All right . Thank you ! I got only some of them covered through watching Walter Lewin's lectures , like proving the Work-Energy Theorem , P= h(rho)g , etc . I learned the general proof for derivatives too , but not the SPM way where you just add delta x and delta y to x and y respectively . I prefer the usual method , those that are available online .

It takes the knowledge of Density of Steel, Density of Helium, Newton’s Law, Archimedes’ principle (Buoyancy), Impulse of Force, and a few assumptions to understand the tricky physics problem.

Statics ::

(1) The boxcar is at rest on a frictionless surface. Between the two objects inside the boxcar, a solid steel ball is much more dense than a helium-filled balloon. The steel pendulum and helium balloon are rigid bodies.

(2) Newton’s Law tells us the downward force of gravity pulls the steel pendulum and helium balloon, which results in their respective weights, W↓ = mg.

(3) Archimedes’ principle tells us the upward buoyant forces exerted on the steel pendulum and helium balloon “immersed fully” in the air, are equal to the respective weights of the air that the objects displace, F↑ = ρVg.

(4) For the heavier-than-air steel ball, because W↓ > F↑, the steel ball is pulled downward by the Net force.

(5) For the lighter-than-air helium balloon, because F↑ > W↓, the helium balloon is pushed upward by the Net force.

(6) Because the freedom of movements of the steel pendulum and helium balloon are constrained in the direction of the respective Net forces (vertical) axes, Newton’s Law tells us their respective normal Reaction forces are generated, so that they appear to be held in their places.



Dynamics ::

(7) For the impulse of force, in which we assume the force exerted on boxcar acts for a very short time δt but is much greater than any other force present. Imagine flicking your finger to apply the impulse of force on the outer left side of the boxcar! The duration of the collision is very short and bring the boxcar to an abrupt STOP.



(8) Now, because the steel pendulum and helium balloon are not constrained horizontally, any applied horizontal force will cause the objects to swing left and right.

(9) Since the boxcar comes to an abrupt STOP, when we reapply the Newton’s Laws and Archimedes’ principle from (1)−(5), the most probable positions of the steel pendulum and helium balloon caused by the impulse is depicted in Image C. What do you think?

My first respond to this question is thinking about inertia and my answer goes to C.

Reedit : second thought, answer is B.

---

My explanation:

1. An opposite force, f occurs at the string to resist F. The direction of f is to the left.

2. Resultant force of L is pointing south west.

3. Resultant force of B is pointing north west.

This post has been edited by work_tgr: Aug 15 2013, 11:38 PM

Probably could use general relativity for that question. Einstein's famous thought experiment (physicist in a box):

http://www.pitt.edu/~jdnorton/teaching/HPS...hway/index.html
(Principle in equivalence section)

Lets imagine the force exerted causes the box to accelerate with acceleration = g. Assuming that this experiment is not conducted in vacuum, we can say that the condition of the inside of the box will be exactly the same as if we were to rotate the box 90 degree anticlockwise.

Using this principle of equivalence, we know that in normal gravity condition (with air), the steel ball will move downward and the helium balloon will move upward. Now lets rotate this image back again 90 degree clockwise and we will get the answer.

The steel ball should move to the left and the helium to the right. So actually there is no answer?

This post has been edited by VengenZ: Aug 16 2013, 02:32 AM

Would you be so kind as to provide the reference link?

But are we not looking for the moment when the box is "pushed"?

Have you discovered this link?

Possible, if it is an “open” boxcar.

In the physics of general relativity (non SPM), we can certainly describe the motion of the steel pendulum and helium balloon by the equivalence principle.

True, but since the impulse acts for a very short time δt ~ 0 sec, the closest image that describes the positions (during δt) of the steel pendulum and helium balloon caused by the impulse that brings the boxcar to an abrupt STOP would probably be Image C. Just my 2 cents.

Refer to Tensile testing and Stress–strain curve.

