hi good evening ! please solve this problem ,A circular turn on a racing track has a radius of 150m . what must be the banking angle if the optimum speed for negotiating the turn is to be 108 kmph? please fast
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Answered by
2
hey dear___ ^^_^^
ur answe is here...
Optimum speed = v₀ = [RgtanΘ]¹⁻₂
R= 300
theta = 15⁰
g= 10
so, v₀= √300 x 10 x tan 15
= 28.06 ms⁻²
2. max speed= Vm= √Rg u+tan theta/ 1- utan theta
as u=0.2
= √300 x 10 0.2+tan15/1- 0.2tan 15
= 38.13 ms⁻¹
✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅
hope its hlp uh!!!
ur answe is here...
Optimum speed = v₀ = [RgtanΘ]¹⁻₂
R= 300
theta = 15⁰
g= 10
so, v₀= √300 x 10 x tan 15
= 28.06 ms⁻²
2. max speed= Vm= √Rg u+tan theta/ 1- utan theta
as u=0.2
= √300 x 10 0.2+tan15/1- 0.2tan 15
= 38.13 ms⁻¹
✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅✅
hope its hlp uh!!!
AryadhaSharma2002:
u are answer is right
or my i don't know
but thank you
for your help
ok bye
take care
good night
keep smiling
Thanks for the wishes
Sorry, wrong comment section
Answered by
1
v= √RgtanQ. (Q= theta)
=> v²= RgtanQ
Convert 108kmph into m/s.
Put the values of v, R and g. Get the value of Q.
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