Finding Capacitor Values in RC Bandpass?
First off, V1 is considered to be a perfect voltage source, meaning it has zero impedance. Thus, C1 and R1 are not affected by the other components. C2 and R4 are not affected by the other components. Same for C2,R3,R5,C4, and for the op amp and its components. For C1 and R1, a high-pass filter, the corner frequency fc is given by: fc = 1/(2 pi R1 C1) http://www.electronics-tutorials.ws/filt... You are given fc = fhp = 300Hz, R1 = 100K. Solve for C1. C1 = 1 / (2 pi R1 fc) = 5.31nf = 5nf (rounded to 1 digit) For R4 and C2, a low-pass filter, the corner frequency is given by: fc = 1/(2 pi R4 C2) (Note that it is the same format as above.) You are given fc = flp = 20kHz, R4 = 10K. Solve for C2. C2 = 1 / (2 pi R4 fc) = 796pf = 1nf (rounded to 1 digit) Jumping down to the op-amp line, the op amp places neglible loading on the high-pass circuit that is at its input, so it doesn't change the output of the high-pass circuit. The op amp isolated and buffers the voltage and provides a low impedance output to drive the low-pass circuit on its output, so the op amp doesn't shift the filter response of the output circuit. Thus, the input and output filters are independent, and you use the same calculations above. So C5 = C1, R6 = R1, R7 = R4, and C6 = C2 The third circuit down (C2, R3, R5, C4) is much more difficult to calculate because the low-pass filter (R5, C4) loads down the high-pass filter (C3, R3). R5 is only 1/10 the value of R3, so C3 is no longer looking into 100K but rather 100K in parallel with 10K. That will shift its corner frequency (the high-pass corner frequency) much lower. It is approximately calculated as follows: fc = 1 / (2 pi R3||R5 C3) but R3||R5 = 100K||10K = 9.091K fc = 1 / (2 pi 9091ohm 5nf) = 3501Hz = 3.5KHz Note that 3.5 KHz is far higher than 300 Hz. Thus, C3 can't be 5nf if the corner frequency is to be 300 Hz and R3 is to be 100Kohm. Similarly, C4 will have to shift but not as much since R3 (100K) shunts the value of R5 (10K) by about 10%.
Help with finding the force?!?
By a smooth horizontal surface I assume it is considered frictionless. The 4.94N force will make the combination of two boxes accelerate with an acceleration given by: a = F / (m1 + m2) = 4.94 / (5.75 + 7.56) = .373 m/s For part a) look at the forces on the large box. The only force is the force of contact (ie, the force that the small box exerts on the large one). Let's call this force Fc. Newton's 2nd law tells us that Fc = M_large x a = 7.56 * .373 = 2.82 N. For part b) look at the forces on the small box. The only force is the contact force Fc. Newt's 2nd tells us that in this case Fc = M_small x a = 5.75 * .373 = 2.14 N
Finding orbital speed, answer given but don't get explanation!?
Find the orbital speed of an ice cube in the rings of Saturn, if the mass of Saturn is 5.67 × 10^26 kg and the rings have an average radius of 100,000 km. The correct answer is 19.5 km/s I approached this problem using GM1M2/(r^2) and even converted everything in the right units and still end up with a different answer, any help appreciated on this!
For the sequence An, find a number "r" such that An/r^n has a finite non zero limit?
freond1 is indeed correct. Since 7^(3n) = (7^3)^n = 343^n, then make n = 1/343 Then you get An/r^n = (6+7^n)^(-3) / (1/343)^n . . . . . = 343^n / (6+7^n)^(3) Now numerator is dominated by 343^n, while denominator is dominated by 7^(3n) = 343^n, so limit as n approaches infinity is 1 -------------------- Use same logic for other problems: 2) An = (3^2n) / (n+5^n) I'll assume that 3^2n means 3^(2n) = 9^n, and NOT (3^2)n = 9n Since dominant term in numerator is 9^n and dominant term in denominator is 5^n, then overall dominant term of An is (9/5)^n. Therefore, r = 9/5 limit[n→∞] [(3^(2n)) / (n+5^n) ] / (9/5)^n = 1 http://www.wolframalpha.com/input/?i=lim... Note: If we had 0 < r < 9/5, then (9/5)/r > 1, and limit would be infinity If we had r > 9/5, then (9/5)/r < 1, and limit would be 0 -------------------- 3. An = (7^n + n^3 + 7) / (15^(7n) + 5^n + 7) is dominated by 7^n / (15^(7n)) = 7^n / (15^7)^n = (7/15^7)^n r = 7/15^7 = 7/170,859,375 lim[n→∞] [(7^n + n^3 + 7) / (15^(7n) + 5^n + 7)] / (7/170,859,375)^n = 1 http://www.wolframalpha.com/input/?i=lim... Mαthmφm
Tension in a non-conducting wire attached to two point charges.?
