How can I make my face look more chiseled (besides by losing body fat)?
Why "besides by losing body fat"? This:And this:are the results of:Low body fatGeneticsGood lightingPhotoediting (mostly to bring out the shadows on the face)Additionally, it looks like a lot of models getting pictures like this will bite their cheeks when the picture is taken, this gives their face a longer look and pulls the cheeks in to make them look more chiseled.Some people do have rounder and wider bone structures in their faces. When you lose your body fat, you'll lose fat from your face and you can see exposed cheek bones, etc giving you a "chiseled" look even with a wider or rounder face.Outside of cutting down on your body fat (sometimes fat can linger in the face as well -- everyone accumulates and loses fat a little differently), you can get plastic surgery to change some bone structure and get your chin, jawline and cheekbones more pronounced.Update due to added detailsMost of a "chiseled" look, I'm convinced, is people sucking their faces in or pronounced cheek bones. Most models that are "chiseled" have a tight little smiles because they're unable to smile big while biting their cheeks. Bigger cheek bones will naturally cause the face to have the caves and shadows attributed to a chiseled look.Additionally, to help your face look thinner, get a haircut that works with your face shape. There isn't a one size fits all. I'm not an expert on this, but I know it makes a difference.Tanning can help, also. Don't go nuts, just get a little darker. Be sure to use sunscreen. I can also make you a bit happier if you experience SAD.For men, a 5 O'Clock shadow can either help or hurt. Depends on your face. For me, it helps. Make up can also help bring out the cheek bones, frame the jaw, and bring out the eyes. It can be done subtly if you don't think make up is your thing.
How long does it take you to fall asleep? Any tips?
My ice hockey team play quite late and it's often 11pm when I get home, but I used to struggle as I was so full of adrenalin that I was too wired to relax. I found this really helped: * Get yourself as comfy as you can in bed and regulate your breathing to match how you would expect it to be when you sleep. Drop your breathing to what you think it is like to be asleep. Once you've got your breathing low and shallow, like being asleep, I found I drop off to sleep much much easier. * Avoid coffee, sports drinks, anything with sugar, glucose or caffeine in from about 7pm onwards. * Take a shower and take your time. Don't dash in and out. The heat and water tends to be pretty calming * Don't list things you have to do tomorrow or run through things in your head at night. Once you're home, it's turnoff time. * On your way home, if you drive, sing your freaking lungs out all the way home to a CD or your favourite radio station. I cannot stress how much this really helps. By the time you get home, all the pent up frustration or stress of the day left when you were belting out Matchbox 20 songs (in my case!). * Ring a friend or family member that you need a catch up with, and chat to them for 30-45 minutes. Simply talking to someone you want to talk to can help relax you and help you vent any frustraion you might have, with someone who understands. Hope this helps and hope you get some good rest!
If the second hand of a clock has length of 10 cm then how much distance and displacement it will cover in 15 seconds?
When you said, seconds hand has a length of 10 cm, that means the seconds hand rotates along the circle with a radius of 10cm.So, in 15 seconds, it travels quarter of the circle or 90 degrees. Forming a right isosceles triangle with sides of 10 cm and an unknown Hypotenuse x.Using Pythagorean Theorem,x^2 = 10^2 + 10^2 [Side^2 + Side^2 = Hypotenuse^2]x = 10 Sqrt(2) = 14.1421 cmDistance is the actual distance traveled through the path taken. And Displacement is the distance in the shortest path.So distance traveled by seconds hand is the perimeter/4 = 2(Pi)*Radius/4= 2 (3.1415)*10/4 = 5*3.1415 = 15.7075 cmAnd Displacement = 14.1421 cmGood LuckThanks for A2A
How long will it take me to adapt to my new sleep schedule?
If you are talking about changing the timing of your sleep-wake cycle, that usually takes at least 2 weeks, sometimes more. For example, if you are needing to get up earlier but haven't yet convinced your body and mind to shut down and go to sleep earlier to compensate, that will usually reset within 2-3 weeks. Just make sure you get up at the same time every day, and you should find it easier to fall asleep earlier. However, it sounds like you are asking if you can/will adjust to getting less sleep per night on an ongoing basis, which is more complicated. You may, possibly, become used to sleeping 6.5 hrs, or even less, and therefore may no longer notice feeling tired. However, that doesn't mean it would be good for you in the long run. On average, it takes an adult about 5-6 hours of sleep to take care of basic body maintenance and repair (cell replacement, dealing with physical impact of stress, fighting off infection, etc) and an additional 2 hours for cognitive consolidation (integrating that day's experiences into existing memory networks, processing and storing new information/new learning so that it can be recalled, unconscious continuation in solving ongoing isses or problems, etc. Therefore, most adults really do need at least 7-8 hours of sleep a night. In fact, many of us do better with 8-9 hours of sleep. This last couple of hours of sleep is particularly important for students, and insufficient sleep frequently does cause problems with concentration (as you noticed), as well as with memory/recall of newly learned information, efficient use of time, and other aspects of higher cognition. I'd recommend trying to get back on a schedule of 7-9 hours of sleep rather than trying to adapt to 6.5 hrs. You won't be doing yourself any favors trying to get by on less sleep.
Calc Word prob - Finding max and mins?
I really need help with the last part to this question. A patient's temperature change, T, due to a dose, D, of a drug is given by the following equation. C is a positive constant. T=(C/2 - D/3)* D^2 (a) What dosage maximizes the temperature change? --> I found D=C (b) If the sensitivity of the body to the drug is defined by dT/dD, what dosage maximizes sensitivity? (??????) No idea on part b!! Help??
The length of the seconds hand in a watch is 1 cm. What is the change in velocity of its tip in 15 seconds?
I'll keep it simple. And understandable.Get off your ass. Grab a clock. Watch it.Pun not intended, mind you.The direction of motion of an object at an instant is given by drawing a tangent to the point on the circle it occupies/corresponds to at that instant.Now, envision that with your clock.You'll see that every 15 seconds, the direction becomes perpendicular to the original one. Just try envisioning those tangents. Trust me.As you know it for a fact now, it'll hold true for everywhere on the clock.So, let us start at the instant the seconds hand hits 12. You can envision the velocity to be sharply towards the right.15 seconds later, the needle will be horizontal and the velocity sharply downwards. Follow me?Next thing. For a clock to be accurate, every minute has to be 60 seconds. And for that, the seconds hand has to navigate it's circle in that exact time. Assuming that our clock is pretty accurate (hush, nothing is perfect), we can say that it's tangential velocity is uniform.The circumference of the path traced by the seconds hand is given by2πr = 2π cmIf this distance is covered in 60sec by the tip, the tangential velocity is given by2π/60 = π/30 cm/sLet's denote this by v.Now, add a bit of vector geometry into the mix. Start envisioning a coordinate axis, where horizontal goes along the X and vertical along the Y axis.If you're unfamiliar, or find yourself lost, do consider reading this: Vectors - Mechanics - Maths A-Level RevisionThereforeThe velocity when the needle hits 12 will be v along the X axis. Or, the velocity V = viThe velocity when the needle covers 15sec and becomes horizontal is v along the -Y axis. Or, V' = -vjNow, all we have to do is calculate the change in velocity.∆V = Final Velocity - Initial Velocity∆V = V' - V∆V = -vj - vi or -vi - vjNow, the magnitude of this change is simply calculated by taking the magnitude of the vector ∆V|∆V| = √{(-v)^2 + (-v)^2 +2(v)(v)cos90°} = v√2We know that v = π/30 cm/sSo|∆V| = π√2/30 cm/s