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Delbarre
Answers (9)
conservation of energy mgh + (1/2)m(Vi)² = (1/2)m(Vf)² 2(gh + (1/2)(Vi)²) = (Vf)² square root( 2gh +(Vi)²) = (Vf) square root( 256.76 + 132,25) = (Vf) square root( 389,01) = (Vf) square root( 389,01) = (Vf) 19.7m/s = Vf your kinetic energy is
the magnitude of the force is 2630.6 N
this exercise is conservation of linear momentum. initial momentum: (0.130kg)(35.5)(i) "remember linear momentum is vectorial. i is vector direction and only the movement is in X" = 4.615(i) Final momentum is : = (0.130)(50.5)(-i) =6.565(-i) then I=impulse
this case the energy of sistem is conservation of energy: (1/2)kx² = (1/2)mv² then you can find the energy of springs, this energy would be the energy of total sistem because no exist friccion. the maximum velocity of the mass v= square root((kx²)/ m)
W=F.d , W= (198.0N)(3.6m)cos 0, W=712.8 N.m this is your work. W=(151.0N)(3.6)cos 180, W=(-1)543,6N.m this work is negative because the force of friccion is opposite a the direction of movement y therefore the angle is 180 for last the total work is the
solution:you must be solved for the y of way implicit -2x^2+16x-224=-24y 2x^2-16x+224=24y (1/12)x^2-(2/3)x+(224/24)=y and solve it
for a body to be in equillibrium, exist two condition: The sum total of forces in the space vectorial must be zero sum of forces en x equal zero sum of forces en y equal zero sum of forces en z equal zero and the other condition is: The sum total of torque
for a body to be in equillibrium, exist two condition: The sum total of forces in the space vectorial must be zero sum of forces en x equal zero sum of forces en y equal zero sum of forces en z equal zero and the other condition is: The sum total of torque
none, because the cylinders not depend of your mass then both arrived in the same time: (only if the cylinders were uniforms, with the same radius and high) we used theorem of conservation of energy: Energy initial = energy final mgh = (1/2)mV² +
