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3 years ago

True or false : In improper integrals infinte intervals mean that both of the integration limits are should be infinity

Engineering

1 answer:

Mashcka [7]3 years ago

3 0

Answer:

An improper integral is a definite integral that has either or both limits infinite or an integrand that approaches infinity at one or more points in the range of integration

Explanation:

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How do I find v0 using Kirchhoff's laws and Ohm's law? The answer is 25V but I'm confused about how to get that.

telo118 [61]

Answer:

25 V

Explanation:

It is convenient to use Kirchoff's current law (KCL), which tells you the sum of currents into a node is zero. The node of interest is the top left node.

The currents into it are ...

20 mA + (-5 -Vo)/(2kΩ) -(Vo/(5kΩ)) = 0

20 mA -2.5 mA = Vo(1/(2kΩ) +1/(5kΩ)) . . . . add the opposite of Vo terms

(17.5 mA)(10/7 kΩ) = Vo = 25 . . . volts . . . . divide by the coefficient of Vo

_____

You will notice that the equation resolves to what you would get if you drew the Norton equivalent of the voltage source with its 2k impedance. You have two current sources, one of +20 mA, and one of -2.5 mA supplying current to a load of 2k║5k = (10/7)kΩ. KCL tells you the total current into the node is equal to the current through that load (out of the node).

5 0

3 years ago

if you’re disassembling a master cylinder and the front piston assembly is stuck in the master cylinder you can

lianna [129]

Answer:

Put water in the cylinder then push the piston inward

Explanation:

When water is put it displaces all the air making it possible for the piston to be pulled easier and without distraction

6 0

3 years ago

You are working in a lab where RC circuits are used to delay the initiation of a process. One particular experiment involves an

Ymorist [56]

Answer:

$t'_{1\2} = 6.6 sec$

Explanation:

the half life of the given circuit is given by

$t_{1\2} =\tau ln2$

where [/tex]\tau = RC[/tex]

$t_{1\2} = RCln2$

Given $t_{1\2} = 3 sec$

resistance in the circuit is 40 ohm and to extend the half cycle we added new resister of 48 ohm. the net resitance is 40+48 = 88 ohms

now the new half life is

$t'_{1\2} =R'Cln2$

Divide equation 2 by 1

$\frac{t'_{1\2}}{t_{1\2}} = \frac{R'Cln2}{RCln2} = \frac{R'}{R}$

$t'_{1\2} = t'_{1\2}\frac{R'}{R}$

putting all value we get new half life

$t'_{1\2} = 3 * \frac{88}{40} = 6.6 sec$

$t'_{1\2} = 6.6 sec$

7 0

3 years ago

1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem

BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁ = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

$\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

$\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}$

$P_2= \dfrac{1060}{5^{1.33}} = 124.65 \ kPa$

$\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2} \right ]^{\dfrac{k}{k-1}}$

$\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2} \right$

$5^{0.33} = \left \dfrac{T_1}{T_2} \right$

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3× $c_v$ ×(T₂ - T₁)

$c_v$ = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86 kJ/(kg·K)

We use cv given by (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3× $c_v$ ×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = $U_g = h_g -pV_g$

$h_g$ = 2777.12 kJ/kg

$V_g$ = 0.194349 m³/kg

p = 1000 kPa

$U_{g1}$ = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, $U_{g1}$ + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0

3 years ago

A. Name the major strengthening mechanisms in metals and explain the working principle under each mechanism.Give the relevant eq

Sever21 [200]

Answer:

a) Solid solution strengthening and alloying, Precipitation hardening, work hardening

b) Absence of enough crystallographic misalignment in the grain boundary region for a small-angle

Explanation:

<u>A) strengthening mechanism</u>

i) Solid solution strengthening and alloying:

In solid solution strengthening and alloying mechanism there is an addition of one atom of solute to another during this process, there might be substitution of interstitial point defect in crystal

also the shear stress required can be represented as: Δz = Gb√Ce^3/2

where : C = solute concentration , e = strain on material

ii) Precipitation hardening:

During precipitation hardening the alloying above the concentrate will lead to the formation of a second phase also under precipitation hardening a second phase can also be created via thermal treatments

particle bowing cab be written as : Δz = Gb / L-2x

iii) work hardening :

Dislocation caused by stress fields been generated hardens metals under the work hardening mechanism

dislocation can be represented as ; Gb √ p

where : G = shear modulus , b = Burgess vector, p = dislocation density

B) The small angle grain boundaries are not effective enough because there is less crystallographic misalignment in the grain boundary region for a small-angle

3 0

3 years ago