Muscular system homework help?
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Muscles.  Front View.\n\nhttp://www.infovisual.info/03/008_en.html\n\nSide View.\n\nhttp://www.infovisual.info/03/009_en.html\n\nBack View.\n\nhttp://www.infovisual.info/03/010_en.html\n\n\nHuman Anatomy and Physiology Web Sites\n\nhttp://www.infovisual.info/03/pano_en.html\n\nAnatomy of the Human Body Henry Gray.\n\nThe Bartleby.com edition of Gray’s Anatomy of the Human Body features 1,247 vibrant engravings—many in colour—from the classic 1918 publication, as well as a subject index with 13,000 entries ranging from the Antrum of Highmore to the Zonule of Zinn.\n\nhttp://education.yahoo.com/reference/gray/;_ylt=AsCEJkqrowKCe6irtRntROb0HYkC\n\nhttp://www.bartleby.com/107/\n\nBegin your tour by choosing a system...\n\n\nhttp://www.innerbody.com/htm/body.html\n\nGood luck.\n\nKevin, Liverpool, England.
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well whatever force that is being exerted on the rope, that is going to be the tension. that means that all you have to do is calculate the force.  you just have to realize that not only is the gymnist accelerating, producing force, she is also overcoming the force of gravity. therefore, if you were to draw your free body diagram, you would have 2 forces pointing down, the force of gravity(mass times gravity), and this acceleration force(which would just be mass times the acceleration.  if you add these two forces together, you get the total force being exerted on the rope which is equal to the tension. here is the equation:\n\nT=F + Fg  (where T is tension, F is the force caused by the acceleration, and Fg is the force due to grabity.)\nT= ma + mg\nT= (74)(.11) + (74)(9.8)\nT= 733.3N.\n\nthis can also be done by just adding the acceleration to the acceleration due to gravity then multiplying this new acceleration by the gymnist's mass. either way you will get T=733.3N
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1. At the neuromuscular juncture, ACh released by the synaptic terminal binds to receptors on the sarcolemma.\n\n2. The resulting change in the transmembrane potential of the muscle fiber leads to the production of action petential that spreads across the entire surface of the muscle fiber and along the T tubules.\n\n3.The sarcoplasmic reticulum releases stored calcium ions, increasing the calcium concentration of the sarcoplasm in and around the sarcomeres.\n\n4. Calcium ions bind to triponin, producing a change in the orientation of the troponin-tropomyosin complex that exposes active sites on the thin filaments (actin). Cross-bridges form when myosin heads bind to the active sites on F actin.\n\n5. The contraction begins as repeated cycles of cross-bridge binding, pivoting, and detatchment occur, powered by the hydrolysis of ATP. These events produce filament sliding, and the muscle fiber shortens.
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Look at how the forces break down. For block 1\n\nSFx = m1*a = F*cos(q) - T1\n\nwhere SFx is the sum of the forces in the x-direction, a is the acceleration, and T1 is the tension from the string. For block 2\n\nSFy = m2*a = T2 - m2*g\n\nwhere SFy is the sum of the forces in the y-direction, a is the acceleration, and T2 is the tension from the string. Considering the pulley now\n\nSt = I*a/r = r*T1 - r*T2\n\nwhere St is the sum of the torque, I is the moment of inertia of the pulley, r is the radius of the pulley. Combining these three equations will let you solve for the acceleration.\n\nI*a/r^2 = T1 - T2\n\nT1 = F*cos(q) - m1*a\nT2 = m2*a + m2*g\n\nso we have\n\nI*a/r^2 = F*cos(q) - m1*a - m2*a - m2*g\n\nwhich rearranges to\n\n(m1 + m2 + I/r^2)*a = F*cos(q) - m2*g\n\nSolving for a yields\n\na = (F*cos(q) - m2*g)/(m1 + m2 + I/r^2)
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maybe what your teacher is trying to say here is the body temperature. now search for that answer. it's your homework.
