We know that `log_(10)100000 = 5`
Show that you get the same answer by writing `100000 = 1000 xx 100` and then using the product rule. Verify the answer.

We know log_(2)32 = 5 . Show that we get the same answer by writing 32 as 64/2 and then using the product rule. Verify your answer .

We have log_(2)32 = 5 . Show that we get the same result by writing 32 = 2^(5) and then using power rules. Verify the answer.

When a jet of liquid strikes a fixed or moving surface, it exerts thrust on it due to rate of change of momentum. F=(rhoAV_(0))V_(0)-(rhoAV_(0))V_(0)costheta=rhoAV_(0)^(2)[1-costheta] If surface is free and starts moving due to thrust of the liquid, then at any instant, the above equation gets modified based on relative change of momentum with respect to surface. Let, at any instant, the velocity of surface be u . Then the above equation becomes F=rhoA(V_(0)-u)^(2)[1-costheta] Based on the above concept, as shown in the following figure, if the cart is frictionless and free to move in the horizontal direction, then answer the following. Given that cross sectional area of jet =2xx10^(-4)m^(2) velocity of jet V_(0)=10m//s density of liquid =1000kg//m^(3) ,mass of cart M=10 kg . Initially ( l=0 ) the force on the cart is equal to

There is another useful system of units, besides the SI/mKs. A system, called the cgs (centimeter-gram-second) system. In this system Coloumb's law is given by vecF =(Qq)/(r^(2)).hatr where the distance is measured in cm (= 10^(-2) m), F in dynes (= 10^(-5) N) and the charges in electrostatic units (es units), where 1 es unit of charge 1/(3) xx 10^(-9) C . The number [3] actually arises from the speed of light in vacuum which is now taken to be exactly given by c = 2.99792458 xx 10^8 m/s. An approximate value of c then is c = [3] xx 10^8 m/s. (i) Show that the coloumb law in cgs units yields 1 esu of charge = 1 (dyne) 1/2 cm. Obtain the dimensions of units of charge in terms of mass M, length L and time T. Show that it is given in terms of fractional powers of M and L. (ii) Write 1 esu of charge = x C, where x is a dimensionless number. Show that this gives: 1/(4piepsilon_(0)) = 10^(-9)/x^(2) (Nm^(2))/C^(2) with x=1/[3] xx 10^(-9) , we have 1/(4pi epsilon_(0)) = [3]^(2) xx 10^(9) (Nm^(2))/C^(2) or 1/(4pi epsilon_(0)) = (2.99792458)^(2) xx 10^(9) (Nm^(2))/C^(2) (exactly).

We can write log"" x/y = log (x.y^(-1 )) Can you prove that log"" x/y = log x - logy using product and power rules.

Consider a closed loop C in a magnetic field as shown in figure. The flux passing through the loop is defined by choosing a surface whose edge coincides with the loop and using the formula phi=vecB_1 . dvecA_1 + vecB_2 . dvecA_2 +…. Now if we choose two different surfaces S_1 and S_2 having C as their edge, would we get the same answer for flux. Justify your answer.