Ajit

This page lists questions and answers that were posted by visitors named Ajit.

Questions

The following questions were asked by visitors named Ajit.

A bottle of ink was 2/3 full.When 7 pens were filled with ink from the bottle and 2 pens full of ink was poured into it, it was 1/4 full.How many pens can be filled with the full bottle of ink?

Answers

The following answers were posted by visitors named Ajit.

Least number of weights to weigh any integral weight from 1 to 40 is 4 according to me. The weights would be-1,3,9 and 27 and may be placed on either side.

We've x = 2t+3 and y = 4t-1. Eliminate t between these two equations to obtain y = 4(x-3)/2 -1 or y = 2x -7

Please note the first equation should be y=-3x/4-1 if it is to represent a stright line. Now the two given lines are: y= - 3x/4 ¨C 1 and 3x+4y=20 or y=-3x/4 + 20. For us to calculate the distance between the two lines, they have to be parallel which they...

Hav u heard about "Carbon Dating"?

Given 2y^2+xy+x=6 -------(1). Differentiate both sides to obtain 4ydy/dx+xdy/dx+y+1 = 0 or dy/dx = -(y+1)/(x+4y) = -1/3. This gives us x+4y-3(y+1)=0 -----------(2). Solve equations (1) & (2) simultaneously to obtain the two points as (2,1)&(6,-3)

Centroid is given by: [(x1+x2+x3)/3, (y1+y2+y3)/3] or (33/3,36/3) or (11,12)

Assume u = Initial velocity of the bullet and v= its vel after leaving block A. V = Vel. of block A after the impact = 2 met/sec M = Mass of block A = 2 kg & m = mass of the bullet = 0.1 kg By the principle of conservation of momentum: m*u = m*v + M*V Or...

If you wish to remember a general equation for projectile motion, we can write: y=xtan(θ) –gx^2/[2(ucos(θ))^2] where u = initial velocity, θ = launch angle and the origin is the point of launch. Putting y=0 and solve for x to get: x=(u^2)sin(2θ)/g which w...

Using the usual connotation, we can say that: y = usin(θ)t-gt^2/2 = vsin(Φ)t-gt^2/2 or usin(θ)=vsin(Φ). Use the given values to obtain: 60*1/2 =50sin(Φ). In other words, sin(Φ)= 3/5. The second projectile must therefore be launched at angle Φ~=36.87° to t...

What I understand is as follows: mgH = (mv^2)/2 + (Iw^2)/2 In this case, I = (mR^2)/3. hence, mgH = (mv^2)/2 +m(Rw)^2/3. But Rw = v so mgH =5(mv^2)/6 or v^2 = 6gH/5 = 9.81*1.81*6/5 or v ~= 4.616 met/sec

H=1.31 met and not 1.81met. Sorry for the error. Accordingly, v~= 3.927 met/sec.

Archimedes's principle states that the upward buoyant force exerted on a body immersed in a fluid is equal to the weight of the fluid the body displaces. In this case. 2.8 cc of water is displaced and its weight 2.8 g. Hence the net buoyant force = 2.8 gf...

I think the question needs to be rephrased as follows: The angles in triangle ABC satisfy 6sin∠A=3√3sin∠B=2√2sin∠C. If sin2∠A, calculated correctly to 2 decimal places,= a/b where a and b are co-prime positive integers, what is the value of (a+b)? Let 6si...

2.94 up and 2.94 down