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alexdok [17]
3 years ago
6

Undisturbed samples from a normally consolidated clay layer were collected during a subsoil exploration. Drained triaxial tests

showed that the effective friction angle was φ’ = 25°. The unconfined compressive strength, qu, of a similar specimen was found to be 133 kN/m2. Find the pore pressure at failure for the unconfined compression test.
Physics
1 answer:
kipiarov [429]3 years ago
3 0

Answer:

pre water pressure = \sigma_1 = 208.76 kN/m2

Explanation:

given data:

frictional angle  25°

unconfined xompressive strength  qu = 133 kN/m2

for clay, unconfined compressive strenth is cu

cu =\frac{qu}{2}

cu = 133/2 =66.5 kN/m2

cell pressure \sigma_3 is zero in uncomfined compressive strenth test

from principle effective stress we have

\sigma_1 = \sigma_2 tan^{2}(45+ \frac{\theta}{2}) +2cu tan(45+ \frac{\theta}{2})

\sigma_1 = 2cu tan(45+ \frac{\theta}{2})

\sigma_1 = 2*66.5 tan(45+ \frac{25}{2})

\sigma_1 = 208.76 kN/m2

as this test is quick test i.e. undrained test therefore

pre water pressure = \sigma_1 = 208.76 kN/m2

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Answer:

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Explanation:

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Light energy causes the removal of an electron from a P680 molecule that is part of Photosystem II, the electron is transferred to an acceptor molecule (primary acceptor), and then passes downhill to Photosystem I through a conveyor chain of electrons The P680 requires an electron that is taken from the water by breaking it into H + ions and O-2 ions. These O-2 ions combine to form O2 that is released into the atmosphere.

The light acts on the P700 molecule of Photosystem I, causing an electron to be raised to a higher potential. This electron is accepted by a primary acceptor (different from the one associated with Photosystem II).

The electron goes through a series of redox reactions again, and finally combines with NADP + and H + to form NADPH, a carrier of H needed in the independent phase of light.

Electron of photosystem II replaces the excited electron of the P700 molecule.

There is therefore a continuous flow of electrons (non-cyclic) from water to NADPH, which is used for carbon fixation.

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7 0
3 years ago
A 0.5 kg ball moves in a circle that is 0.4 m in radius at a speed of 4.0 m/s. Calculate the force exerted on the ball.
blsea [12.9K]

Answer:

Explanation:

Given a ball of mass m= 0.5kg

The ball moves in as circle of radius r= 0.4m

Speed of the ball is v = 4m/s

Centripetal force is exerted on ball and it is given as

Fc = m•ac

ac is centripetal acceleration and it is given as

ac = v²/r

Then,

Fc = mv²/r

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Fc = 20N.

The force exerted on the ball is 20N

5 0
3 years ago
Suppose you are measuring double‑slit interference patterns using an optics kit that contains the following options that you can
svetlana [45]

Answer:

3.6 m

Explanation:

\lambda_R = 650 \ nm\\\\\lambda_R = 650*10^{-9} m\\\\L \ should \ be \ minimum \\\\i.e \  0.25 \ mm\\\\= 0.25 *10^{-3} m

\lambda_R = 700 \ nm\\\\\lambda_R = 700*10^{-9} m\\\\

Also

\beta = 1 \ mm \ fringe \  width

D_{min} = \frac{\beta d}{\lambda}\\\\D_{min} = \frac{10^{-3}*0.25*10^{-3}}{700*10^{-9}}\\\\D_{min} = 3.57 \\D_{min} =  3.6 m

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4 0
3 years ago
Two cylinders each contain 0.30 mol of a diatomic gas at 320 K and a pressure of 3.0 atm. Cylinder A expands isothermally and cy
Svetllana [295]

Answer :

(a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

Explanation :

Given that,

Number of mole n = 0.30 mol

Initial temperature = 320 K

Pressure = 3.0 atm

Final pressure = 1.0 atm

We need to calculate the initial volume

Using formula of ideal gas

P_{1}V_{1}=nRT

V_{1}=\dfrac{nRT}{P_{1}}

Put the value into the formula

V_{1}=\dfrac{0.30\times8.314\times320}{3.039\times10^{5}}

V_{1}=2.62\times10^{-3}\ m^3

(a). We need to calculate the final temperature of the gas in the cylinder A

Using formula of ideal gas

In isothermally, the temperature is not change.

So, the final temperature of the gas in the cylinder A is 320 K.

(b). We need to calculate the final temperature of the gas in the cylinder B

Using formula of ideal gas

T_{2}=T_{1}\times(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}-1}

Put the value into the formula

T_{2}=320\times(\dfrac{3}{1})^{\frac{1}{1.4}-1}

T_{2}=233.7\ K

(c). We need to calculate the final volume of the gas in the cylinder A

Using formula of volume of the gas

P_{1}V_{1}=P_{2}V_{2}

V_{2}=\dfrac{P_{1}V_{1}}{P_{2}}

Put the value into the formula

V_{2}=\dfrac{3\times2.62\times10^{-3}}{1}

V_{2}=0.00786\ m^3

V_{2}=7.86\times10^{-3}\ m^3

(d). We need to calculate the final volume of the gas in the cylinder B

Using formula of volume of the gas

V_{2}=V_{1}(\dfrac{P_{1}}{P_{2}})^{\frac{1}{\gamma}}

V_{2}=2.62\times10^{-3}\times(\dfrac{3}{1})^{\frac{1}{1.4}}

V_{2}=0.0057\ m^3

V_{2}=5.7\times10^{-3}\ m^3

Hence, (a). The final temperature of the gas in the cylinder A is 320 K.

(b). The final temperature of the gas in the cylinder B is 233.7 K.

(c). The final volume of the gas in the cylinder A is 7.86\times10^{-3}\ m^3

(d). The final volume of the gas in the cylinder B is 5.7\times10^{-3}\ m^3

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