What is the difference between microfinance and commercial banks?
Commercial banks are Financial institutions started by/under some banking regulation act of the country. The aim of the commercial bank/s is to accept deposits and lend money to customers. Commercial banks also have nonbanking activities such as insurance, security brokerage, stocks and funds and so on. Commercial banks motive is to earn profits and expand their business and branches. registration, tax payment and dividends are key features of commercial banksMicrofinance is something of a group activity. The investors are not looking for profit or rather profit is not the main aim. The aim is to lend money to a fellow group member form the savings the members put together (predecided). Slowly but surely the group makes sure the money is being utilized and the group member is able to earn profit form an income generating activity (dairy, piggery, sheep/goat rearing, carpentry etc). Every group member gets a loan and thus the group moves towards self sustenance. It is what you call “Social banking”. No registration, no taxation and no dividends. One for all and all for one
When is instantaneous velocity equal to avg velocity, answer given but need help with explanation!?
You have already obtained for the instantaneous velocity ( I will plug in values for a and b at the end): v(t) = dx/dt = 3 b t² - 2 c t The average velocity over a time t is: the distance covered in time t, divided by time t ! v_avg(t) = x(t) / t = b t² - c t Equating these gives: 3 b t² - 2 c t = b t² - c t 2 b t² - c t = 0 t = 0 or 2 b t - c = 0 The other than t=0 solution thus is t = c/(2b) = 3.6 m/s² / ( 8.4 m/s³ ) = 0.43 s
The position of a toy locomotive moving on a straight track along the x-axis is given by the equation?
The position of a toy locomotive moving on a straight track along the x-axis is given by the equation x = bt^3 - ct^2 + dt, where b = 5 m/s^3, c= 240 m/s^2, and d= 3843 m/s, x is in meters and t is in seconds. Find the time t when the net force on the locomotive is equal to zero. Answer in units of s. At what time when the net force on the locomotive is equal to zero, the velocity of the locomotive is either positive, negative, or zero?
The position of a particle moving along the x axis depends on the time according to the equation?
x = ct^2 - bt^3, where x is in meters and t in seconds. (a) What are the units of constant c? a. s^2/m b. m^2/s c. m/s^2 d. s/m^2 (b) What are the units of constant b? a. m/s^3 b. s/m^3 c. s^3/m d. m^3/s For the following, let the numerical values of c and b be 3.8 and 1.0, respectively. (For vector quantities, indicate direction with the sign of your answer.) (c) At what time does the particle reach its maximum positive x position? s (d) From t = 0.0 s to t = 4.0 s, what distance does the particle move? m (e) From t = 0.0 s to t = 4.0 s, what is its displacement? m (f) Find its velocity at t = 1.0 s. m/s (g) Find its velocity at t = 2.0 s. m/s (h) Find its velocity at t = 3.0 s. m/s (i) Find its velocity at t = 4.0 s. m/s (j) Find its acceleration at at t = 1.0 s. m/s2 (k) Find its acceleration at at t = 2.0 s. m/s2 (l) Find its acceleration at at t = 3.0 s. m/s2 (m) Find its acceleration at at t = 4.0 s. m/s^2
The position y of a particle moving along the y axis depends on the time t according to the equation y=at-bt^2?
y is a distance. The units are length (L). So both terms being added, at and bt^2, have to have units of length. Since a*T has units of L, since you get L by multiplying a by a time (T), then a must have units of L/T. You can do similar reasoning for bt^2. You get L out by multiplying b by a time squared. So b has to have units of L divided by time^2.
The position of a particle moving along the x axis depends on the time according to the equation x = ct^2 - bt^6,?
where x is in meters and t in seconds. Let c and b have numerical values 2.7 m/s2 and 1.9 m/s6, respectively. From t = 0.0 s to t = 1.2 s, (a) what is the displacement of the particle? Find its velocity at times (b) 1.0 s, (c) 2.0 s, (d) 3.0 s, and (e) 4.0 s. Find its acceleration at (f) 1.0 s, (g) 2.0 s, (h) 3.0 s, and (i) 4.0 s.
The velocity of a particle moving along the x axis?
is given by vx = a t − b t^3 for t > 0 , where a = 28 m/s2, b = 1.7 m/s4, and t is in s. What is the acceleration of the particle when it achieves its maximum displacement in the positive x direction? Answer in units of m/s2.