Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15. The

Question

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15. The sand enters a pipe h= 4.3m below the end of the conveyer belt, as shown in the figure (Figure 1) . What is the horizontal distance between the conveyer belt and the pipe?

Sand moves without slipping at 6.0 m/s down a conveyer that is tilted at 15. The sand enters a pipe h= 4.3m below the end of the conveyer belt, as shown in the figure (Figure 1) . What is the horizontal distance between the conveyer belt and the pipe? no title provided

Explanation / Answer

Uy=6sin15=1.553m/s. Ux=5.8 m/s.

d=Ux*t

h=Uy*t + 0.5gt^2

So t=d/Ux.

putting this in other equation

h=dUy/Ux + 0.5g(d/Ux)^2

putting values and solving for d

we hav d=4.59 m.

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