Setting reference frame so that the x axis is along the incline and y is perpendicular to the incline
<span>X: mgsin65 - F = mAx </span>
<span>Y: N - mgcos65 = 0 (N is the normal force on the incline) N = mgcos65 (which we knew) </span>
<span>Moment about center of mass: </span>
<span>Fr = Iα </span>
<span>Now Ax = rα </span>
<span>and F = umgcos65 </span>
<span>mgsin65 - umgcos65 = mrα -------------> gsin65 - ugcos65 = rα (this is the X equation m's cancel) </span>
<span>umgcos65(r) = 0.4mr^2(α) -----------> ugcos65(r) = 0.4r(rα) (This is the moment equation m's cancel) </span>
<span>ugcos65(r) = 0.4r(gsin65 - ugcos65) ( moment equation subbing in X equation for rα) </span>
<span>ugcos65 = 0.4(gsin65 - ugcos65) </span>
<span>1.4ugcos65 = 0.4gsin65 </span>
<span>1.4ucos65 = 0.4sin65 </span>
<span>u = 0.4sin65/1.4cos65 </span>
<span>u = 0.613 </span>
Hallar el valor numerico de 24m2n3p
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s .The amplitude of the subsequent oscillations 48.13 cm/s
a 1.25 kilogram block is fastened to a spring with a 17.0 newtons per meter spring constant. Given that K is equal to 14 Newtons per meter and mass equals 10.5 kg. The block is then struck with a hammer by a student while it is at rest, giving it a speedo of 46.0 cm for a brief period of time. The required energy provided by the hammer, which is half mv squared, is transformed into potential energy as a result of the succeeding oscillations. This is because we know that energy is still available for consultation. So access the amplitude here from here. He will therefore be equal to and by. Consequently, the Newton's spring constant is 14 and the value is 10.5. The velocity multiplied by 0.49
Speed at X equals 0.35 into amplitude, or vice versa. At this point, the spirit will equal half of K X 1 squared plus half. Due to the fact that this is the overall energy, square is equivalent to half of a K square or an angry square. amplitude is 13 and half case 14 x one is 0.35. calculate that is equal to initial velocities of 49 squares and masses of 10.5. This will be divided in half and start at about 10.5 into the 49-square-minus-14. 13.42 into the entire square in 20.35. dividing by 10.5 and taking the square as a result. 231 6.9 Six centimeters per square second. 10.5 into 49 sq. 14. 2 into a 13.42 square entire. then subtract 10.5 from the result to get the square. So that is 48.13cm/s.
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This is incomplete question Complete Question is:
a 1.25 kg block is attached to a spring with spring constant 17.0 n/m . while the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46.0 cm/s . what are The amplitude of the subsequent oscillations?
15) a
16) b
17) a
Hope this helps
Kinematics : Study of motion
Fluid kinematics : study of how fluid flows and how to describe its motion.
There are two ways to describe fluid motion
one is Eulerian, where the variations are described at all fixed stations as a function of time.
the other is Lagrangian, in which one follows all fluid particles and describes the variations around each fluid particle along its trajectory.
<u>DIFFRENCE BETWEEN LAGRANGIAN AND EULERIAN:</u>
1.Both Lagrangian and Eulerian describes time variation.
2. Eulerian describes the rate of change in one point of space
Lagrangian descries rate of change of a property of material system.
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