How do you handle sexual tension at work?
I can't seem to get past this sexual tension at work. I am trying to ignore this guy who makes subtle passes at me. I notice he tries to stand close to me when we're talking, like within a foot away from me, and he finds reasons to talk to me about trivial workplace issues. Now, I am starting to feel attraction to him, when I didn't before. The thing is, I don't like the feeling...it's NOT positive, it's a NEGATIVE feeling, as if I'm doing something wrong. :( I can't work with guys who try to make passes at me. I don't know why, but I don't find it flattering at all. It makes it harder for me to concentrate on my job. Yesterday, I avoided this guy like the plague. I am getting headaches just thinking about having to converse with him. I need some thoughtful advice. How do you handle sexual tension when it becomes a negative thing?
What is surface tension work?
Surface tension arises because there is an unbalanced force acting on the molecules near the surface of a liquid. The attractions between molecules in the body of the liquid are balanced on all sides, but near the surface the molecules are being attracted more strongly toward the interior of the liquid than toward the surface. As a result, there is tension. The amount of surface tension is proportional to the strength of the attraction between molecules in the particular substance. Water has a fairly high surface tension because it consists of strongly polarised molecules which strongly attract each other. Oil has a low surface tension because it consists of covalent molecules with very low polarisation, so there is little attraction between those molecules.
Sexual tension at work..?
Alrighty, here's a little background informatiom: I'm 18, never had a boyfriend (er, never really cared for high school guys, only one thing on their minds), and a virgin. That being said, consider this: Sexual tension at work..well, there's one guy, he's my manager, and I know he likes me; he's nice to me sometimes, then acts irritated, and plays with me, teases me, gets way too close to me, etc. I'm not really worried about him; he's married, I'm pretty sure he won't do anything. Now this other guy, he's a senior in high school, and exhibits the typical high school guy qualities (kind of arrogant for no reason other than he's good looking, kind of immature). Now, I know there's sexual tension; he's made it VERY clear in numerous ways what he wants to do (for the sake of time and decency, I won't list them here), along with the "nonverbals"; the licking of the lips, passing as close to me as he can. Sometimes things can be awkward; I avoid him. He's kind of hot and cold like the other guy; he'll be rude one minute, and kind the next. He's got a girlfriend, by the way (typical high school guy). So...ah, what do I do? He has the potential to be a good friend, but I'm afraid the sexual tension is a bit too much for that. Advice?
Work done by tension in the rope?
You have correctly identified what is missing, namely, the distance moved. With that information the answer is trivial. Without it the problem is impossible. So what you have to do is either declare the answer is impossible or you give an algebraic answer with a symbolic distance moved. Resolve the tension in the rope into vertical and horizontal components. Vertical component is 80 sine(20) = 27 Newtons. This is less than the weight of a 10kg sledge and so the sledge will not be lifted off the ground. This means that the vertical component does no work and can be ignored. The horizontal component is 80 cosine(20) = 75 Newtons The work done in moving the sledge h metres is 75h Joules Notice that this is greater than the resistance so the sledge will accelerate. This, however, makes no difference. Work is simply force times distance.
Why is work done by tension always zero?
Net work done by tension of an ideal thread on its adjoining bodies is always zero. Ideal thread means it is massless and inextensible. So for such a thread, tension at each and every point of the thread would be same. Also since the thread is inextensible, so for it to remain always taut, the displacements of both the ends of the thread (along the thread) in any time interval should be equal. Further the tension force applied by the thread on its adjoining bodies are equal in magnitude but are relatively opposite in direction (tension is a force of pulling nature, so it always pulls the objects attached to it).The above discussion concludes following two things:Tension at both the ends of the thread are equal but they act relatively opposite to each other.Displacements of both the ends of the thread along the thread are equal.Therefore work done ( F. ds ) at one end is negative of the work done at the other end. In other words work done (by the tension) at one end is positive and the same at the other end is negative but magnitude-wise both are same. Hence total work done by tension of an ideal thread/ rope is always zero. The statement applies to all the physical systems involving ideal string(s).Regards!
Physics(Work and Tension related)?
This and the next question refer to the following situation: A 6 kg box is pulled across a rough floor by a rope. There is friction between the box and the floor. The tension in the rope is T = 5 N. Consider a time interval during which the box moves a distance of 2 m, and its velocity decreases from 1.5 m/s to 0.8 m/s. How much work is done on the box by the rope? (a) 0 J (b) 10 J (c) -10 J Calculate the work done on the box by friction? (a) -4.83 J (b) -6.91 J (c) -8.11 J (d) -9.25 J (e) -14.83 J
Please help me with tension and work problem!?
