Question

An Open rectangular tank 5 m * 4 m*3 m high containing water upto a height of 2 m is accelecrated horizontally along the longer side.

(a) =Determine the maximum acceleration that can be given without spilling the water.Ans=4 m/s2

(b)=Calculate the percentage of water spill over ,if this acceleration is increased by 20%. Ans=10%.

Answer 1

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%

Explanation:

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%

Answer 2

Answer:

Maximum acceleration that can be given without spilling the water a_{max} = 4(\frac{m}{s^{2}})amax=4(s2m) and percentage of water spill over is 10%

Explanation:

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

1. From equation

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

  where g =10(\frac{m}{s^{2}})(s2m)

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}tanθ=10a=52

     From here we got a_{max} = 4(\frac{m}{s^{2}})amax=4(s2m)

3. On increase acceleration 20%

 New acceleration a= 1.2×4 = 4.8 (\frac{m}{s^{2}})(s2m)

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

\tan \theta = \frac{4.8}{10}= \frac{h}{5}tanθ=104.8=5h

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

4. Old volume of water V_{o}Vo = 5×2×4= 40(m^{3})(m3)

 New volume of water V_{N}VN = 5×(3-2.4)×4 + (5×2.4×4)÷2=36(m^{3})(m3)

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

Explanation:

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