{"id":32391,"date":"2026-07-01T14:04:09","date_gmt":"2026-07-01T14:04:09","guid":{"rendered":"https:\/\/academicwritersbay.com\/solutions\/unit-transient-unit-2-4002-engineering-maths-assignment-1-mathematical-systems-and-statistical-ways-assignment-1-of-3-introduction-job-1-a-the-energy\/"},"modified":"2026-07-01T14:04:09","modified_gmt":"2026-07-01T14:04:09","slug":"unit-transient-unit-2-4002-engineering-maths-assignment-1-mathematical-systems-and-statistical-ways-assignment-1-of-3-introduction-job-1-a-the-energy","status":"publish","type":"post","link":"https:\/\/academicwritersbay.com\/solutions\/unit-transient-unit-2-4002-engineering-maths-assignment-1-mathematical-systems-and-statistical-ways-assignment-1-of-3-introduction-job-1-a-the-energy\/","title":{"rendered":"Unit Transient Unit 2\/4002: Engineering Maths Assignment 1: Mathematical Systems and Statistical Ways Assignment #1 of 3 INTRODUCTION Job 1) a) The energy"},"content":{"rendered":"<p>Unit Transient Unit 2\/4002: Engineering Maths Assignment 1: Mathematical Systems and Statistical Ways Assignment #1 of 3 INTRODUCTION Job 1) a) The energy (P) dissipated in a resistor (R) which is subjected to a voltage (V) across its terminals is given by the successfully-identified system;<\/p>\n<p>P = V^2\/R <\/p>\n<p>Because you are new to the corporate your line supervisor is doubtful of your capabilities, and has requested you to make insist of the dimensions of P and V to resolve the dimensions of R.<\/p>\n<p>Voltage (V) = 25v   <\/p>\n<p>Resistance (R) = 20\u03a9<\/p>\n<p>Present (I) = ?<\/p>\n<p>Ohms Legislation is V=IR (Unicourse Ltd, 2025) we are capable of transpose the above for I:<\/p>\n<p>I=V\/R <\/p>\n<p>Which strategy of this truth:<\/p>\n<p>I=25\/20 <\/p>\n<p>I = 1.25 A (Amperes, in general shortened to Amps, A)<\/p>\n<p>b) A guitar string, made by your Musical Devices division, has mass (m), Size (l) and tension (F). It is proposed by even handed one of your junior colleagues that a system for the length of vibration (t) of the string will most certainly be;<\/p>\n<p> t=2\u03c0\u2008\u221a\u00a0 m3l\/F<\/p>\n<p>Use dimensional evaluation to prove your colleague that this system is wrong.<\/p>\n<p>As we know from above the dimensions of every variable, we are capable of deem that when the equations are labored, they must always be of equal system both facet to be correct.<\/p>\n<p>From the general portions table encountered in engineering we are capable of deem the next: &#8211;<\/p>\n<p>T = time (T)<\/p>\n<p>M = mass (M)<\/p>\n<p>L = Size (L)<\/p>\n<p>F = Power (mass x acceleration) [MLT]<\/p>\n<p>Dimensional evaluation Expression beneath the sq. root <\/p>\n<p>[m3l\/F]=(M^3 L)\/(M\u2008LT-^2 )=M^(3-1)\u2008L^(1-1\u2008) T^2\u2008=M^2 T^2 <\/p>\n<p>Square root the equation<\/p>\n<p>(_^([\u221a\u2008(m^3 l)\/F]\u2008=\u2008\u221a(M^2 ) T^2\u2008=M\u2008T)) <\/p>\n<p>As we are capable of search the equations don`t equal the identical spoil consequence and with their calculation notion and the system is wrong to cloak a length of vibration.<\/p>\n<p>c) An analogue-to-digital converter (ADC), manufactured by you indicators division, takes 15 voltage samples of ramp waveform, measured in mV, as follows &#8230; 3, 6, 9, 12, 15 &#8230; 45.<\/p>\n<p>Your Check colleague has requested you to relief by utilizing a system to calculate the sum of these 15 voltage samples.