crushing strength question?
486807
You will get 3 different answers until you specify which surface of each brick contacts the one above/below.  The answer comes out of the equation for the pressure each brick adds to the pile. (P=F/A)  Don't forget the lowest brick; it's own weight will cause the bottom of it to be under more pressure than the top!
761342
We did something like this in my physics lab with an Atwood machine.  It was a pully with weights hanging from it.  I can't exactly remember how it worked, but maybe this will inspire you.  We also did something with an air track and these two masses colliding and we'd see which one would go in which diretcion.  You need a nearly frictionless surface, like an air hockey table or ice slab.
948476
I won't show in details since it's hard to type out the work, but you'll get the rest, here it goes:\n\na)\nWork = mgd  *d=change in height\nW = (10kg) (9.8m/ss) (sin15) (-7.5m) = -190J\n\nb)\nW=(friction coef)(natural force)(distance traveled)\nnatural force = (cos15)(10kg)(9.8m/ss)\nW= (0.4) (cos15) (10kg) (9.8m/ss) (-7.5m) = -280J\n\nc) \nW=Force x distance\nW=(100N)(7.5m)=750J\n\nd)\n(initial KE)+ (puller work)=(gravity work) + (frictional work) + (final KE)\n(final KE) - (initial KE) = 750J - 280J - 190J = 280J\n\ne)\n(final KE) - (initial KE) = 280J\n*solve for final v\nv = squareroot (2 x [280j+ (1/2)(10kg)(1.5squared)] / 10kg) =7.63m/s
97685
Assuming the blocks were made from the same batch of iron, they would have the same density. So a 2-kg block of solid iron would have twice as much mass, weight, volume, and potential energy (if the block are the same distance off the ground).
747946
It's a bit hard to tell without the drawing, but I think I can see how this is set up.  Start by calculating the applied force, based on what you know about the spring.  F = -kx, so the applied force is 0.053 m, the compression of the spring, multiplied by 470 N/m, the spring constant; the negative sign can be ignored in this case, because we're dealing with magnitudes, not vectors.  Now, if the block doesn't slip, the upwards friction force must equal the downwards gravitational force.  Call the spring force F_spring.  The friction force, then, would be F_friction = u*F_spring, where u is the unknown coefficient of static friction.  The force of gravity, the weight, would be W = mg, where m is the 1.9 kg mass of the block and g is gravity, 9.8 m/s^2.  So F_friction = W ==> u*F_spring = mg ==> ukx = mg, and u is your only unknown.
743532
.  Concrete is much stronger against compression than it is against tension.  When concrete is bent, the tension on one side will increase, while the compression on the opposite side increases.  The concrete will fail when the tension reaches a few hundred pounds per square inch.  It will not fail in compression until it reaches 3000 to 4000 or more pounds per square inch.\n.  Post tensioned cables, installed in the concrete before placing the concrete and pulled to high tension after the concrete has hardened, put the whole piece of concrete in compression. Lets call this pre-compression compression on one side, and negative tension on the other side of the concrete. When this concrete is bent, tension will not develop in the side which would normally go into tension until the negative tension placed on it is overcome by the bending force (called the bending moment) increasing the tension on that side to a positive value.  So more bending moment can be applied to the concrete before it starts to crack.
931774
.  I do not know what you mean by resistance.  It is something I have not heard of measuring in concrete.\n.  There is no meaningful relationship between compressive strength and density.  Concrete all comes out to about 145 to 150 pounds per cubic foot, for ordinary concrete, less for light weight concrete, which is made light weight by using porous aggregate.  Lightweight concrete, depending on the mix designs, can be stronger than normal weight concrete.\n.  The strength of concrete is determined by the cement content.  The cement is not the same thing as concrete.  The cement is the stuff that makes it all stick together.  Concrete also contains sand gravel, rock dust and water.  Too much water weakens the concrete by reducing the density of the crystals which grow from the cement as the concrete sets.\n.  The strength of the concrete is also affected by the quality of the cement.  A concrete batch plant has to make trial samples with every delivery of cement to determine how much cement has to go into the mix to achieve a certain strength.  I will vary form one cement delivery to the next.
