Question

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Answer 1

Answer:

Let water container is moving with acceleration along length 5(m) is a.

1. From equation

\tan \theta = \frac{a}{g}= \frac{h}{L} ...1)

where g =10(\frac{m}{s^{2}})

and h is difference in position of top water level between front and back end of container along length 5(m).

here maximum of change in top level to allow to avoid water spilling = 2m.

2. \tan \theta = \frac{a}{10}= \frac{2}{5}

From here we got a_{max} = 4(\frac{m}{s^{2}})

3. On increase acceleration 20%

New acceleration a= 1.2×4 = 4.8 (\frac{m}{s^{2}})

Now from equation \tan \theta = \frac{a}{g}= \frac{h}{L}

\tan \theta = \frac{4.8}{10}= \frac{h}{5}

So new change in top water level height (h) = 2.4(m)

4. Old volume of water V_{o} = 5×2×4= 40(m^{3})

New volume of water V_{N} = 5×(3-2.4)×4 + (5×2.4×4)÷2=36(m^{3})

5. Percentage of volume spoil =(\frac{V_{N}-V_{o}}{V_{o}})\times 100 = (\frac{40-36}{40}})\times 100 =10%

Maximum acceleration that can be given without spilling the water a_{max} = 4(\frac{m}{s^{2}}) and percentage of water spill over is 10%