The formula to calculate the “apparent” engineering stress, σ, is given by σ = F / Ao, assuming that the original cross-sectional area Ao is UNCHANGED (not reducing) through which the force is applied. In fact, as deformation continues, cross-sectional area decreases.

Similarly, the formula to calculate the engineering strain, ε, is given by ε = ΔL / Lo. Some questions give the elongation measurement L instead, and if that is the case, use ΔL = L − Lo.



Refer to Properties of Water, Continuity equation, Bernoulli's principle and Venturi effect.

Water is treated as an incompressible fluid because the changes in relative volume ΔV / V is negligible as the pressure increases.

In fluid mechanics, the continuity equation states that, in any steady state process, the rate of change of mass within the control volume is equal to the rate at which mass leaves the surface S of volume V.



In other words, mass inflow = mass outflow. However, mass is also given by the product of ρV (density × volume). Because water is essentially incompressible, therefore the density ρ of water is treated as constant throughout the streamline from the hose to the head. For that reason, we can rewrite the continuity equation as follows:

Volumetric inflow rate, Q1 = Volumetric outflow rate, Q2[/B][/COLOR]



Because Volumetric flow rate, Q is defined as Q = dV / dt,

and the volume of a mass, V is given by V = A*l (cross-sectional area × traversed length), therefore

A1 × (dl1 / dt) = A2 × (dl2 / dt)

Since dl / dt is defined as the velocity, υ, we can simplify the equation to

A1 × υ1 = A2 × υ2

I wonder where does we learn this
So deep

If the gas is continuously supplied at a constant pressure, the new level of “A” would probably look like this:

Ok thanks! Im impressed how well you edited the pic. I also want to know that,when an object fully immersed in water and its weight is then lets say, 5N , it will still be 5N when it is immersed further right as the vol of water displaced is still the same. Am I correct?

At first glance, you maybe tempted to solve ODE using Laplace Transforms. But it is not easy to find the Laplace Transform of 1/(1+t²). Since the ODE is a nonhomogeneous equation y'' + p(t)*y' + q(t)*y = g(t), your Lecturer probably has taught you about the Variation of Parameters formula.



This post has been edited by Critical_Fallacy: Nov 15 2013, 01:58 PM

Because the gas undergoes expansion from L to J

W(L,J) = (0.005 − 0.003)m³ × (100)kPa = 200 J

and then the gas undergoes compression from J to L

W(J,L) = ½ × {(0.005 − 0.003)m³ × (600 + 100)kPa} = 700 J

Therefore, the net work of the system is:

W(L,J) − W(J,L) = 200 − 700 = -500 J

Ans (A)...

This post has been edited by Critical_Fallacy: Nov 14 2013, 12:39 AM

Technically NO because you missed out the Matrix B.

Given a system of linear equations in the form of Ax = B:



Transform the linear system into augmented matrix:



Then, perform Gaussian Elimination to put the augmented matrix into the upper triangular form:

hi tanks for ur reply! i can understand finally!!

how to do this???? i need ur help!!!!!


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|-2/(x+1)| < 1

-1 < -2/(x+1) < 1

0 < -2/(x+1) + 1 or -2/(x+1) - 1 < 0

(x-1)/(x+1) > 0 or -(x+3)/(x+1) < 0

By number line or sign table, you'll get

{x| x<-3 or x>1}



This post has been edited by ystiang: Nov 14 2013, 06:35 PM

This belongs to the topic called Absolute-Value Inequalities.

To solve this problem, let's understand what is absolute value.

The absolute value of a real number R (denoted “|R|”) may be defined as the distance of R from zero.

For example, both –1 and 1 are one unit from zero, and therefore we have |–1| = |1| = 1.

The absolute value of R can be written as a function of the positive square root: |R| = √(R²).

Come back to the problem: |–2 / (x + 1)| < 1.

Let's define f(x) = –2 / (x + 1), and it follows that |f(x)| < 1.

By the above definition, we can intuitively imagine that f(x) = –1 and f(x) = 1.

