3/2 v² e-Mv²/2R derive the expression for (a) Vavg, (b) (v²) avg, (c) M Starting with (v) = 4T 2πRT Vrms, and (d) vp showing all steps necessary to validate each result.
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- (Please refer to Fig. 1) W = FAx (Equation 1) F = qE (Equation 2) APEelec = -W (Equation 3) Substituting Eq. 1 in Eq. 3, APE elec Then subs. Eq. 2 in Eq. 4, 4. What does q means in the derived equation? = -FAx (Equation 4) APE etec = -qEAx (Eq. 2.1) Which is what we need to derive.If In(x) 3.37, what is the value of x?Additional Data: VCOM = Vn,LAB/(A + 1) n,COM = A Vn,LAB/(A + 1) VN,COM = - Vn,LAB/(A + 1)
- 4. The one-particle speed distribution, Ø₁(u) is given by an expression $1 ₁(v) = 4π =4(2mm) Derive an expression for (i) most probable speed. (mp) (ii) ensemble average speed. (iii) root mean square speed. 3/2 v²e-mu²/2kT 2k (iv) compare these three speed values. 8kT MIT (v) mean inverse speed. VThe curve is described by V(t)=Vm*sin(w*t+theta). Given V1=21, V3=7.518 volts, V2=-V1, and T=5.5 msec. Find Vm (volts), w (rad/sec), theta (degrees), and t4 (ms).Problem 2: At its peak, a tornado is 65 m in diameter and has 260 km/h winds. Randomized Variables d = 65 m s = 260 km/h What is its frequency in revolutions per second? f= cos() tan() 7 9 HOME sin() IT cotan() asin() acos() EAA 4 6. atan() acotan() sinh() 2 3
- A student wants to calculate the value of Newton's constant G using the following relation: GM g= R where g is the acceleration due to gravity near the surface of the Earth, M is the mass of the Earth and R is the radius of the Earth. She has the following values and uncertainties : g±Ag= (9.8±0.2)m/s², M ± AM = (6.0000 ±0.0004) × 1024 kg, R±AR=(6380 ± 20)km Which option below correctly expresses the resulting value of G and its uncertainty AG? Select one: 3 G+AG = (6.67±0.02) × 10-11 m² kg-s? G±AG= (6.7±0.2) × 10-11 _m² kg-s² 3 G±AG = (6.67±0.2) × 10-10 m² kg-s 2 G±AG = (6.7±0.02) × 10" kg-s²The radius of circle has been measured as r = (4.19 ± 0.3) cm. Calculate the uncertainty δA (in cm²) of its area (A = πr²). Enter the number only, without unit.I1:00:00 PM Problem 3: Suppose the speed of light were only 3000.0 m's. A jet fighter moving toward a target on the ground at 810 m/s shoots bullets, each having a muzzle velocity of 1050 m's. Randomized Variables v-810 m/s v2 1050 m's ed What is the velocity of the bullets relative to the target in km's? Grade Sommary Deductions 0% Potential 100%