<\/p>\n<p>There are a few ways this is also finished; the principle approach is to total the system for your total numbers as follows with Sigma notation: &#8211;<\/p>\n<p>(_^([\u221a\u2008(m^3 l)\/F]\u2008=\u2008\u221a(M^2 ) T^2\u2008=M\u2008T))<\/p>\n<p>We are capable of search that every new term is constructed by including 3 the old amount and is identified because the <code>common ratio<\/code><\/p>\n<p>No. of samples n = 15<\/p>\n<p>1st Voltage sample a1 = 3mV<\/p>\n<p>Final Voltage sample a15 = 45mV<\/p>\n<p>Sn\u2008=\u2008n\/2 (a1+an)<\/p>\n<p>S\u200815\/2\u2008=\u2008(3+45)<\/p>\n<p>S\u200815\/2 (forty eight)<\/p>\n<p>7.5\u2008x\u2008forty eight\u2008=\u2008360\u2008mV<\/p>\n<p>d) A digital chip, made by your microelectronics division, counts continuously in the sequence: 512, 1024, 2048, 4096, &#8230;<\/p>\n<p>You had been requested to make insist of a system to calculate the 11th depend of the chip<\/p>\n<p>an\u2008=\u2008a\u2008r\u2008(n-1) <\/p>\n<p>a11\u00a0= 512 * 2(11-1)<\/p>\n<p>a\u2008=\u2008512\u2008x\u20082^10<\/p>\n<p>a11 = 524,288<\/p>\n<p>e) A chain electrical circuit which you are checking out beneficial properties a capacitor (C) charging by diagram of a resistor (R) and a DC provide (Vs). The voltage across the capacitor (Vc) could per chance well additionally be described by the equation &#8230;<\/p>\n<p>vc = vs(1-e-t\/_\/rc) <\/p>\n<p>Where t represents time<\/p>\n<p>Assuming Vc is 1V after a time of 4 seconds, resolve the approx label of the capacitor.<\/p>\n<p>system working:<\/p>\n<p>vc\/vs=1-e^(-t\/rc)<\/p>\n<p>e^(-t\/RC)=1\u2008-\u2008Vc\/Vs<\/p>\n<p>C\u2008=\u2008(-t)\/(R1n(1-Vc\/Vs))<\/p>\n<p>Voltage (V) = 25v  <\/p>\n<p>Resistance (R) = 20\u03a9  <\/p>\n<p>Vc\/Vs = (1 &#8211; e^(-4(20 x 10^(-6) C)))<\/p>\n<p>1\/25 = 1 &#8211; e^(-4(20 x 10^6 C))<\/p>\n<p>e^(-4(20 x 10^6 C)) = 1 &#8211; 1\/25<\/p>\n<p>e^(-4(20 x 10^6 C)) = 24\/25<\/p>\n<p>C = -4 \/ (20 x 10^6 ln(24\/25))<\/p>\n<p>ln(24\/25) = -0.0416666667<\/p>\n<p>C = -4 \/ (20 x 10^6 (-0.0416666667))<\/p>\n<p>C = -4 \/ (20 x 10^6 x -0.0416666667)<\/p>\n<p>C = 4 \/ (0.0833333334 x 10^6)<\/p>\n<p>C = 4 \/ 83333.334 = 4.8 x 10^(-5)<\/p>\n<p>C = 4.8 x 10^(-5)<\/p>\n<p>C = 4.8 \u03bcF<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Unit Transient Unit 2\/4002: Engineering Maths Assignment 1: Mathematical Systems and Statistical Ways Assignment #1 of 3 INTRODUCTION Job 1) a) The energy (P) dissipated in a resistor (R) which is subjected to a voltage (V) across its terminals is given by the successfully-identified system; P = V^2\/R Because you are new to the corporate [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-32391","post","type-post","status-publish","format-standard","hentry","category-solutions"],"_links":{"self":[{"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/posts\/32391","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/comments?post=32391"}],"version-history":[{"count":0,"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/posts\/32391\/revisions"}],"wp:attachment":[{"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/media?parent=32391"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/categories?post=32391"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/academicwritersbay.com\/solutions\/wp-json\/wp\/v2\/tags?post=32391"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}