Physics Question Work #11?
922769
Okay so start off by looking at the definition of work.\n\n"Work is a force exerted on an object that causes it to have a displacement in the same direction.\n\nW = Fd\n\nW = work (Joules), F = force (Newtons), d = displacement (metres)\n\nBy definition, 1 J of work is done by applying 1 N of force to move an object 1 m.\n\n    * Work has no direction, so it is a scalar quantity.\n    * Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done."1\n\nSo there are two things you could look at here.\n\nW = Fd\n\nHow much displacement is there in one full cycle of something moving in a circle?  Displacement is the change in position so look at the position at the beginning and ending of the cycle.  There is no change in position so d=0 which gives\n\nW=F(0)\nW=0.\n\nNo work done.\n\nAnother thing you have to look at is "Also, the direction of motion must be in the same direction (parallel) as the direction of the force, otherwise no work is being done"\n\nIn uniform circular motion the movement of the object at any point in time is tangential to the circle while the force is always directly towards the centre of the circle so the movement and force are never in the same direction.  The definition of work demands that they be in the same direction for any work to have been done and, since they are not, no work has been done.\n\nShort answer: c. 0 J
827118
We can't use conservation of energy because this is an inelastic collision. Conservation of momentum does hold, so\n\nm1*u1 + m2*u2 = (m1+m2)*v\n\nu2 is zero. So \n\nv = m1*u1/(m1+m2) = .24 kg*u1/2.85 kg = .084*u1\n\nThe projectile's incident kinetic energy \nKEp = (1/2)*m1*u1^2 = 0.12*u1^2 joules\nThe final kinetic energy \nKEf = (1/2)*(m1+m2)*v^2 = 1.42*(.084*u1)^2 = .0100*u1^2 joules.\n\nPercentage = 100%*KEf/KEp
734494
(a) Work done by gravity = change in potential energy of the book. Now change in potential energy = m*g*h2 - m*g*h1=2*10*(10-1.5) =  170J. So, work done by gravity is 170J\n(b)Change in gravitational potential energy = work done = 170J as explained above.\n(c) Potential energy when textbook is released = m*g*h1 = 2*10*10=200J\n(d) Potential energy when it reaches the hands = m*g*h2 = 2*10*1.5=30J\n(e) If U =100J at ground level, work done by gravity = 170J, the work done does not change with change in reference.\n(f) change in potential energy also remains 170J\n(g) U at release point = 100 + 2*10*10 = 300J\n(h) U at hands = 100 + 2*10*1.5 = 130J
986389
Energy = force * distance\n= 100 N * 6 m = 600 J\n\nAnswer is d
748368
first get the time from the horizontal motion  t = .51 / 1.2 s (uniform motion, no acceleration here)\n\nthen use h = 1/2gt^2 for the vertical motion (with initial velocity zero)
800064
For these problems, it would be helpful to draw a free-body diagram to represent all the forces on this problem.  Here you have 4:\n\n1) Weight in the vertical direction going down, let's denote it W\n2) Force, moving in the x-direction, denote it F_1\n3) frictional force, moving opposite of Force, denote it F_2\n4) Normal force, perpendicular place of contact (opposite direction of W), denote it N\n\nNow use Newton's 2nd law to write equations in each direction:\n\ny-direction: \nN - W =0 (because the block does not move in the y-direction).  Therefore:\n\nN = W, where W = m*g\n\nx-direction:\nF_1 - F_2 = 0 [because the block is just about to move when F_1 is applied].  Therefore:\n\nF_1 = F_2, where F_2 = u*N\n\nMaking the necessary substitutions for F_1 and N, you get:\n\n30 = u*(4.4)(9.8).  Solving for u, you get --------->\n\nu = 0.7\n\n--------\n\nHope this helps
351440
well ok. for part a, the acceleration is in the horizontal direction, because its moving horizontally. so the force is just mass x acceleration which is 70*6=420N. but thatz only in the horizontal direction.\n\nfor part b, you have to account for gravity. so gravity is pulling down at 9.45m/s^2, but youre accelerating upwards at 6m/s^2. so that means that youre overcoming the gravity and accelerating at 6m/s^2 MORE than the gravitational acceleration.\nso 9.45+6 = 15.45. this is your actual acceleration.\nso now you just do F=ma again and you get 70*15.45=1081.5N. \n\nhope this helps. let me know if you need more help. and this isnt thermodynamics.
188376
I was a rock star in college physics!   Top of the class!  But I can't remember any of it now.  My teacher always said to start each solution by drawing a picture, then making a list of the varriable you know, and writing down the equations you thik you might need to solve.  It worked for me!\n\nPost your problem, maybe I can help.  Traditional physics problems have not changed much in the last 5 years, maybe I can remember.
741402
a) F = ma \nYou're given F = -6070 N (negative because it's in the opposite direction) and m = 2600 so a = -2.33 ms^-2. Then over 1.76s, the change in velocity is (-2.33)(1.76) = -4.10 m/s. Car velocity is then 34.9 - 4.10 = 30.8 m/s.\n\nb) v^2 = u^2 + 2as\nv = final velocity = 30.8 m/s (from above)\nu = initial velocity = 34.9 m/s\na = acceleration = -2.33 ms^2 (from above)\nRearrange to get s = (v^2 - u^2)/2a = 155.61 m\n\nc) v = u + at\nYou want v = 0. So if the force is constant then acceleration is constant at 2.33ms^-2. \nu = initial velocity = 34.9 m/s\nSo rearrange to solve for time: (v-u)/a = t = 14.98 seconds.\n\nHope that helps. And learn your equations!
