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• Hello mmmeee, let me derive the right expression to get the required velociy. If I be the current through the rod of length l to whose length B cos 25 magnetic field induction is acting then the force on the rod up the plane will be B cos 25 l sin…

  in physics problem : Faraday's law? Comment by trueprober June 2021
• First reciprocate on both sides you get, 4(7+3g) = - 8/g Divide both sides by 4. Now you will get 7+3g = -2/g Now multiply both by g there by you get 7g + 3g^2 = -2 Bring -2 to the left side. So 3g^2 + 7g + 2 = 0 Factorising, 3g^2 + 6g + g + 2 =…

  in algebra problem.....? Comment by trueprober June 2021
• First reciprocate on both sides you get, 4(7+3g) = - 8/g Divide both sides by 4. Now you will get 7+3g = -2/g Now multiply both by g there by you get 7g + 3g^2 = -2 Bring -2 to the left side. So 3g^2 + 7g + 2 = 0 Factorising, 3g^2 + 6g + g + 2 =…

  in algebra problem.....? Comment by trueprober June 2021
• Hello Kristen Jacobs, 2 + 2 sin @ = 4(1 -sin^2 @) So cancelling 2 through out, we have 1 + sin@ = 2 - 2 sin^2@ OR 2 sin^2@ + sin@ -1 = 0 (2 sin@ - 1 )(sin@ + 1) = 0 So two options, sin@ = 1/2 or sin@ = -1 So @ = pi/6 or @ = 3 pi/2

  in factor 2+2sintheta=4cos^2theta? Comment by trueprober May 2021
• Hello Kristen Jacobs, 2 + 2 sin @ = 4(1 -sin^2 @) So cancelling 2 through out, we have 1 + sin@ = 2 - 2 sin^2@ OR 2 sin^2@ + sin@ -1 = 0 (2 sin@ - 1 )(sin@ + 1) = 0 So two options, sin@ = 1/2 or sin@ = -1 So @ = pi/6 or @ = 3 pi/2

  in factor 2+2sintheta=4cos^2theta? Comment by trueprober May 2021
• Hello T, Energy given out in joule is 16000 x 4 x 10^9 J So the energy in eV will be 16000 x 4 x 10^9 / 1.6 × 10−19 = 4 x10^32 eV For 208x 10^6 eV energy we need one fissionble atom So for 4 x 10^32 eV the number of fissions will be (4000/208) …

  in physics problem trouble? Comment by trueprober May 2021
• Hello T, Energy given out in joule is 16000 x 4 x 10^9 J So the energy in eV will be 16000 x 4 x 10^9 / 1.6 × 10−19 = 4 x10^32 eV For 208x 10^6 eV energy we need one fissionble atom So for 4 x 10^32 eV the number of fissions will be (4000/208) …

  in physics problem trouble? Comment by trueprober May 2021
• Hello Kristen Jacobs, 2 + 2 sin @ = 4(1 -sin^2 @) So cancelling 2 through out, we have 1 + sin@ = 2 - 2 sin^2@ OR 2 sin^2@ + sin@ -1 = 0 (2 sin@ - 1 )(sin@ + 1) = 0 So two options, sin@ = 1/2 or sin@ = -1 So @ = pi/6 or @ = 3 pi/2

  in factor 2+2sintheta=4cos^2theta? Comment by trueprober May 2021
• First reciprocate on both sides you get, 4(7+3g) = - 8/g Divide both sides by 4. Now you will get 7+3g = -2/g Now multiply both by g there by you get 7g + 3g^2 = -2 Bring -2 to the left side. So 3g^2 + 7g + 2 = 0 Factorising, 3g^2 + 6g + g + 2 =…

  in algebra problem.....? Comment by trueprober May 2021