The units you give are not clear, you should spell them out. I assume the charges are 8.75 micro coul and -6.50 micro coul. In the absence of the field, the force on the charges is Fc = -K*q1*q2/L² where L is the length of the wire. When the field is applied, a force is added to this which is the difference between the forces from the field on each charge: F1 = E*q1; F2 = - E*q2 the difference F = E*(q1 + q2). The net force is the wire tension, is F + Fc = E*(q1 + q2) - K*q1*q2/L² If both charges are negative, then F1 = -E*q1, F2 = -E*q2, and Fc = +K*q1*q2/L². The tension is then -E*(q1 - q2) + K*q1*q2/L²
What kind of relocation assistance does Amazon offer new employees?
You have two options:1 ) you can take a lump sum relocation package (e.g. $15K) and you figure out the relocations details yourself. or2) you use Amazon’s 3rd party relocation service that will offer you30 day of corporate housing (negotiable) — these are furnished apartment.rental car for a period of timea moving service that will take everything from your house and move to Seattle (or whatever location you’ll be working at)pick up your car and move to SeattleFlights from your current relocation to SeattleAndrew FranklinAuthor: How to Get a Job at Amazon & How to Get a Job at Google
Why is the diagram for strain linear and stress a curve in RCC?
The figure given is as per the theory of LIMIT STATE DESIGN. Following is the stress-strain diagram for a concrete block under compression test. The curve represented here is the true picture of what would happen if the concrete block was taken to failure. The curve is almost parabolic for a certain limit and after that plastic behavior of concrete is represented by a straight line (not the true curve though).Any structure however, is designed for specific load conditions which is usually much smaller than the ultimate load. This minimizing of design loads by applying factor of safety led to two prominent theories:Working stress methodLimit state methodWorking stress method assumed very small working load conditions and hence the diagram given above was limited to a small portion which could be considered linear. Hence the stress variation w.r.t. to distance from NA remained linear. Strain diagram remained linear in profile as per the assumption* that plane section remains plane after bending.The diagrams look like this:The beam stress & strain diagrams are as follows:Limit state method however was a refinement of the working stress method which used the concept of actual stress-strain curve. Here also the design loads were reasonably reduced, however the stress curve employed was similar to the true experimental curve. This is the curve that is being used both by the IS code and the diagram that you have mentioned. Strain diagram remained linear in profile as per the assumption* that plane section remains plane after bending.The curve looks like this:(A similar diagram is given in IS 456, pg 69, Section 38) The beam stress-strain diagram is the same as given by you.*Assumption:Plane sections normal to the axis remain plane after bending.” (IS 456, Pg 69)This implies that the bending is negligible as compared to the length of the beam, hence the beam deformations can be ignored.Hence the strain diagram w.r.t. to 'y' of beam is LINEAR.The tensile strength of concrete is ignored (which although is fcr= 0.7√fck) This is because as soon as the concrete bends the portion of beam under tension develops micro cracks being a brittle material, thereby any contribution by it becomes zero at even very small values of loads.Hence the stress diagram for concrete below the Neutral Axis (tension zone) is completely ignored.
What is the relationship between Celsius, Fahrenheit and Kelvin?
Generally there are three type of temperature unitsKelvin (K)Celsius (C)Fahrenheit(F)Mostly Kelvin and Celsius are used in thermal or i say physical science related calculations.Now, Let's Talk about relationsKelvin & Celsius K=C+273Celsius & Fahrenheit F = 9/5(C) + 32LET'S MAKE IT SIMPLE TO REMEMBER AND CONVERT: C/5 = (F-32)/9 = (K+273)/5
What is the meaning of 1.65 in concrete mix design?
We know from statistical studies, in a normal distribution curve (the bell shaped curve shown above), if the mean value of a number of samples is say 100 and standard deviation is calculated as, say, 4, then68.26% of all the results fall between 96 and 104.90% of the test results fall between -1.645σ and +1.645σ i.e 93.42 & 106.58Similarly, 95.45% of samples fall between -2σ and +2σ, ie between 92 and 108 and so on.By definition,characteristic strength, required for concrete mix design, is equal to that value of the strength of the concrete below which no more than 5% of all cubes tested from the chosen concrete mix will fall.As indicated above, 1.645s gives a confidence level of 90%, and out of the balance 10%, 5% may be below the lower limit (fck-1.645s) and 5% may be above the upper limit (fck+1.645s). Since we are only worried about the value falling below the required strength, we conclude that 95% of the test results are ABOVE the lower limit set, ie characteristic strength. Hence, for getting a characteristic strength, the mean strength should be 1.645 σ. That is how the target mean strength for concrete design has become fck + 1.645σ (or fck +1.65σ, rather).Taking example of M20 concrete for which as we know, characteristic strength will be 20 N/mm2.From IS 456: 2000 we can “assume” standard deviation ‘s’ as 4.Then the mean target strength will be calculated as : 20 + 1.65*4 = 26.6 N/mm2which means that 90% of the cubes tested for M20 concrete mix are will have compressive strengths between 20.6 (26.66–1.65*4) and 33.26 (26.66+1.65*4) N/mm2Since we are only worried about the value falling below the required strength i.e 20 N/mm2, we have ensured that 95% of the test results are ABOVE 20 N/mm2 by taking a value of “1.65” in calculation.