Project

Problem 1:

m= 1.2 kg, x= 90m

PE= mgh

PE= 1.2*9.8*90=1058.4 J

Given,

KE= 3* PE= 3*1058.4= 3175.2

we know, by law total energy of system remains constant PE=KE=3175.2=mgh

h=3175.2/(9.8*1.2)= 270; therefore, h=270m

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Problem 2:

T2= 22 Deg C

T1= -10 Deg C

A= 140m^2

t1= 2.5 cm = 0.025m

t2= 2 cm= 0.02m

t3= 25 cm= .25m

Q= Delta T/ (Summation (L/kA))

Delta T= 22-(-10)= 32 deg C

Q= 32/( (0.025/.125)+(.025/1.5)+(.25/1))

therefore Q= 69.0647 W or 0.06906 kW

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Problem 3:

Equvalent resistance R=>

R1= 15

R2= 30

R3=15

R4=30

R5=50

R= ((R1+R3)*(R2+R4)*R5 + (R1+R2)*R3*R4+ R1*R2(R3+R4))/ ((R1+R2+R3+R4)R5 + (R1+R2)*(R3+R4))

R= ((15+15)*(30+30)*50 + (15+30)*15*30+ 15*30*(15+30))/ ((15+30+15+30)*50 + (15+30)*(15+30))

R = 130500/ 6525

R= 20 Ohm

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