Calculate the energies of the n = 1, n = 2, and n = 3 levels for an electron in a box with a length of 305 pm.?
h = 6.626*10^-34 m^2 kg / s (Planck's constant) m = 9.11*10^-31 kg (electron mass) L = 305 pm = 305*10^-12 m For a one dimensional box: E_1 = n^2 * h^2 / 8 * m * L^2 = 6.4758E-19 J E_2 = n^2 * h^2 / 8 * m * L^2 = 2.5903E-18 J E_3 = n^2 * h^2 / 8 * m * L^2 = 5.8282E-18 J
No.of sides of polygon..?
The sum of the interior angles of a polygon is 180(n-2) if there are n sides. Assume there are n sides and solve for n Sum = n/2 [ 2a+(n-1) d] Since the interior angles are in arithmetic progression with first term a=52 and common difference d=8, 180(n-2) = n/2 [ 2(52)+(n-1) 8] Multiply both sides by 2 360(n-2) = n[ 104 +8n-8] 360n-720 = 104n +8n^2-8n Rearrange the equation as a quadratic. 8n^2+96n-360n-720=0 8n^2-264n-720=0 n^2-33n-90=0 n^2-30n-3n-90=0 n(n-30)-3(n-30)=0 (n-30)(n-3) = 0 n=30 or n=3
Solve the following trig equation for 0 < x < 360°?
one way divide both sides by 2 cos(3x) = 1/2 so you want the angle measures that result in cosine of that angle equalling 1/2. Using the unit circle as a reference (or one cycle of the cosine graph... whichever you were taught) 3x = 60° + 360°n or 3x = 300° + 360°n, where n is any integer (integer being any positive or negative whole number) divide by 3 to get x = 20° + 120°n ..or.. x = 100° + 120°n start with n = 0: x = 20° ..or.. x = 100°, if n = 1 then x = 140° ..or.. x = 220° if n = 2 then x = 260° ..or.. x = 340° if n = 3 then x is no longer less than 360° (and if you started with n < 0 then x would also be less than 0) so your solution set is x in {20°, 100°, 140°, 220°, 260°, 340°} hope that helps
How do you solve n(n-1) (n-2) (n-3) =1680?
Let’s answer it simply :Do prime factorisation of 1680.1680= 2x2x2x7x2x3x5Now you can see 2x2x2=87=72x3=65=5We can clearly see form as: 8 x (8–1) x (8–2)x (8–3).n=8Thanks :)
Evaluate the expression. If P(n, 4) = 360, find n.?
P(n, 4) = 360 so (n!) / (n - 4)! = 360 n(n -1)(n - 2)(n - 3)(n - 4)! ------------------------------------- = 360 .......... (n - 4)! n(n - 1)(n - 2)(n - 3) = 360 (n² - n)(n² - 5n + 6) = 360 n⁴ - 5n³ + 6n² - n³ + 5n² - 6n = 360 n⁴ - 6n³ + 11n² - 6n - 360 = 0 solving for n and taking only real positive solutions (b/c n must be a positive integer) ... so n = 6
How can I find the value of n if 20/27= (2n C n) 3^-n?
(2n C n) = (2n)!/(n!)^2Therefore,20/27 = (2n)!/[(n!)^2 * 3^n]Due to the factorial component, I will be assuming we are looking for a solution in the natural numbers. I will from there conclude that since 20 has 5 as a prime factor, that 2n >= 5, and from the 27 in the divisor that n >= 3I will then attempt:n = 36!/[(3!)^2 * 3^n]6*5*4/[(3*2)*27]5*4/2720/27Therefore, n=3.
If n = (√7+√5) (√7-√5), what is 1/n?
The answer to the question is 2.Solution(√7+√5)(√7-√5) is in the form of(a+b) (a-b)=a^2 - b^2So let a=√7+√5 and b=√7-√5a^2 - b^2= (√7)^2 - (√5)^2=7–5=2So n=2Therefore 1/n= 1/2So the answer is 1/2
If nC3 = 35. What is n?
If nC3 = 35, since nCr= (n!)/((n-r)! r!)nC3=(n!)/((n-3)! 3!) =35[n(n-1)(n-2)(n-3)! ] / [(n-3)! 3!] ]= 35[n(n-1)(n-2) ] = 35 * 3!n³ - 3n² + 2n - 210 =0By Factorising, we get....(n-7) (n² + 4n + 30) = 0i.e n-7 = 0 or (n² + 4n + 30) = 0But... Since ’n' cannot be either negative or complex number, we ignore the quadratic equation (n² + 4n + 30)=0 which whose DISCRIMINANTS D=b² - 4*a*c <= 0 ( it has an imaginary roots when solved).Therefore...., (n-7) = 0, n = 7So, 7C3 = 35
What is the difference between [n>>3-n&7] and [(int) (n>>3)-(int) (n&7)] in Java?
First Case: (n >> 3 - n & 7)In this case, following the operator precedence, the first to get evaluated is3 - n = -9.Now, the next operator is the right shift operator. The right shift operator with a negative shift count is masked with 0x1f [1]. -9 & 0x1f = 23.Now, we evaluate the rest according to operator precedence.n >> 23 == 0 and0 & 7 = 0.Second Case: (n >> 3) - (n & 7) (the int typecasting is unnecessary)The explicit braces will ensure their first evaluations.n >> 3 = 1n & 7 = 4and 1 - 4 = -3The valuable lesson to learn here is to always use braces, don't rely on language quirks for any evaluation.[1] Chapter 15. Expressions
What is the answer of limit n tending to infinities (5^n + 3^n/5^n- 3^n)?
= Limit x tending to infinity { 5^n(1+ 3^n/5^n)/5^n(1–3^n/5^n)}= Limit x tending to infinity { (1+ 3^n/5^n)/(1–3^n/5^n)}=( 1+0)/(1–0) since 3^n/5^n = (3/5)^n tends to 0 as n tends to infinity= 1Therefore, Limit x tending to infinity { (5^n+3^n)/(5^n-3^n)} = 1