To satisfy the condition of the inequality |f(x)| < 1, let's do four possible logic tests:

Test 1: f(x) < –1 :: Let f(x) = –1.1, and |f(x)| = |–1.1| = 1.1 < 1 ~ FALSE

Test 2: f(x) > –1 :: Let f(x) = –0.9, and |f(x)| = |–0.9| = 0.9 < 1 ~ TRUE

Test 3: f(x) < 1 :: Let f(x) = 0.9, and |f(x)| = |0.9| = 0.9 < 1 ~ TRUE

Test 4: f(x) > 1 :: Let f(x) = 1.1, and |f(x)| = |1.1| = 1.1 < 1 ~ FALSE

From the results, because Test 2 and 3 are TRUE, we can conclude that given the inequality |x| < a, the solution is always of the form –a < x < a.

Therefore, we have two inequalities: –1 < |f(x)| < 1.

Solve inequality #1: –1 < –2 / (x + 1) ... → ... we have x > 1.

Solve inequality #2: –2 / (x + 1) < 1 ... → ... we have x < –3.

Once you understand the concept, you can use the same simple approach as ystiang did.

Question: What if you were given the inequality |x| > a?

Even though it is a complex number, you can solve this problem using basic algebraic manipulation skill that you have already learned in Form 2 Factorization & Form 3 Simultaneous equations. Let me show you:

Find √(16 − 30i).

Since (16 − 30i) is a complex number, your developed mathematical intuition should tell you that the expected answer must be also a complex number (a + bi). Therefore it can be written as

√(16 − 30i) = √[(a + bi)²]

Your task now is to find a and b. As simple as that, regardless of the square root sign.

16 − 30i = (a + bi)² = a² + 2abi + b²i².

Since i² = −1, you can rearrange the equation to

16 − 30i = (a² − b²) + (2ab)i.

Eq.(1): a² − b² = 16

Eq.(2): 2ab = −30 ... → ... b = −15/a

You can solve the simultaneous equations by expressing b in term of a and substitute it into Eq.(1):

a² − (−15/a)² = 16

Multiply both sides with a² and turn it into a standard quadratic equation:

(a²)² − 16a² − 225 = 0

(a² − 25) (a² + 9) = 0

By the definition of a complex number (a + bi) where a and b must be Real Numbers, it renders a² = −9 invalid and therefore

a² − 25 = 0 ... → ... a = 5

Back-substituting a = 5 into Eq.(2) to find b:

b = −15/5 = −3

With the values found for a & b, and for all complex z with Re(z) > 0, √(z²) = z, the answer for

√(16 − 30i) = 5 − 3i

This post has been edited by Critical_Fallacy: Nov 15 2013, 01:59 PM

Given a system of linear equations with a set of variables x, y, z. Find k.

R1: 1x − 2y + 2z = 6
R2: 2x + 1y + 3z = 5
R3: 4x + 7y + 5z = k

Let's do some elementary row operations: You're lucky because element a11 is 1. See R1.

To eliminate coefficient of x (2) in R2, therefore 2 × R1: 2x − 4y + 4z = 12
R2 − 2R1:
R2': 5y − 1z = −7 ... eliminated x

To eliminate coefficient of x (4) in R3, therefore 4 × R1: 4x − 8y + 8z = 24
R3 − 4R1:
R3': 15y − 3z = k − 24 ... eliminated x

If you compare carefully R2' with R3', you'll notice that R2' & R3' are the same equation when scaled by a factor of 3, therefore 3 × R2': 15y − 3z = −21

Because it is stated that the equations are consistent, only then you can perform the next operation:
R3' − 3R2': 0 = k − 3

Solve for k and you have k = 3.

P.S. If you analyze the coefficient matrix, its determinant is 0, and thus the matrix is said to be singular, or it doesn't have a matrix inverse. If you plot the three equations in 3D, you'll discover that the linear system has infinitely many solutions because the equations are linearly dependent (see above R2' and R3'). If the system has a single unique solution, the three planes will intersect at a single point.

why this is B why not A???


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