7. A car of mass 1000 kg moving with a velocity of 72 km/h comes to a half in 10 sec; find the retardation and force applied by the breaks of the car?
Answers
Given mass of car =1000kg
Given mass of car =1000kgu=speed=90km/hr=90×185m/sec
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/sec
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5sec
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+at
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000M
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by break
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)v2=u2+2as
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)v2=u2+2as0=(25)2−2×5×s
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)v2=u2+2as0=(25)2−2×5×s=s=10(25)2=62.5m
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)v2=u2+2as0=(25)2−2×5×s=s=10(25)2=62.5mwork done by break =F×s
Given mass of car =1000kgu=speed=90km/hr=90×185m/secu=25m/secit have to stop in 5secthen final velocity =0⇒v=u+ato=25+a×5=a=−5m/sec2then the force applied by break is ma=1000X−(5m/sec2)=−5000Mforce =5000N by breakdistance travelled to stop the car after break is applied (s=distance)v2=u2+2as0=(25)2−2×5×s=s=10(25)2=62.5mwork done by break =F×sw=5000×62.5=312.5KJ
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