840601
(a) Gravity exerts a downward force on the block of 9.8 Newtons/kg * (0.25 kg) = 2.45 Newtons. Work done by gravity = force * distance = 2.45 Newtons * 0.12 m = 0.294 Joules\n\n(b) The spring is a bit tougher since the upward force increases as the spring is compressed. Force = 250 N * (C) where C is the distance the spring has been compressed (in meters). Work is then the integral of Force over distance, so Int(250 * C (dC)) from C = 0 to 0.12 M . The integral is 125 * C^2. Evaluation at 0 gives 0. Evaluation at 0.12 gives 125 (0.12)^2 =  1.8 Joules. So the work done by the spring is -1.8 Joules . Note that the work is negative, because the object is moving against the applied force.\n\n(c) When the block strikes the spring, it will have a kinetic energy of (1/2)mv^2 = (1/2)(0.25kg)v^2 = 0.125v^2.\nThis kinetic energy is absorbed by the spring as it compresses. Note that the work done by gravity is also absorbed by the spring.\nTherefore, we have the following equation:\n0.125v^2 + 0.294 = 1.8\nv^2 = 8(1.8-0.294) = 8(1.506) = 12.048\nSpeed of block just before it hits = SQRT(12.048) = 3.47 meters/sec downward.\n\n(d) I'll leave this as an exercise for you. Just apply the same techniques as above. Note that the block hits the spring with a velocity of 2(3.47) = 6.94 meters per second...
995132
The slip load is 2.63 imperial tons\nThe breaking load if required is legally not less than 20kn(2000kg)
767488
Divide the 1000 pounds by the 7 inches.\n\nThis will tell you how many pounds is required for each inch of compression.  Now multiply that by the additional 5 inches.  This will tell you how many more pounds of compression is required for the extra 5 inches of spring.\n                                              \nEdit:  I see this as being 714.285714
427779
Slow it down.\n\nIf you mean 'break' then you will have a job as it is a solid lump of steel - you can break the shaft fairly easy though (personal experience).\n\n\nEnjoy
612059
Consider two objects having masses 1kg and 5 kg are in a horizontal plane.\n\nWe apply a force of 10 newton to the 1 kg mass and 50 newton to the 5kg mass. \n\nThat is, we apply less force for less mass and greater force for greater mass.\n\nNow we find the acceleration of the two bodies.\n\nThe1 kg mass will have an acceleration of 10/1 = 10 m/ s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nSimilarly the 5 kg mass will have an acceleration of 50/5 = 10 m/s^2. That is its speed will be going on increasing by 10meter / second in every second. \n\nBoth have the same acceleration and both have the same speed at any time after start.\n\nNow returning to gravitational pull, let us place the 1kg mass in one pan of the balance and the 5kg mass in another pan. \n\nThe pan having 5kg mass will come down and the pan having 1kg mass will go up.\n\nThis is because the earth is pulling the 5 kg mass with a force of 50 newton and 1kg mass with a force of 10 newton.\n\nIf  we drop the two objects from the same place, as we have seen already the 1 kg mass will fall down with an acceleration of 10 m/ s^2 and the 5kg mass will also fall down with 10 m/ s^2. But the forces acting on them are different.\n\nThe Earth’s pull is such that it gives less force for less mass and greater force for greater mass.\n\nTo be more specific it gives a force of 10 newton for every kilogram of mass and hence the acceleration is for any mass is the same as approximately 10m/s^2.
257076
dude u need to start doin your own hw. ill do it this time\n5. well you have to check all the forces on the bag. the first force is gravity, which is downwards and its 9.8m where m is the mass (Force = mass x acceleration) so gravity force is 20.5kg * 9.8 = 200.9N. then the 2nd force is the force of your own movement. you accelerate it upwards at 5m/s^2. so you apply a force of 5x20.5 = 102.5N on the bag upwards. but this is the thing. it gets confusing. the force is applied on the bag, not on the items in the bag. the items in the bag  is what  makes the bag weigh so much. so if you pull the bag up with 102.5N force, the items in the bag pull downwards with the same force (another one of newtons laws) which means the total downwards force is 102.5 + 200.9 = 303.4N downwards. which is more than 245N so the bag will rip. \n\n6. well in general F  = ma. but the F is total force. in this case, there is a friction force along with the applied force.\nso Fa + Ff = ma\n   37 + Ff = 5 x 5.8\n   Ff = 29 - 37\n   Ff = -8N     its negative because its opposing the applied force. in this case i can choose which direction i want to be positive, i chose the direction that the object is moving. you could put negatives in front of the applied force, friction force and acceleration   and youll get the same answer.  so friction force is 8N.\nb. the coefficient of friction is calculated by Ff = u x Fn\nwhere Fn is the normal force and u is the coefficient. you already know Ff. the normal force is easy its basically  the force tha topposes gravity. if the object isnt moving up or down (if its on a surface). so then gravity is 5 kg x 9.8 = 49N \nso then Ff = u x 49\n    8 = u x 49\n u = 8/49\n u = 0.1632..... coefficient of friction is 0.1632\n\n7.\na.  the force is 107N and the mass is 5.1kg. so acceleration is 20.98m/s^2\nb. so acceleration = Vf-Vi/t\n  20.98 = Vf - 0 / 10s\n  Vf = 10 x 20.98 = 209.8m/s\nfinal velocity is 209.8m/s\nc. the net force that acts on the object is only the downward gravity force which is 9.8x 5.1 = 49.98.\nd. the direction of their velocity goes down when the velocity becomes 0. so you can do this by saying acceleration is gravitational and it is -9.8. initial velocity is 209.8 and final velocity is 0. and now u gotta find time. so  a = Vf-Vi/t\n-9.8 = -209.8 / t\nt = 209.8/9.8\nt = 21.4s   so it starts going down   21.4s after 10s after its been released (the first 10 s  its accelerating upwards)   so it starts going down 31.4s after its been realeased.
