plz answer this question guys
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Answer:
F= ∆P/∆t , F=m∆v/∆t , f=m(v2-v1)/∆t here v1 and v2 is same but direction different and angle between them 74° 144km/hr is convert into 40m/sec then mode of v2-v1 calculate through vector formula |v2-v1|^2 = v1^2+v2^2-2v1×v2cos74° put the value in equation v2-v1 = 40^2+40^2-2×40^2cos74° then common 40 among them ( here v2-v1 ka square tha vo dusari side Jake root main convert ho gya h issiliye 40^2 common lene par root ke vgh se 40 common lena) then v2-v1= 40 × √1+1-2cos74° then v2-v1= 40√2(1-cos74°) common liya 2 ko but vo abhi Root main hi h here cos 74° if we assume it is cos2¢ then cos74°=cos2×37° and cos2¢= 1-2sin^2¢ (1-2sin^2×37°) put the value v2-v1=40√2{1-(1-2sin^2×37°)} cut 1 and -1 then v2-v1=40√4sin^2× 37° solve the Root v2-v1=40×2×sin37° (sin37°=3/5) then v2-v1= 48m/sec ,,,,,F=m(v2-v1)/∆t ,F=(1/3×48)/0.02 then after solving F=800N