play a game of Kahoot! With the recent publication of PHYSICS IS. Click on the book images below for information on the content of the.

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Swedish zoologist Carl Linnaeus first described the genus Elephas and an elephant from Sri Lanka (then known as Ceylon) under the binomial Elephas maximus in 1758. Even more » Account Options. Sign in; Search settings. Retrouvez toutes les discothèque Marseille et se retrouver dans les plus grandes soirées en discothèque à Marseille.

After the collision. To. find v. 1, use momentum conservation: 2x. Next, imagine we have a 1. After the. collision, the two move together with a speed of v. But. suppose that you were the putty ball.

During the collision you feel a. Do you get hurt as. What. determines the force you feel is the acceleration you experience during.

For the. putty ball moving initially, (vfinal- vinitial)/t=(0. For the. bowling ball moving initially, (vfinal- vinitial)/t=(0- 4. You could go. through through the exact same process to find that the bowling ball. A. physicist would say that you were right, but the ambiguity of your. As far as physics.

If the putty ball were. QUESTION. My question is about the maximum tension experienced by a bow string. I'm. specifically concerned with a traditional or recurve bow NOT a compound bow.

I want to know the max tension compared to the draw weight so. I have an idea how strong to make my strings. So here's the scenario, what's. I'm assuming the maximum tension is when. I think this. because not only does the string have to oppose the restoring force of the.

I can. tell you that the tension will be at the maximum draw for a simple bow, not. It Is braced at 6 inches and has tips that are 3 inches recurved behind the handle. Its draw length is 2. This is a reposting, correct.

I hope!). When researching the physics of archery I discovered that this can be a. You, however, require only a rough calculation for estimating. I can do that and it is more appropriate for. This problem requires facility with trigonometry.

Hooke's law, and application of Newton's first law. The. simple model I will use was one used before the advent of computers; the. With this simple model, most of the. When the string is braced.

So, the maximum will be at the maximum. The figure shows, roughly to scale, the situation.

Using simple. trigonometry (law of cosines), I find . I think that your concern about the string having to . Not so. if you draw and release without an arrow, though; what I have read is.

I am working on a general solution which I will later add here. I would post the part of the solution which answers your. GENERALIZED. SOLUTION: To get a better understanding of this problem it is worthwhile to. My research showed me that a traditional or recurve bow. Hooke's law), the draw weight being proportional to the draw distance. In this case, since W=7. Using the law of cosines, cos.

Therefore T(0)=k. L/2. At full draw, . It was interesting to me that in order for the. The value was. R=3.

L=3. 0. QUESTION. Suppose a block is moving with constant velocity towards right on a frictionless surface and during its motion another block of slightly smaller mass lands on top of it from a negligible height.

My friends can't accept my reasoning. Please help! This is actually a simple.

Call the masses of the upper and lower. M, respectively. Before they come. Mv where v is the incoming. M. When the masses come in contact they will slide on. M+m)u. Conserving momentum, u=. So, choosing +x to the.

It would be a good exercize. And then, if this is greater than the size of the bottom. QUESTION. I do not want a theoretical answer, but has any experimentalist ever put a very sensitive weight balance below a vacuum chamber before and after vacuating it? Does it get lighter or.. I do not have a sensitive balance nor a vacuum chamber. I simply do not know. Let us assume the.

Envision just a simple string. M and volume V. from the string. Besides the string, there are two forces on the object.

Mg and the buoyant force B=. The scale will read W=Mg- B, an.

Putting the whole device in a vacuum. B to zero because the air is gone, so W=Mg. To get an idea of how important this is, consider. The density of iron. However. there are certainly examples where the effect of buoyancy would be very.

For example, consider an air- filled balloon. I did a rough. calculation and estimated that the volume of an inflated balloon is. N. But if you weighed it in air you would only measure half that amount. I stopped for a break. Since you were estimating, the pendulum was swinging with.

I. would conclude that either the pendulum got swinging somehow and the. If the building was swinging with a period. QUESTION. Let's say you have three 2. Same stimp meter (pace on the green). Let's say in all cases, the green is flat after the hole. In that case vhole=.

Since the frictional. But, what matters is not the force but its. Wf=- fs where. the negative sign indicates that the friction takes energy away from the.

When the ball is moving forward along along the segment L2. L2/cos. The total work done by. Wf=. Choosing the man plus pulley above him as the body. Newton's second law is.

T- mg=ma or T=. Finally sum forces for the weight Mg which has the. Mg- T=Ma, so. M=T/(g- a)=5. You may want to keep in. Install Hp Printer Officejet 6700 Premium. Refer to the figure above.

I have made the boat on the trailer invisible!): I have. S as the distance between the wheels, L the. COM . The normal forces Ni. N1+N2+N3=W. I have also assumed that the. COM is centered laterally. The geometry and algebra are a bit tedious. I give only final answers.

In each case I imagine lifting one point. N'3. is the easiest since it does not depend on . As an example, suppose. S=3 ft and q=5 ft and you lift a wheel h=4. The culprit here is that the center of gravity is so. To get a more accurate measurement, block the. N1=N2 and N3. does not depend on its height.

FOLLOWUP QUESTION: Also, in your table weight example, wouldn't elevating one end of the table with a 2x. Co. G) somewhat toward its other end and lessen the measured weight of the weighed end, and error be particularly noticeable if the span of the table is short and the scale is several inches high?

If I could remember my trig I could pose this question more intelligently, but the extreme, if hypothetical, case would be if the table end were elevated all the way to vertical, where it would weigh nothing. I must admit that I was thinking small angle here and that is. And, again, I did not. COM is not. colinear with the other two forces until I started puzzling over your. You can see an example of a case where this is. The correct. answer to the table weight. W'=W. In particular, I have assumed that.

COM is a distance q above the floor and equidistant from. Show Me Love America Torrent. I have neglected the weight w of the 2x.

Here are the numbers. Second slope : 2.

Meter high, 2. 5 degrees, friction. Obviously the final speed will not be the total of both speeds, so how. I add the speed of the first slope to the second slope to obtain. ANSWER: Well, I started to work this out and found that the length of the two. I therefore surmise that this is. At 5. 1, I am surely not going back to school : ) In fact the lenght I gave you are not precisely the one I will use for the slides.

I want to understand the way to make the right calculations because the first incline has to be steeper to reduce the overall length of the slide. I am a conceptor mostly in transportation but this project is for a waterpark. Total energy of a mass. E=. For a. given system the energy is conserved (the same everywhere) as long as. If there. were no friction in your slides, energy would be conserved. In your case, v=. Alas, there is friction and the.

The amount. which a force changes the energy if it acts over a distance s. Wf=- . Now, instead of writing E2=E1. E2=E1+W—the. energy you end up with equals the energy you started with plus what you. Now we can address your specific. Solving for the velocity, v=7. The formula you seek if you are.