508539
A) F = mg = 50 kg x 9,81 m/s2 = 0,49 kN.\nB) F = GMm/R2 = 6.67E-11 x 5.98E24 x 50 / (6.370E6)^2 N\n   = 0,49 N
481398
sum the forces in the direction that they are applied and it will be equal for a static object. \n\nRule of thumb on this one is that you have 100N down ward, then you need 100 N's up ward.\n\nGivens, legs are equal lengths, same material, and no angles to calculate.\n\n4 legs - 4 forces - each at 25 Newtons.  \n\nNow if you want to get fancy, break out the strain gauges and let us see what we really find.  But it helps if you have preformed a few material tests and experimenters first.
860005
I finished the problem.  The answer is s = 0.48517m.  It was hard, although now that I've worked it, it doesn't seem as difficult anymore.  Had to do some serious thinking though.\n\nThe collar A will tend to slide <down> the bar CD.  The vector CD (from C to D) is [-0.2, -0.3, 0.25] with magnitude 0.43875 and direction cosines [-0.45584, -0.68376, 0.56980].  The weight of collar A, directed downward, is 8g = 78.4N and is represented as [0, -78.4, 0].\n\nWe don't have to worry about the collar being supported by bar CD (the normal force).  The component of the collar directed down the bar is the inner (scalar, dot) product of the weight vector and the direction cosines.  This force is 78.4 x .68376 = 53.60706N.\n\nLet E be the point on the bar CD such that BE is perpendicular to CD.  Then BEA is a right triangle, and by the Pythagorean Theorem, the force along BE is sqrt(150^2 - 53.60706^2) = 140.09384N.  (I'm using numbers stored in the calculator and showing five decimal digits.  All this can be rounded off at the end.)\n\nSince the force triangle BEA is similar to the distance triangle BEA, we just have to get one of the legs to find our distance s = AB.  We will get the distance BE.\n\nTriangle BEA is in the plane defined by the points B, C, and D.  Vector CB = [-0.4, 0.2, 0.15].  The vector cross product CB x CD (see above) is normal to the plane defined by points B, C, and D.  CB x CD = [0.095, 0.07, 0.16].\n\nNow, the cross product CD x (CB x CD) lies in the plane defined by B, C, and D; is perpendicular to CD; and is parallel to the line segment BE.\n\nCD x (CB x CD) = [-0.0655, 0.05575, 0.0145] with magnitude 0.08723 and direction cosines [-0.75091, 0.63914, 0.16623].\n\nNow we can get parametric equations of the lines CEAD and BE, and find the point of intersection E.  Starting with point C and using vector CD, the first line is\n\nx = 0.4 - 0.2t\ny = 0.3 - 0.3t\nz = 0 + 0.25t\n\nStarting from point B and using the direction cosines from vector BE, the second line is\n\nx = 0 + 0.75091u\ny = 0.5 - 0.63914u\nz = 0.15 - 0.16623u\n\nSetting like terms equal to each other, we have three equations in two unknowns.  The three equations are not linearly independent.  Solving the x and y equations for t and u, we get\n\nt = 0.29871 and u = 0.45313\n\n(and these results are confirmed in the z equations).  Plugging these results into the x, y, and z equations, we find that the intersection point E is (0.34026, 0.21039, 0.07468).\n\nWe know the coordinates of point B, so by the Pythagorean Theorem, the distance BE is 0.45313m.\n\nComparing the force and distance vectors in similar triangles BEA, we have\n\ns/150 = 0.45313 / 140.09384\n\nand the length s of the cable AB is 0.48517m.  (Answer)\n\nA difficult, but fun, problem.
281995
Is it elasticity? Try to solve the problem with some formula below:\nYoung's Modulus(Y)=Stress/Strain\n=(F/A)/(dl/L) \nwhere, F=force applied, \nA=cross sectional area, \ndl=leangth increase and L=total lenght.\nBulk Modulus(B)\n=dP/(dV/V) \nwhere P=pressure and V=volume.