TutorMe homepage
Subjects
PRICING
COURSES
Start Free Trial
Aashish R.
Tutor Satisfaction Guarantee
Physics (Newtonian Mechanics)
TutorMe
Question:

A ball is thrown vertically above form the ground. The displacement of ball from the ground at the end of 8th second and 12th second is equal. Find the maximum height reached by the ball. (Assume acceleration due to gravity as \$\$ 10\$\$ \$\$ m/sec^2.\$\$ )

Aashish R.

Since the displacement of ball is equal at 8th and 12th second, therefore by symmetric nature of motion under gravitation, the ball reached maximum height at the end of \$\$(8 + 12)/2 = 10th\$\$ second. Therefore, again by symmetric nature of motion under gravitation, the ball will take 10 second to reach ground from maximum height, let's call it 'h'. So, to calculate 'h', we have initial velocity, \$\$ u = 0 m/sec \$\$ acceleration due to gravity, \$\$ g = 10 m/sec^2 \$\$ total time taken, \$\$ t = 10 sec \$\$ therefore, using second equation of motion: \$\$ s = ut + (1/2)at^2 \$\$ substituting the value of 's', 'u', 'a' and 't' gives us: \$\$ h = (1/2)\$\$ X \$\$10\$\$ X \$\$10^2 \$\$ this gives us \$\$h = 500 m \$\$ which will be our answer. Therefore, maximum height reached by ball will be 500 meters.

Physics (Electricity and Magnetism)
TutorMe
Question:

Consider and 'H' shaped arrangement of capacitors, with one capacitor each on the solid line of 'H', making 5 capacitors in total. Somewhat like this (Please ignore the bad drawing) : | ----------| |-----------|------------| |-----------| | 10uF | B 5uF | | | | | --- | ------------| A --- 5uF C |----------------- | | | | 5uF | D 10uF | | ----------| |-----------|------------| |------------| The value of each capacitor is given alongside them in the figure. Find the equivalent capacitance between nodes A & C.

Aashish R.

Since, non of the capacitance in the figure are in series or parallel, the figure cannot be simplified further. Therefore, to solve the question, we'll have to look into the basics and apply Kirchoff's Loop Rule. To do this, suppose a potential difference of 'V' volts is applied to nodes A and C. And we'll assume charge on each capacitance as given in the figure below. \$\$ (All \$\$ \$\$ the \$\$ \$\$ charges \$\$ \$\$ given \$\$ \$\$ are \$\$ \$\$ in \$\$ \$\$ Columb. \$\$ \$\$ '+' \$\$ \$\$and \$\$ \$\$ '-' \$\$ \$\$ sign \$\$ \$\$ represent \$\$ \$\$ the \$\$ \$\$ higher \$\$ \$\$ and \$\$ \$\$ lower \$\$ \$\$ potential, \$\$ \$\$ respectively.) \$\$ +(Q - q)- +(q)- | ----------| |-----------|------------| |-----------| | 10uF | B 5uF | | | | | + --- | ------------| A (Q - 2q) --- 5uF C |--------------- | | - | | | | | 5uF | D 10uF | | | | ----------| |-----------|------------| |------------| | | +(q)- +(Q - q)- | | | |__________________+ (V) -___________________| (Figure 1) If, the whole system is replaced by an equivalent Capacitance, the figure will look something like this: + (Q) - |--------------------| |----------------| | Ceq uF | | | |_________+ (V) -_______| (Figure 1) So, the Kirchoff's Loop equation in second figure would be: \$\$ Q = Ceq V \$\$ ...(1) So, we would apply Kirchoff's Loop Rule in (Figure 1) and try to an equation similar to equation (1). Since, we have two variable in consideration ( Q and V), so we'll need two equations to solve those. We'll do so by applying Kirchoff's Loop Rule in two different loops in (Figure 1). \$\$Loop\$\$ \$\$1:\$\$ Loop through ABC \$\$( (Q - q)/10 ) - (q/5) = V \$\$ ....(2) \$\$Loop\$\$ \$\$2:\$\$ Loop through ADBC \$\$ (q/5) - ( (Q - 2q)/5 ) + (q/5) = V \$\$ .....(3) Solving (2) and (3), we get: \$\$ Q = 7 V \$\$ Comparing the result with equation (1), we get \$\$ Ceq = 7\$\$ \$\$ uF \$\$ And that is our answer.

Calculus
TutorMe
Question:

Solve the differential equation: \$\$ dy/dx = (2x - y +3)/(x - y + 2) \$\$

Aashish R.

Solving this kind of differential equation may seem troublesome as its not an homogeneous differential equation. So, to solve the differential equation, we will try to manipulate the equation and convert it into a linear differential equation. To do that, we'll use a simple substitution: \$\$ x = X + h \$\$ \$\$ y = Y + h \$\$ where, h and k are constants. This implies: \$\$ dx = dX \$\$ \$\$ dy = dY \$\$ Now, the substituted differential equation is: \$\$ (dY/dX) = ( 2(X + h) - (Y + k) +3 ) / ((X + h) - (Y + k) + 2) \$\$ \$\$ => (dY/dX) = ((2X - Y) + (2h - k +3)) / ((X - Y) + (h - k +2)) \$\$ So, to convert this into an homogeneous differential equation, we'll equate the constants part in both numerator and denominator to zero. That is: \$\$ 2h - k + 3 = 0 \$\$ and \$\$ h - k + 2 = 0 \$\$ on solving the above two equations, we find that: \$\$ h = -1 \$\$ and \$\$ k = 1 \$\$ Therefore, our initial substitution becomes : \$\$ x = X - 1 \$\$ ...(1) \$\$ y = Y + 1 \$\$ ...(2) So, the differential equation now becomes: \$\$ (dY/dX) = (2X - Y) / (X - Y) \$\$ which is a Homogeneous differential equation and can be solved easily. To solve this, we'll divide both numerator and denominator in RHS by X, so the equation becomes: \$\$ (dY/dX) = (2 - (Y/X)) / (1 - (Y/X)) \$\$ now, we'll substitute: \$\$ (Y/X) = t \$\$ ...(3) this implies: \$\$ (dY/dX) = t + X (dt/dX) \$\$ So, our differential equation becomes : \$\$ t + X (dt/dX) = (2 - t) / (1 - t) \$\$ with a bit of rearranging, the equation looks like : \$\$ ((1 - t) / ( (1-t)^2 + 1) ) dt = (dX/X) \$\$ Now, this can be solved with our basic knowledge of differential equations, the solution looks like: \$\$ (1/2) ( log | (1 - t)^2 + 1 | ) = log CX \$\$ Now, using equation (3) and substituting back the value of t: \$\$ (1/2) ( log | (1 - (Y/X))^2 + 1 | ) = log CX \$\$ =>\$\$ (1 - (Y/X))^2 + 1 = (CX)^2 \$\$ now, using equation (1) and equation (2) to substitute back the value of Y and X: \$\$ (1 - ((y -1) / (x + 1)))^2 + 1 = (C(x + 1) )^2 \$\$ which will be our answer. Of course, it could be further simplified, with a bit of simple Algebra, but it is not the need of this solution.

Send a message explaining your
needs and Aashish will reply soon.
Contact Aashish