What number comes next in this sequence 5,14,30,55,91?
Look at the difference between the consecutive terms of the series.14 - 5 = 930 - 14 = 1655 - 30 = 2591 - 55 = 36You can see it clearly that the difference of the terma is a square series. So next term of this difference series is 7² = 49So next term of the series will be 91 + 49 = 140I hope it was clear enough. Want some more tough questions then check this one out…For excelling in quantitative aptitude section of the competitive exams, I would heartily invite you to download our Android app “APTITUDE GURU”. We also provide daily dose of aptitude which will keep your flow of study on and every day you will learn something new.Aptitude Guru : Tricks & Tips - Android Apps on Google PlayIt provides daily dose of aptitude which contains :-> 1 quantitative aptitude question-> 1 reasoning question-> 1 English word to learn-> 1 inspirational thought-> Calculation booster which is a rapid fire quiz of calculation questions.Besides this, it covers 15 most often asked topics of aptitude in offline mode.WE have provided detailed explanation of all the questions along with some easy tricks and there are three levels - Easy, medium and miscellaneous.For the beginners, easy level is the best place to start. Sincere efforts have been made to make the users understand the concepts.We have been receiving positive reviews from our users.Just check it out once and you won't regret for it…Aptitude Guru : Tricks & Tips - Android Apps on Google PlayThanks for reading..
The reaction of 50 mL of N2 gas with 150 mL H2 gas to form ammonia via the equation:?
the ratio between N2 and H2 is 1 : 3 we have 50 mL of N2 so we needed 3 x 50 = 150 mL of H2 because there is not a limiting reactant we can get the volume of ammonia : the ratio between N2 and NH3 is 1 : 2 volume NH3 = 2 x 0.50= 100 mL
What will be the next resonance after this?
closed pipe: f=v(2n-1)/4L n=1: 30=v(2-1)/4L = v/4L , fundamental frequency n=2: v(4-1)/4L= 3v/4L= 3*30= 90Hz , 1st.overtone n=3: v(6-1)/4L= 5v/4L= 5*30= 150Hz , 2nd.overtone n=4: v(8-1)/4L= 7v/4L= 7*30= 210Hz , 3rd.overtone
A body travels 2m in the second second and 6m in the next four seconds. What is the distance traveled in the ninth second?
Let u, v and a be the initial velocity, final velocity and acceleration of the body respectively.As we know some formulas that I’ll use here:a) Sn=u+a/2 * (2n-1) ………(1)where Sn= distance in nth second and it is only one term here not S*n it should be remind.b) S= ut+a/2 * t^2 ………….(2)So, we are going to solve our problem.as we have to find the distance in 9th second so it is clear here that we will use the first formula (a) here. but the main problem is that we are not given u and a. so now as there are 2 variables that we have to find and this is possible only when we have two simultaneous equation involving only two variable that is a and u.so now read question again and find the equations to find the a and u.As in question it is clear that body covers 2 meter distance in 2nd second.so by using first formula (a)here n=2 & Sn=2 by putting values we get2=u+a/2 * {2(2)-1}2=u+a/2 *32=u+3a/2multiply by 2 on both side4=2u+3a …………..(3)and for the second equation now read again the statement that distance covered in next four second is 6 meter.so we know that it (distance in next four seconds) is the distance that is covered by body in 2 to 6 seconds interval. and we can say this same statement that it is the distance that is the “ Difference of the distances covered in 6 seconds and 2 seconds.”so in mathematical statement you can write it as:S(2 to 4)=S6-S26= (S in 6 seconds) - (S in 1st 2 seconds)by using the second formula (b) to find distances in 6 and first 2 seconds.6=(ut+a/2 * t^2) - (ut+a/2 * t^2)6=[{u(6)+a/2 * (6)^2} - {u(2)+a/2 * (2)^2}]6={6u+a/2 * (36)} - {2u+a/2 * (4)}6=(6u+18a) - (2u+2a)6=6u+18a-2u-2a6=4u+16adividing by 2 on both side we get3=2u+8a …………..(4)so now (3) and (4) are two simultaneous equation by solving, we get.u=2.3 m/sa= -0.2 m/s^2so now using our above formula (a) we are able to calculate distance in 9th second.Sn=u+a/2 * (2n-1)here S9 means distance in 9th second.S9=2.3+(-0.2)/2 *{2(9)-1}S9=2.3-(0.2)/2 *{2(9)-1}S9=2.3-(0.1)(17)S9=(2.3–1.7)S9= 0.6 Answer.I hope now you will be able to understand it and can solve such types of problems in Physics.
What is the formula for the nth term in this sequence?
The formula for any arithmetic sequence, the formula is: a_n = a_1 + (n - 1)d where a_n = nth term, a_1 = 1st term, d = distance, and n = number of term. To get the formula for this one, we can just solve for the value of d, since we have 1st and nth terms. For instance, in the 2nd term... 105 = 100 + (2 - 1)d 105 - 100 = 1d 5 = d So, substituting the value of d, we'll get the formula for this sequence. a_n = 100 + (n-1)*5 Hope I helped. (:
Enthalpy problem. What is the next step?
