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3 years ago

Match each titration term with its definition.

Engineering

1 answer:

Illusion [34]3 years ago

3 0

Answer:

1) titration

2) titrand

3) equivalence point

4) titrant

5) Burette

6) Indicator

Explanation:

The process in which a known volume of a standard solution is added to another solution so that the standard solution can react with the solution of unknown concentration such that its concentration is determined can be referred to as titration.

The solution which is added to another solution is called the titrant. The titrand is the solution of unknown concentration

A burette is a glassware used to slowly add a known volume of the titrant to the titrand.

The indicator used signals the point when the reaction is complete by a color change. At this point, a stoichiometric amount of titrant has been added to the titrand. This is also referred to as the equivalence point.

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The unit weight of a soil is 14.9kN/m3. The moisture content of the soil is17% when the degree of saturation is 60%. Determine:

Serggg [28]

Answer:

a) 2622.903 N/m^3

b) 1.38233

c)4.878811765

Explanation:

Find the void ratio using the formula:

$y = \frac{G_{s}*y_{w} + w*G_{s}*y_{w} }{1+e} ....... Eq1$

Here;

$G_{s}$ is specific gravity of soil solids

$y_{w}$ is unit weight of water = 998 kg/m^3

$w$ is the moisture content = 0.17

$e$ is the void ratio

$y$ is the unit weight of soil = 14.9KN/m^3

Saturation Ratio Formula:

$w*G_{s} = S*e ..... Eq2$

S is saturation rate

Substitute Eq 2 into Eq 1

$y = \frac{(\frac{S*e}{w}) * y_{w} + S*e*y_{w} }{1+e}$

$14900 = \frac{3522.352941*e + 598.8*e }{1+e} = \frac{4121.152941*e}{1+e}\\\\ e= 1.38233$

Specific gravity of soil solids

$G_{s} = \frac{S*e}{w} = \frac{0.6*1.38233}{0.17} = 4.878811765$

Saturated Unit Weight

$y_{s} = \frac{(G_{s} + e)*y_{w} }{1+e} \\=\frac{(4.878811765 + 1.38233)*998 }{1+1.38233}\\\\= 2622.902571 N/m^3$

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3 years ago

Determine the angular acceleration of the uniform disk if (a) the rotational inertia of the disk is ignored and (b) the inertia

lukranit [14]

Answer:

α = 7.848 rad/s^2 ... Without disk inertia

α = 6.278 rad/s^2 .... With disk inertia

Explanation:

Given:-

- The mass of the disk, M = 5 kg

- The right hanging mass, mb = 4 kg

- The left hanging mass, ma = 6 kg

- The radius of the disk, r = 0.25 m

Find:-

Determine the angular acceleration of the uniform disk without and with considering the inertia of disk

Solution:-

- Assuming the inertia of the disk is negligible. The two masses ( A & B ) are hung over the disk in a pulley system. The disk is supported by a fixed support with hinge at the center of the disk.

- We will make a Free body diagram for each end of the rope/string ties to the masses A and B.

- The tension in the left and right string is considered to be ( T ).

- Apply newton's second law of motion for mass A and mass B.

ma*g - T = ma*a

T - mb*g = mb*a

Where,

* The tangential linear acceleration ( a ) with which the system of two masses assumed to be particles move with combined constant acceleration.

- g: The gravitational acceleration constant = 9.81 m/s^2

- Sum the two equations for both masses A and B:

g* ( ma - mb ) = ( ma + mb )*a

a = g* ( ma - mb ) / ( ma + mb )

a = 9.81* ( 6 - 4 ) / ( 6 + 4 ) = 9.81 * ( 2 / 10 )

a = 1.962 m/s^2

- The rope/string moves with linear acceleration of ( a ) which rotates the disk counter-clockwise in the direction of massive object A.

- The linear acceleration always acts tangent to the disk at a distance radius ( r ).

- For no slip conditions, the linear acceleration can be equated to tangential acceleration ( at ). The correlation between linear-rotational kinematics is given below :

a = at = 1.962 m/s^2

at = r*α

Where,

α: The angular acceleration of the object ( disk )

α = at / r

α = 1.962 / 0.25

α = 7.848 rad/s^2

- Take moments about the pivot O of the disk. Apply rotational dynamics conditions:

Sum of moments ∑M = Iα

( Ta - Tb )*r = Iα

- The moment about the pivots are due to masses A and B.

Ta: The force in string due to mass A

Tb: The force in string due to mass B

I: The moment of inertia of disk = 0.5*M*r^2

( ma*a - mb*a )*r = 0.5*M*r^2*α

α = ( ma*a - mb*a ) / ( 0.5*M*r )

α = ( 6*1.962 - 4*1.962 ) / ( 0.5*5*0.25 )

α = ( 3.924 ) / ( 0.625 )

α = 6.278 rad/s^2

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Xelga [282]

Answer:

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