833368
find the force of her weight. \nthat should be the answer
858861
1) The gravitational potential energy of water at the top of a waterfall is first converted to kinetic energy (falling) and then converted to both acoustic (sound) and thermal (temperature) energy when it hits the pool of water at the bottom. The temperature of the water in the pool will be slightly higher than the temperature of the water at the top of the waterfall.\n\n2) The thing to remember for this problem is that the energy in a system before some action = the energy in that system after some action:\n\nP0 + K0 + Th0 = P1 + K1 + Th1\nWhere P is gravitational potential energy (mgh), K is kinetic energy (0.5mv^2), and Th is thermal energy (leftover).\n\n22*9.81*4.5 + 0 + 0 = 0 + 0.5*22*2.5^2 + Th1\nSolve for Th1:\nTh1 = 22*9.81*4.5 - 0.5*22*2.5^2 =  902.44 Joules of thermal energy were generated due to friction.\n\n3) Split this problem into two cases, the first 10m and the second 10m. Find the velocity of the crate after the first 10m:\na = F/m = 200/70 = 2.86 m/s^2\nv = sqrt(2*a*d) = sqrt(2*2.86*10) = 7.56 m/s\n\nFor the second 10m, the force of friction on the crate is:\nFf = 0.3*70*9.81 = 206 N\n\nThe net force on the crate is 6N in the direction opposite of movement. The acceleration for the second 10m is:\na = F/m = 6/70 = 0.086 m/s^2\nvf = sqrt(vi^2 + 2*a*d) = sqrt(7.56^2 + 2*-0.086*10) = 7.45 m/s\n\n5) Since watts (W) are measured in J/s and joules (J) are measured in N*m, a watt is the same as N*m/s. Convert lb to N and ft to m to go from horsepower to watts:\n\n550 ft*lb/s * 0.3048 m/ft * 4.45 N/lb = 746 N*m/x = 746 W\n\nPS - You have in your problem description that horsepower is equal to 764 W but that should be 746 W.\n\nI hope this helps!
735831
Did you put your entire homework assignment on here? I noticed you posted a bunch of science questions just now.\nIf you don't have your textbook handy [you really should be looking up these answers so that you can remember and understand them when you have your test/exam], go to the links below and look them up. The 1st is a down loadable physics textbook. The 2nd has animations of physics.
917347
It sounds like you know what to do.  From your description, it looks like a basic statically determinate problem.\n\nYou have a known applied force P, and two unknown reaction forces, lets say A and B for the top and bottom reactions respectively.  You also have two equations of static equlibrium, specifically the sum of the forces in the "x" direction, and sum of the moments about any point you wish.\n\n(Ok, really you also have the force of gravity and the normal force, but those are the only forces in the "y" direction so that equation won't help you, especially since the mass or weight isn't given).\n\nAssuming to the right is positive, assume A is to the left and B is to the right (this is intuitive, but if directions are incorrectly assumed the answer will just have a negative sign, thats all).\n\n                    _ ------------> P\n                    |   |\n                    |   |\n                    |   |     10.5 m\n                    |   |\n                    |   |\n                    |   |\n                    |   |<----------A\n                    |   |\n                    |   |       1.5 m\n                    |_|--------------> B\n                    \n                 (FBD!)\n\nSum of forces in "x" gives: P - A + B = 0\nSum of moments about B: A(1.5) - P(12) = 0\n\nI took moments about an unknown reaction, B, to eliminate it.  Also, I assumed counter-clockwise to be positive.\n\nSubbing in the value of P = 5.40 N into the second equation:\n\nA = (5.40*12)/1.5 = 43.2 N\n\nThen, put that into the first equation:\n\nB = A - P = 43.2 - 5.40 = 37.8 N\n\nThat should do it.  Both come out positive because their assumed directions were correct, however on the Free-Body-Diagram they are in opposite directions necessary for equilibrium.\n\nCheers,
748800
1)c because the forces are equal (a also for the same reason 0 is constant)\n2)b because the box has an unbalanced force in the horizontal direction and to the right\n3)d because the unbalanced force is to the right\n4)d because there is more force to the right\n5)b there is an unblanced force to the right
288780
From the following link, look for "Chapt22, Problem-24".\n\nThis problem is similar, except the speed is 10% instead of 11%. I'm sure you can use the same formulae though.\n\nhttp://www.jca.umbc.edu/~george/html/courses/phys112/2004fall/homework/04fall_phys112_hk8.pdf
822266
Conservation of energy:  KE=PE\n1/2*Mb*Vb^2= (Mb+Mw)*g*h\nVb^2=2*(Mb+Mw)*g*h/Mb\nVb^2=2*(0.01+2)*10*0.1/0.01\nVb^2=402\nVb=20.05 m/s
744862
F(Y)=117.6N becasue W=mg\n\nm=12kg\ng=9.8m/s/s\n12*9.8=117.6N\n\nbut i disagree with the second part of f(X)=55N. though i could be wrong i guess\n\nstart with some trig.\n\nsin^-1(2.4/2.8)=58.99 deg.\n\nyou know your Y component is 117.6(opposite side in a trangle from the angle just found) and you are looking for your X component(adjacent side to your angle) and that tan=opp/adj\n\nso:\n117.6/tan(58.99)=70.68N\n\ni'm not saying your teacher is wrong, but if he is just getting the numbers out of the answer book then the book could be wrong. i see this all the time even in my college books.