Asked by Adekunle Adebola
The resistance of the motionof a vechincle is partly constant and partly constant and partly varies to the square of its speed.At 30km/h the
resistance is 496, and at 50km/h is 656N.Find the resistance at 60km/h
resistance is 496, and at 50km/h is 656N.Find the resistance at 60km/h
Answers
Answered by
scott
R = c + k s^2
... 496 = c + (k * 30^2) = c + 900 k
... 656 = c + (k * 50^2) = c + 2500 k
subtracting equations ... 160 = 1600 k ... k = .1
... substituting ... c = 406
... R = 406 + .1 s^2
plug 60 into the equation to find the resistance at 60 kph
... 496 = c + (k * 30^2) = c + 900 k
... 656 = c + (k * 50^2) = c + 2500 k
subtracting equations ... 160 = 1600 k ... k = .1
... substituting ... c = 406
... R = 406 + .1 s^2
plug 60 into the equation to find the resistance at 60 kph
Answered by
Abraham David
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Answered by
Peter
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Answered by
Peter
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Answered by
Gbemisola
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Answered by
Fatima Ahmad Umar
Thanks
Answered by
Komolafe Tofunmi
(i) R=496N
S=30km/h
(ii) R=656N
S=50km/h
R=a+s²k
496=a+900k........(I)
656=a+2500k......(ii)
Subtract equation (ii) from (I)
K=0.1
Putting our k in equation (I)
496=a+900*(0.1)
496=a+90
a=496-90
a=406
When s=60km/h, find R.
R=406+(60)²(0.1)
R=406+3600(0.1)
R=406+360
R=766
Final Answer is 766
S=30km/h
(ii) R=656N
S=50km/h
R=a+s²k
496=a+900k........(I)
656=a+2500k......(ii)
Subtract equation (ii) from (I)
K=0.1
Putting our k in equation (I)
496=a+900*(0.1)
496=a+90
a=496-90
a=406
When s=60km/h, find R.
R=406+(60)²(0.1)
R=406+3600(0.1)
R=406+360
R=766
Final Answer is 766
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