ENGINEERING COLLEGE

A site is compacted in the field, and the dry unit weight of the compacted soil (in the field) is determined to be 18 kN/m3. Determine the relative compaction if the maximum dry unit weight was measured to be 17 kN/m3. Express your answer as a percentage (but do not write the percentage sign in the answer box).

Answers

Answer 1

Answer:

Answer:

the relative compaction is 105.88 %

Explanation:

Given;

dry unit weight of field compaction, $W_d_((field))$ = 18 kN/m³

maximum dry unit weight measured, $W_d_((max))$ = 17 kN/m³

Relative compaction (RC) of the site is given as the ratio of dry unit weight of field compaction and maximum dry unit weight measured

Relative compaction (RC) = dry unit weight of field compaction / maximum dry unit weight measured

$RC = (W_d_((field)))/(W_d_((max)))$

substitute the given values;

$RC = (18)/(17) = 1.0588$

RC (%) = 105.88 %

Therefore, the relative compaction is 105.88 %

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Q/For the circuit showm bellow:a) find the mathematical expression for the transient behavior of ve and ic after closing the switeh,
b) sketch vc and ic.

Answers

Answer:

hello your question is incomplete attached is the complete question

A) Vc = 15 ( 1 - $e^(-t/0.15s)$ ) , ic = $1.5 mAe^(-t/0.15s)$

B) attached is the relevant sketch

Explanation:

applying Thevenin's theorem to find the mathematical expression for the transient behavior of Vc and ic after closing the switch

$R_(th)$ = 8 k ohms || 24 k ohms = 6 k ohms

$E_(th)$ = $(20 k ohms(20 v))/(24 k ohms + 8 k ohms)$ = 15 v

t = RC = (10 k ohms( 15 uF) = 0.15 s

Also; Vc = $E( 1 - e^(-t/t) )$

hence Vc = 15 ( 1 - $e^(-t/0.15)$ )

ic = $(E)/(R) e^(-t/t)$ = $(15)/(10) e^(-t/t)$ = $1.5 mAe^(-t/0.15s)$

attached

Determine the voltage across a 2-μF capacitor if the current through it is i(t) = 3e−6000t mA. Assume that the initial capacitor voltage is zero g

Answers

Answer:

$v = 250[1 - {e^(-6000t)}]$ mV

Explanation:

The voltage across a capacitor at a time t, is given by:

$v(t) = (1)/(C) \int\limits^(t)_(t_0) {i(t)} \, dt + v(t_0)$ ----------------(i)

Where;

v(t) = voltage at time t

$t_(0)$ = initial time

C = capacitance of the capacitor

i(t) = current through the capacitor at time t

v(t₀) = voltage at initial time.

From the question:

C = 2μF = 2 x 10⁻⁶F

i(t) = 3 $e^(-6000t)$ mA

t₀ = 0

v(t₀ = 0) = 0

Substitute these values into equation (i) as follows;

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + v(0)$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt + 0$

$v = (1)/(2*10^(-6)) \int\limits^(t)_(0) {3e^(-6000t)} \, dt$

$v = (3)/(2*10^(-6)) \int\limits^(t)_(0) {e^(-6000t)} \, dt$ [Solve the integral]

$v = (3)/(2*10^(-6)*(-6000)) {e^(-6000t)}|_0^t$

$v = (-3000)/(12) {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)}|_0^t$

$v = -250 {e^(-6000t)} - [-250 {e^(-6000(0))]$

$v = -250 {e^(-6000t)} - [-250]$

$v = -250 {e^(-6000t)} + 250$

$v = 250 -250 {e^(-6000t)}$

$v = 250[1 - {e^(-6000t)}]$

Therefore, the voltage across the capacitor is $v = 250[1 - {e^(-6000t)}]$ mV

At its current short-run level of production, a firm's average variable costs equal $20 per unit, and its average fixed costs equal $30 per unit. Its total costs at this production level equal $2,500. A. What is the firm's current output level? B. What are its total variable costs at this output level? C. What are its total fixed costs?

Answers

Answer and Explanation:

The computation is shown below

Total average cost + total variable cost = total cost

Let number of output be x

So,

Total fixed average cost = x × $30

Total variable cost = x × $15

Total cost = $2,500

Therefore,

$20 × x + $30 × x = $2,500

50 × x = $2,500

x = 50

Now the total variable cost is

= 50 × $20

= $1,000

And, the fixed cost is

= 50 × $30

= $1,500

Answer:

12 4

Explanation:

because the production average is variable

Indicate whether the following statements are true or false for an isothermal process: (A) Q=T(∆S). (B) ∆U=0.(C) The entropy change of the system is always zero. (D) The total entropy change of the system and the surroundings is always zero. (E) The entropy change of the surroundings is negative. (F) Q=W.

Answers

Answer:

A=False

B=False

C=False

D=False

E=False

F=False

Explanation:

A. In an isothermal process, only the reversibly heat transfer is 0, $Q_(rev)=T (\Delta S)$

B. Consider the phase change of boiling water. Here, the temperature remains constant but the internal energy of the system increases.

C. This is not true even in reversible process, as can be inferred from the equation in part A.

D. This is only true in reversible processes, but not in all isothermal processes.

E. Consider the phase change of freezing water. Here, the surroundings are increasing their entropy, as they are taking in heat from the system.

F. This is not true if $(\Delta U)\neq 0$ , like in answer B. One case where this is true is in the reversible isothermal expansion (or compression) of an ideal gas.

On what single factor does the efficiency of the Otto cycle depend?

Answers

Answer:

Compression ratio(r)

Explanation:

Otto cycle:

Otto cycle is an ideal cycle for all working petrol engine.It have four processes in which two are constant volume process and other two are reversible adiabatic or we can say that isentropic processes.All petrol engine works on Otto cycle.

The efficiency of Otto cycle given as follows

$\eta =1-(1)/(r^(\gamma-1))$

Where r is the compression ratio and γ is heat capacity ratio.

So from above we can say that the efficiency of Otto cycle depends onl;y on compression ratio (r).

A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool 5 large watermelons, 10 kg each, to 8 C. If the watermelons are initially at 28 C, determine how long it will take for the refrigerator to cool them.