Problem 19. Must use formula from formula sheet. Not sure which one or where to start. v" for volume no"U" use subscriptsMusH start wlone ofuseThese eguahbasPhysics 104 Equation Sheetk = 9.0 x 10° Nm²/C²Ho = 4M x 107 N/A?C = 3.00 x 10° m/se = 1.6 x 10-19 cEo = 8.85 x 1012 C²/Nm?"-6.63 x 10 34 Js = 4.14 x 10 15 eVs lo = 1012 W/m?me = 9.11 x 10 kgm, = 1.67 x 10 27 kgnair = 1.00nsoap = 1.50nwater = 1.33Vsound = 340 m/s1 eV = 1.6 x 10 19 JWsilver = 4.74 eVF = kqıq2/r²E = kQ/r?F = qEPE = kqı92/rV = kQ/rΔV - -ΕΔxAPEE = -WE = -qEdAPE = qAVC = q/AVC = E0A/dR.Σ R| = Aq/AtF = qvB sinoEMF = NA DB/AtPE = ½ QAVAV = IR1/Req Σ 1/RP = IAVB = Hol/2nrR = PL/AF = ILB sinoOB = BA cosOB = Ho nl¡V1 = 12 V2V1/V2 = N1/N2EMFg = NBAW sin(wt)f = 1/Tv = fAW = 2nf =F = -kAxn = c/vAn = Aair/nPE = ½ KAX²fsfo =V =ni sino1 = n2 sino2B = 10 log10 (1/lo)12/l1 = r/r2?d sino = (m + ½ )A1/o + 1/i = 1/fm = -i/ofo = f.(1 ± °o/p)d sino = mf = R/2P = 1/fAm T = 2.90 x 103 mKEn = -13.6eV/n?KEmax = hf – WE = hfL= nh/2nRn = (0.0529 nm) n²A = h/mv = h/pAxAp, 2 h/4nΔΕΔt h/4πA-N' = (h/mc)(1 – cos©) what factor does the amplitude change?move from y= 1.5 cm to y= 2.5 cm? (d) If thevelocity? (c) How long does itation is 3.5 cm,ofto two springs as it slides along aamplitude? (e) If the spring constant is increased by a factor ofmass is oscillating with a maximum speed of 45 m/s, what is thetwo and the maximum kinetic energy of the mass is the same, byFigum17. * A mass m =2.4 kg is attachedfrictionless floor, while the springsare fastened to two walls as shownin Figure P11.17. The springs bothakmww= 400 N/m and are bothhave kin their relaxed states (unstretchedand uncompressed) when the massis centered between the two walls. What is the frequency of thisFigure P11.17simple harmonic oscillator?19 SSM * RT Estimate the spring constant for a trampoline.Assume a person is standing on the trampoline and oscillating upand down without leaving the trampoline. Hint: Begin by estimat-ing the mass of the oscillator and the period of the motion.29.19. * A mass-on-a-spring system has m =The mass is pulled a distance 0.25 m from its equilibrium posi-tion and then released. (a) What is the maximum acceleration ofthe mass? (b) What is its maximum velocity?50 kg and k200 N/m.%3D30.20. * Consider the mass-on-a-spring system inFigure P11.20. Three identical springs, withthe same spring constant kused to connect the mass (m = 20 kg) to aceiling. What is the frequency of this simpleharmonic oscillator?40 N/m, are%3D%3Dm21. * RT Estimate the maximum accel-eration of a bungee jumper. Is it larger orsmaller than the acceleration due to gravityduring free fall? Hint: Example 11.3 gives values for severalFigure P11.20Simple spring (which is not quite realistic) and that the mass ofthe nerson is 50 kg.of a rock of mass

Question

Problem 19.  Must use formula from formula sheet. Not sure which one or where to start. v&#34; for volume no&#34;U&#34; use subscriptsMusH start wlone ofuseThese eguahbasPhysics 104 Equation Sheetk = 9.0 x 10° Nm²/C²Ho = 4M x 107 N/A?C = 3.00 x 10° m/se = 1.6 x 10-19 cEo = 8.85 x 1012 C²/Nm?&#34;-6.63 x 10 34 Js = 4.14 x 10 15 eVs lo = 1012 W/m?me = 9.11 x 10 kgm, = 1.67 x 10 27 kgnair = 1.00nsoap = 1.50nwater = 1.33Vsound = 340 m/s1 eV = 1.6 x 10 19 JWsilver = 4.74 eVF = kqıq2/r²E = kQ/r?F = qEPE = kqı92/rV = kQ/rΔV - -ΕΔxAPEE = -WE = -qEdAPE = qAVC = q/AVC = E0A/dR.Σ R| = Aq/AtF = qvB sinoEMF = NA DB/AtPE = ½ QAVAV = IR1/Req Σ 1/RP = IAVB = Hol/2nrR = PL/AF = ILB sinoOB = BA cosOB = Ho nl¡V1 = 12 V2V1/V2 = N1/N2EMFg = NBAW sin(wt)f = 1/Tv = fAW = 2nf =F = -kAxn = c/vAn = Aair/nPE = ½ KAX²fsfo =V =ni sino1 = n2 sino2B = 10 log10 (1/lo)12/l1 = r/r2?d sino = (m + ½ )A1/o + 1/i = 1/fm = -i/ofo = f.(1 ± °o/p)d sino = mf = R/2P = 1/fAm T = 2.90 x 103 mKEn = -13.6eV/n?KEmax = hf – WE = hfL= nh/2nRn = (0.0529 nm) n²A = h/mv = h/pAxAp, 2 h/4nΔΕΔt h/4πA-N' = (h/mc)(1 – cos©) what factor does the amplitude change?move from y= 1.5 cm to y= 2.5 cm? (d) If thevelocity? (c) How long does itation is 3.5 cm,ofto two springs as it slides along aamplitude? (e) If the spring constant is increased by a factor ofmass is oscillating with a maximum speed of 45 m/s, what is thetwo and the maximum kinetic energy of the mass is the same, byFigum17. * A mass m =2.4 kg is attachedfrictionless floor, while the springsare fastened to two walls as shownin Figure P11.17. The springs bothakmww= 400 N/m and are bothhave kin their relaxed states (unstretchedand uncompressed) when the massis centered between the two walls. What is the frequency of thisFigure P11.17simple harmonic oscillator?19 SSM * RT Estimate the spring constant for a trampoline.Assume a person is standing on the trampoline and oscillating upand down without leaving the trampoline. Hint: Begin by estimat-ing the mass of the oscillator and the period of the motion.29.19. * A mass-on-a-spring system has m =The mass is pulled a distance 0.25 m from its equilibrium posi-tion and then released. (a) What is the maximum acceleration ofthe mass? (b) What is its maximum velocity?50 kg and k200 N/m.%3D30.20. * Consider the mass-on-a-spring system inFigure P11.20. Three identical springs, withthe same spring constant kused to connect the mass (m = 20 kg) to aceiling. What is the frequency of this simpleharmonic oscillator?40 N/m, are%3D%3Dm21. * RT Estimate the maximum accel-eration of a bungee jumper. Is it larger orsmaller than the acceleration due to gravityduring free fall? Hint: Example 11.3 gives values for severalFigure P11.20Simple spring (which is not quite realistic) and that the mass ofthe nerson is 50 kg.of a rock of mass

Accepted Answer

REcall
For SHM
y=Ymcos(wt)
ω=K/m

19. A mass-on-a-spring system has m = 50 kg and k = 200 N/m. The mass is pulled a distance 0.25 m from its equilibrium posi- tion and then released. (a) What is the maximum acceleration of the mass? (b) What is its maximum velocity? %3D