Ask a New Question
Search
Questions and answers by
Ajit
Questions (2)
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 answer
1,363 views
Professor Merlin has asked you to help him. He has 5 students who have taken four tests. He wants a program where he can enter
0 answers
1,029 views
Answers (15)
Good
2.94 up and 2.94 down
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
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
H=1.31 met and not 1.81met. Sorry for the error. Accordingly, v~= 3.927 met/sec.
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
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
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
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
Centroid is given by: [(x1+x2+x3)/3, (y1+y2+y3)/3] or (33/3,36/3) or (11,12)
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)
Hav u heard about "Carbon Dating"?
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
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
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.
