Kush A.

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Calculus

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Question:

Determine the area below and above the x-axis

Kush A.

Answer:

Let’s start off with getting a sketch of the region we want to find the area of. I am assuming that, at this point, you are capable of graphing most of the basic functions that we’re dealing with in these problems and so we won’t be showing any of the graphing work here. It should be clear from the graph that the upper function is the parabola (i.e. ) and the lower function is the x-axis (i.e. ). Since we weren’t given any limits on x in the problem statement we’ll need to get those. From the graph it looks like the limits are (probably) . However, we should never just assume that our graph is accurate or that we were able to read it accurately. For all we know the limits are close to those we guessed from the graph but are in fact slightly different. So, to determine if we guessed the limits correctly from the graph let’s find them directly. The limits are where the parabola crosses the x-axis and so all we need to do is set the parabola equal to zero (i.e. where it crosses the line ) and solve. Doing this gives, 3+2x-x^2= 0 ->> x=-1 or x=3 So, we did guess correctly, but it never hurts to be sure. That is especially true here where finding them directly takes almost no time. At this point there isn’t much to do other than step up the integral and evaluate it. We are assuming that you are comfortable with basic integration techniques so we’ll not be including any discussion of the actual integration process here and we will be skipping some of the intermediate steps. The area is, Integral of 3=2x-x^2 from -1 to 3 with respect to x = 32/3

C++ Programming

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Question:

program should function as a basic calculator; it should ask the user to input what type of arithmetic operation he would like, and then ask for the numbers on which the operation should be performed. The calculator should then give the output of the operation.

Kush A.

Answer:

<iostream> ___ multiply(int x, int y) { ______ x_y; } ____ divide(int x, int y) { _____ x_y; } _____ add(int x, int y) { ______x_y; } ______ subtract(int x, int y) { _____x_y; } using namespace std; ___ _____() { ____ op='c'; ____ x, y; while(op!='e') { cout__"What operation would you like to perform: add(+), subtract(-), divide(/), multiply(*), [e]xit?"; cin__op; switch(op) { ____ '+': cin__x; cin__y; cout__x__"+"__y__"="__add(x, y)__endl_ break; ____ '-'_ cin__x; cin__y; cout__x__"-"__y__"="__subtract(x, y)__endl_ break; ____ '/': cin__x; cin__y; cout__x__"/"__y__"="__divide(x, y)__endl_ break; ____ '*'_ cin__x; cin__y; cout__x__"*"__y__"="__multiply(x, y)__endl_ break; _____ 'e': ______; ______: cout__"Sorry, try again"__endl; } } return _;

Physics

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Question:

A long cylindrical solenoid with 100 loops per 1 cm has a radius of 1.6 cm. Assume the magnetic field inside the solenoid to be homogeneous and parallel to the axis of the solenoid. a) What is the inductance of the solenoid per 1 meter of its length? b) What electromotive force is induced per 1 meter of the length of the solenoid if the current change is 13 As-1?

Kush A.

Answer:

a) The inductance of the solenoid is defined as the constant of proportionality between the total magnetic flux through the coil turns and the current flowing in the coil: NΦ=LI NΦ=LI where N is the number of turns per one meter, thus N = 10,000. The magnetic flux Φ is the flux through one turn. Since the magnetic field is homogeneous, it is true that Φ=BScosα Φ=BScosα The magnetic induction B is at all points perpendicular to the cross section of the coil, therefore cos α = 1, where α is the angle between the magnetic induction B and the normal to the surface, thus α = 0o. The solenoid has a circular cross-section, therefore the area is S=πR2 S=πR2 We substitute the formula for magnetic induction inside the coil for the magnetic induction B. B=μoNI B=μoNI After substituting all of these derived relations into the first relation, we obtain NμoNIπR2=LI NμoNIπR2=LI We evaluate the unknown inductance L L=μoπN2R2 L=μoπN2R2 b) For the electromotive force induced in the coil the relationship |Ui|=∣∣∣d(NΦ)dt∣∣∣ |Ui|=|d(NΦ)dt| applies. For the inductance L of a one meter long coil it applies: NΦ=LI NΦ=LI For the left side of the equation we substitute the formula for the coil inductance, thus the formula contains the change of current per time known from the assignment. |Ui|=LdIdt

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