(A) N 2(g) + 3 H 2(g) -> 2 NH3(g) H* = -91.8 kJ (B) 4 NH 3(g) + 5 O 2(g) -> 4 NO (g) + 6 H20(g) H* = -906.2 kJ (C) H 2(g) + 1/2 O 2(g) -> H20( g) H* = -241.8 kJ Step 1: You want to get rid of water in (B), so flip equation (C). This will flip the H*. H20( g) -> H 2(g) + 1/2 O 2(g) H* = +241.8 kJ 4 NH 3(g) + 5 O 2(g) -> 4 NO (g) + 6 H20(g) H* = -906.2 kJ Step 2: You need at least 6 waters, so multiply (B) by 6. The put the two equations together, canceling out like terms. 6 H20( g) -> 6 H 2(g) + 3 O 2(g) H* = +1451 kJ 4 NH 3(g) + 5 O 2(g) -> 4 NO (g) + 6 H20(g) H* = -906.2 kJ - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - 4 NH3 + 2 O2 ----> 4 NO + 6 H2 H* = 1451 - 906.2 = 544.6 kJ Step 3: Now you want to get rid of NH3, so multiply (A) by 2. 2N 2(g) + 6 H 2(g) -> 4 NH3(g) H* = -184 kJ 4 NH3 + 2 O2 ----> 4 NO + 6 H2 H* = 544.6 kJ Step 4: Put the two equations together, canceling out like terms. 2N 2(g) + 6 H 2(g) -> 4 NH3(g) H* = -184 kJ 4 NH3 + 2 O2 ----> 4 NO + 6 H2 H* = 544.6 kJ - - - - - - - - - - - - - - - -- - - - - - - - - - - 2 N2 + 2 O2 ---> 4 NO H* = 544.6 - 184 = 360 kJ Step 5: Divide both sides by 4 to get the target reaction. 1/2 N2 + 1/2 O2 ---> NO H* = 360 / 4 = 90 kJ Close enough. My rounding was different.
The sum of two numbers is 75 and their ratio is 3:2, what are the numbers?
Let n = the smaller number, and ...Let (3/2)n = the larger number since the ratio of the two numbers is 3:2 or 3/2.Since the sum of the two numbers is 75, we can write the following equation:n + (3/2)n = 75Now, clearing the equation of fractions by multiplying both sides by the denominator, 2:2[n + (3/2)n] = 75(2)2n + (2)(3/2)n = 75(2)2n + (2/2)(3n) = 1502n + (1)(3n) = 1502n + 3n = 150Now, collecting like terms on the left side, we have:(2 + 3)n = 1505n = 150Now, dividing both sides by 5 to solve for n, we get:(5n)/5 = 150/5(5/5)n = 150/5(1)n = 30n = 30Therefore, ...(3/2)n = (3/2)(30) = 90/2 = 45CHECK:n + (3/2)n = 7530 + 45 = 7575 = 75Therefore, our two numbers are 30 and 45.
A body starting from rest is moving under a constant acceleration up to 20 seconds. If it moves S1 distance in first 10 seconds, and S 2 distance in next 10 seconds then S 2 will be equal to?
Basic equation of motion S= ut + 0.5a(t*t)S1= 0.5a(10*10)S1 + S2 = 0.5a(20*20) =4*S1S2 = 3S1
A 344 g mass is connected to a light spring?
Assuming the spring obeys Hooke's Law, then the mass will execute simple harmonic motion about the equilibrium position of the spring (see source). The angular frequency of the oscillation is given by w = sqrt(k/m) where k is the spring constant, and m is the mass of object subjected to the force exerted by the spring. In this case, w = sqrt(6 N/m / 0.344 kg) = 4.176 sec^-1 The period is related to the angular frequency by: Period = 2*pi/w so in this case, Period = 2*pi/sqrt(6 N/m / 0.344 kg) = 1.504 sec <- This is the answer to part (a). The potential energy of a spring-mass system is given by: U = 0.5*k*x^2 Initially, the mass is at rest so all the energy in the system is potential energy. We are told that initially, the mass is displaced by 8 cm from the equilibrium position, so U = 0.5 * 6 N/m * (0.08 m)^2 = 0.019 J The mass will have its maxumum speed when the potential energy of the mass is zero, and all the energy in the system is present as kinetic energy (= 0.5*m*v^2). (This occurs when the mass is in the equilibrium position of the spring, x = 0). At that point: 0.5 * 6 N/m * (0.08 m)^2 = 0.5 * 0.344 kg * (v_max)^2 (v_max)^2 = (6 N/m * (0.08 m)^2)/(0.344 kg) (v_max) = 0.334 m/s <- This is the answer to part (b) For part (c), we start with Hooke's Law: F = -k*x and combine it with Newton's second law: F = m*a to obtain: m*a = -k*x a = -(k/m)*x Both k and m are constants, so the acceleration is directly proportional to the displacement of the mass from the equilibrium position of the spring. The maximum acceleration will occur at the maximum displacement, which in this problem is at +/-8 cm (presumably the question is asking about the maximum *magnitude* of the acceleration). We have, then, that: a_max = -(6 N/m / 0.344 kg) * (+/-0.08 m) |a_max| = 1.395 m/sec^2
Consider a monopolistically competitive market with N firms. Each firm's business opportunities are described by the following equations:?
Consider a monopolistically competitive market with N firms. Each firm's business opportunities are described by the following equations: Demand: Q=100/N-P Marginal Revenue: MR=100/N-2Q Total cost: TC=50+Q(squared) Marginal Cost: MC=2Q a. How does N, the number of firms in the market, affect each firms demand curve? Why. b. How many units does each firm produce? (The answer to this and the next two questions depend on N.) c. What price does each firm charge? d. How much profit does each firm make? e. In the long run, how many firms will exist in this market?