W = F x D If you were going to be subtracting weight from the sled, You would be subtracting the force component that is in the vertical direction, which would leave you wanting sin30 * 230. Because the rope does not move the sled in an upward direction, no work is done by the rope in the y component, so we only want to know the work done on the horizontal component which is, as you said cos 30 * F. We do not need to know anything about the weight of the sled, only how much work is done by the rope, and that force is given to us. It is asking how much work is done by the rope, on the sled, and the force that the rope uses, is hte only work the rope will be doing and it is given to us. Distance = 1.5m/s * 15m = 22.5m W = cos30 * 230N * 22.5m
How do you calculate work done by tension on a pendulum?
In Cartesian coordinates: Referring to the diagram in the answer, if [math](x,y)[/math] are the coordinates of the bob, then [math]x^{2}+y^{2}=l^{2}[/math]⟹[math]2xdx+2ydy=0[/math]⟹[math]ydy=−xdx[/math]. The position vector of the bob is [math]\vec{r} =x\hat{i}+y\hat{j}⟹d\vec{r} =dx\vec{i}+dy\vec{j}[/math]. Now, [math]\vec{T} =T_{x}\hat{i}+T_{y}\hat{j}[/math], where, from the diagram, [math]T_{x}=−T\cos\theta[/math], [math]T_{y}=−T\sin\theta[/math]. Finally, [math]\vec{T}.d\vec{r}=T_{x}dx+T_{y}dy=dx(T_{x}+T_{y}\frac{dy}{dx})=dx(T_{x}+T_{y}\frac{-x}{y})=dx(−T\cos\theta+(−T\sin\theta\frac{−lcosθ}{lsinθ})=0[/math],thus proving that the work done by [math]\vec{T}[/math] in an infinitesimal displacement [math]d\vec{r}[/math] is zero.In Polar coordinates: The position vector [math]\vec{r}[/math] of the bob of the pendulum is of constant magnitude equal to the length of the string, [math]l[/math], say. So the dot product [math]\vec{r}.\vec{r}=l^{2}[/math]. Differentiating with respect to time, [math]t[/math], [math]\frac{d}{dt}(\vec{r}.\vec{r})=0\implies 2\vec{r}[/math].[math]\frac{d}{dt}\vec{r}=0\implies \vec{r}.\vec{v}=0[/math], where [math]\vec{v}[/math]=velocity of the bob. Thus, we see that [math]\vec{v}[/math] is at a right angle to [math]\vec{r}[/math]. Now, the tension [math]\vec{T}[/math] is directed along the length of the string, so [math]\vec{T}=-T\frac{\vec{r}}{l}[/math]. The work done per unit time by [math]\vec{T}[/math] is given by the dot product [math]\vec{T}.\vec{v}=-\frac{T}{l}\vec{r}.\vec{v}=0[/math], since as shown previously, [math]\vec{r}.\vec{v}=0[/math].
What is the most tension-free job in India?
I think primary teacher jobs in government schools is the most tension free job in India.1 very handsome salaried 40–50k is the starting salary of a teachers.2 no time bounding a teacher is not bothered about the time when he will start working. Normally it is 8am but you have many excuse and ask to the head teacher why you are late or not coming to to the school. You’re are supposed to be fully authorized for your work. No one can argue you. If you are really your boss.3 no work load. You have some 3- 4 hours to teach your students and you can collect all them and can give some 5–10 questionnaires to learn and rest in rest time. Actually you have full time for you and your family .4 full meal. Government schemes are providing lunch in schools are very important for their teacher. You can ask the cook a good lunch for the staff and teachers.full of joys.if you have time of occasions you can enjoy the teenage girls dance music and thrills of the massoom girls in name of rehearsals on 10–20 days.all these factors show that you’re enjoying and loving your job.
How do you handle sexual tension?
having sex with him wont stop the sexual tension. Sometimes it increases. Atleast with all of my experiences. I LOVEEE sexual tension! Its an awesome high. wwhy ease it??? Do nothing with the guy and let the tension keep going. It feels good, and its not cheating. If the tension gets too much, associate him with disgusting things that disgust you, avoid him, dont think about him, and switch off your feelings. ITs possible.