UI SLIDES & UPDATES

Definite integrals @scholars9ja

Integration 2 @scholars9ja

Logarithms @scholars9ja

*PHY 103 Solutions by Engr. Sileski Jr...* (1) ∆U=nCv∆T, ∆T=0; since T1=T2 =>∆U=0 J (A) (2) Absolute Zero temperature is the temperature at 0 K or –273.15°C, and the pressure is 0. D (3) ∆Tc=5/9 ∆Tf=5/9 × 70 =>∆Tc≈38.9°C (B) (4) Interpolating on the two scales:¹ (17.8–15.5)/(16.75–15.5) =(100–0)/(x–0) 2.3/1.25=100/x x=(1.25×100)/2.3 x≈54.3°C (C) (5) Let ¥, μ, ß represent gamma beta and alpha. ¥=3μ, ß=2μ; =>¥/3=ß/2, ¥=3ß/2 and ß=2¥/3 (D) only (6) L2=L1(1+μ∆T) ∆T= –65°C= –65 K =30[1+(9•10–⁶×–65)] =30[1–0.000585] =>∆T≈29.98 cm (B) (7) V2=V1(1+¥∆T) =V1(1+3μ∆T) =>V1=Area•L=πr²•L =(πd²L)/4 =(1.5²×30×3.142)/4 =>V1≈53.0213 cm³1 Now, V2=53.0213•[1+(3×9×10–⁶×–65)] V2=53.0213•[1–0.001755] =>V2≈52.93 cm³ (A) (8) A is very much valid. (9) N2 is a diatomic linear molecule, hence two rotational degrees freedom (D) (10) Recall that for a Polytropic process: PV^¥=K ---> (1) =>PiVi^¥=PfVf^¥ ---> (2) But, from the ideal gas eqn: PV=nRT V=nRT/P ---> (3) Substitute eqn (3) in eqn (2); Pi(nRTi/Pi)^¥=Pf(nRTf/Pf)^¥ Simplifying; Ti^¥•P^(1–¥)=Tf^¥•Pf^(1–¥) Also, P=nRT/V --->(4) Substitute eqn (4) in eqn (2) similarly; (nRTi/Vi) • Vi^¥=(nRTf/Vf)•Vf^¥ Simplifying; Vi^(¥–1)•Ti=Vf^(¥–1)•Tf Option (C) is valid 🤝✅

PHY 102 TEST ANSWERS (1) PV^¥=C This is a polynomial of degree ¥ depending on the values of ¥, to make this linear, we take normal logarithm of both sides; =>Log(PV^¥)=LogC LogP+¥LogV=LogC *Transposition;* *LogP= –¥LogV+LogC* *Comparing the above eqn with y=mx+c;* *=>A plot of Log P against LogV will give a slope of –¥ and intercept of LogC on the y–axis (D)* ✅🤝 *(2) A, namely fundamental and derived quantities* *(3) C is valid... If a quantity is unitless, it is dimensionless (no dimension or that the dimension is 1), but a dimensionless quantity doesn't mean the quantity has no units (e.g supplementary quantities like angles)* *(4) x=Et²+F (displacement in x–axis)* *y=Ht³–I (displacement in y–axis)* *z=Gt⁴+J (displacement in z–axis; 3 dimensional plane)* *From principle of dimensional homogeneity;* *[x]=[Et²]=[F]* *=>[F]=[L]; [E]=[x/t²]=[L/T²]=[LT–²]* *=>[E]=[LT–²]* *Also, [y]=[Ht³]; [H]=[y/t³]=[L/T³]=[LT–³]* *=>[H]=[LT–³]* *[z]=[Gt⁴]=[J]* *=>[J]=[L]* *[G]=[z/t⁴]=[L/T⁴]=[LT–⁴]* *=>[G]=[LT–⁴]* *: . The dimensions of E, F, J and G are: LT–², L, L and LT–⁴ (B)*✅🙂 *(5) R²=A²+B²+2ABCos∅; where ∅ is the angle between the two vectors, A and B are the forces exerted by the girls on the Toy;* *R²=15²+10²+2(10×15Cos30)=325+259.808* *R=√584.808≈24.18 (D)* *(6) R=F1+F2+F3=7i+6j* *=>R=√(7²+6²)=√(85)* *R≈9.22 N (A)* *(7) VAB=VB–VA=3i+2j+7k* *VAB=√(3²+2²+7²)=√62* *=>VAB≈7.874 ms–¹* *S=VAB • t=7.874×2.5* *=>S=19.69 m (C)* *Note: whether you added VA and VB instead of subtracting, you should arrive at same answer* *(8) VAB..... *(9) Recall that: dw=F•dx* *w=$Fdx* *=>w=$(5+2x)dx from x=0 to x=2* *w=(5x+x²) from x=0 to 2* *w=[5(2)+2²]–[5(0)+0²]* *=10+4=14 J (A)* *(10) W=F•∆x* *∆x=x2–x1=P–(0,0)* *=>∆x=P* *=(5i+3j+2k)•(2i–j)* *=10–3=7 J* *=>W=7 J (B)*

Thermodynamics @scholars9ja

Newton's law if motion @scholars9ja

FORCES @scholars9ja