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Simply keep in mind that summation of forces equal to zero.\n\nSum (Fx)=0\nSum (Fy)=0\nlet th\nF1- force o string supporting W1 (weight on the left)\nF2- force o string supporting W2 (weight on the right)\nF3- force o string between the weights\n\nFx=0=F1cos(54) - F2cos(68) +F3cos (13)\nFy=0= W1+W2-F1sin(54) - F2sin(68) -F3sin (13)\n\nwe have 2 equations and three unknowns.\nSince the dimensions or length of the ropes ar enot given we can't employ moments.\n\nlet's use superposition and solve one side at a time\n\nW1=F1sin(54) then F1=W1/sin(54)=37.73/sin(54)=46.6N\nsimilarly\nF2=W2/sin(68)=90.944/sin(68)=98.08\n\nsince \nF1cos(54) - F2cos(68) +F3cos (13)\n\nthen  F3=(F1cos(54) - F2cos(68))/cos (13)\n\nF3=(46.6cos (54)-98.08cos(68))/cos(13)= - 9.6 N\n\n(Negative sign means that the originally assumed 'positive' direction to the left was wrong. F3 is pointing to the right.)
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hey hope this help it has all the answers to you question good luck http://www.bodybuilding.com/fun/behar11.htm
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I think it's easiest to break the body up into "movement groups", as it's a bit more functional than thinking of the individual muscles.  This is how I do it:\n\n1.  Pushing Muscles (includes vertical and horizontal pushing):\n -BASIC muscles invloved:  pectoralis group, anterior and lateral deltoids, triceps.\n -Examples:\n     bench press (horizontal push)\n     military press (vertical push)\n\n2.  Pulling Muscles (includes vertical and horizontal pulling):\n-BASIC muscles involved:  trapezius, latisimus dorsi, posterior deltoid, rhomboids (major and minor), biceps, brachialis.\n-Examples:\n     seated row (horizontal pull)\n     pull-up (vertical pull)\n\n3.  2-Leg Movements:\n-BASIC muscles involved:  quadriceps group, hamstrings group, gluteals, adductor group, abductor group, calf muscles, muscles of the anterior lower leg.\n-Examples:\n     squat, stiff-legged deadlift\n\n4.  Single-Leg Movements:\n-BASIC muscles involved:  same as 2-leg exercises, more emphasis on gluteals, hamstrings, and the stabilizers of the knee.\n-Examples:\n     lunges, step-ups, split squats, single-leg stiff-legged deadlifts\n\n5.  Trunk Muscles:\n-BASIC muscles involved:  rectus abdominis, internal and external obliques, transverse abdominis, quadratus lumborum, erector spinae group (illiocostalis, multifidus, etc.).\n-Examples:\n     crunches, plank-holds, reverse crunches, quadraplex.
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Its tricky work, i need digrams to explain to you so your better just following this link, I think it will help you out.
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The vertical, horizontal components of all forces should balance.\n       Let the tensions in the ropes be T1 and T2 newtons.\n       T1*cos(90-71) + T2*cos(90-28) = 254.8\n       i.e., T1*sin71 + T2*sin28 = 254.8\n\n       For horizontal componets (which are opp. & equal)\n       T1*cos71 + T2*cos28 = 0\n\nSolve for T1, T2 please.
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Try this:\n\nhttp://education.yahoo.com/reference/weights_and_measures/
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muscle = yes\nmuscle fiber = yes\nmyofibril = yes\nactin filaments = no\nmyosin filaments = no\nsarcomere = yes\nZ line = doesn't shorten...they get closer together (the two z lines get closer)\n\n\nI'm not sure this is all 100% correct...but I hope it helped a little.  You can check our the websites in my sources for more info
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The contraction phenomenon in muscular cells are mostly due to interaction of two elements, ie, Actin and Myosin. Their interaction are powered by some molecules called ATP. The ATP is produced in some cellular structures called mitochondria. The role of food for human is to produce energy. Energy is produced by hydrolyzing of ATP.
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arm muscles mostly\nthey work by contraction and expansion directed by brain synapses
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Another homework question.
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E1=M*g*h, M=(your mb)=32kg, - pot energy of the system at rest,\nh=2m and g=9.8m/s^2.\nE2=kinetic of M=32kg\nE3=kinetic of m=20kg\nE4=kinetic of pulley u=5kg\nE5=potential energy of m at the height h, E5=m*g*h\nThus energy conservation: E1=E2+E3+E4+E5.\nE2+E3=(M+m)*V^2/2, V to be found.\nE4=I*W^2/2, angular speed being W=V/r, r=0.1m, and I=u*r^2/2 for your pulley.\nHence 2*M*g*h=(M+m)*V^2+(u*r^2/2)*(V*r)^2+2*m*g*h,\n(M+m+u/2)*V^2=2*(M-m)*g*h, - does not depend on r.\nV=2.08m/s, don’t believe & check it! – good picture gal – a piece of art.
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http://www.studentquestions.com/homework_ans/4244.html