Is u+a/2(2n-1) is dimensionally correct or not??
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u+a/2(2n-1) is dimensionally correct.
Dahiya22:
Please give the solution
So, S(nth) = distance travelled in nth second
i.e S(nth) = [L]/[T] = [LT^-1] dimensionally
RHS
Now, U+(1/2)*a*(2n-1) = where,
U= initial velocity = [LT^-1]
a= acceleration = [LT^-2]
2n-1= n=time = [T]
1/2=dimensionless = [M°L°T°]
——————————————————
U+(1/2)*a*(2n-1)= [LT^-1]+[LT^-2]*[T]
[LT^-1]+[LT-^1]=[LT^-1] by principle of homogeneity.
As LHS= RHS dimensionally i.e [LT^-1]
Therefore, it is dimensionally correct.
i.e S(nth) = [L]/[T] = [LT^-1] dimensionally
RHS
Now, U+(1/2)*a*(2n-1) = where,
U= initial velocity = [LT^-1]
a= acceleration = [LT^-2]
2n-1= n=time = [T]
1/2=dimensionless = [M°L°T°]
——————————————————
U+(1/2)*a*(2n-1)= [LT^-1]+[LT^-2]*[T]
[LT^-1]+[LT-^1]=[LT^-1] by principle of homogeneity.
As LHS= RHS dimensionally i.e [LT^-1]
Therefore, it is dimensionally correct.
Thanks
wlcm
Answered by
0
Concept:
- Dimensional analysis
- One dimensional motion
Given:
- Distance travelled in the nth second Sn = u+a/2(2n-1)
Find:
- Whether Sn = u+a/2(2n-1) is dimensionally correct
Solution:
Consider the right-hand side
- Distance covered in the nth second Sn = [LT⁻¹]
Consider the left-hand side
- u is the initial velocity, so u =[LT⁻¹]
- a is the acceleration, so a = [LT⁻²]
- n is the time, so n = [T]
- u+a/2(2n-1) = [LT⁻¹]
This is equal to the right-hand side.
Therefore, the equation is dimensionally correct.
#SPJ2
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