ENGINEERING COLLEGE

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 1

Answer:

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.

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Automobiles must be able to sustain a frontal impact. Specifically, the design must allow low speed impacts with little damage, while allowing the vehicle front end structure to deform and absorb impact energy at higher speeds. Consider a frontal impact test of a vehicle with a mass of 1000 kg. a. For a low speed test (v = 2.5 m/s), compute the energy in the vehicle just prior to impact. If the bumper is a pure elastic element, what is the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm? b. At a higher speed impact of v = 25 m/s, considerable deformation occurs. To absorb the energy, the front end of a vehicle is designed to deform while providing a nearly constant force. For this condition, what is the amount of energy that must be absorbed by the deformation [neglecting the energy stored in the elastic deformation in (a)? If it is desired to limit the deformation to 10 cm, what level of resistance force is required? What is the deacceleration of the vehicle in this condition?

Answers

Answer:

Explanation:

The concept of Hooke's law was applied as it relates to deformation.

The detailed steps and appropriate substitution is as shown in the attached file.

What are two advantages of forging when compared to machining a part from a billet?

Answers

Answer:

Less material waste and time.

Explanation:

Two advantages of forging vs machining would be that with forging there is much less waste of material. With machining you remove a large amount of material turning into not so valuable chips.

There is also a time factor, as machining can be very time intensive. This depends on the speed of the machining, newer machines tend to be very fast, and forging requires a lengthy heating, but for large parts the machining can be excessively long.

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

Write a Python program that does the following. Create a string that is a long series of words separated by spaces. The string is your own creative choice. It can be names, favorite foods, animals, anything. Just make it up yourself. Do not copy the string from another source. Turn the string into a list of words using split. Delete three words from the list, but delete each one using a different kind of Python operation. Sort the list. Add new words to the list (three or more) using three different kinds of Python operation. Turn the list of words back into a single string using join. Print the string.

Answers

Answer:

The code is attached.

Explanation:

I created a string s including 6 colors with spaces in between. Then I converted the string into a list x by using split() method. I used three different methods for removing elements from the list. These methods are remove(), pop() and del.

$\n$ Then I used methods append(), insert() and extend() for adding elements to the list.

$\n$ Finally I converted list into a string using join() and adding space in between the elements of the list.

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