Check The Following Equation For
Calculating Displacement Is Dimensionally Correct
Or Not (a) X = X0 + Ut+ (1/2) At2 Where, X Is
Displacement At Given Time T Xo ls The
Displacement At T = 0U ls The Velocity At T = 0A
Represents The Acceleration. (b) P = (pgh)'% Where
Pls The Pressure, p Is The Density G Is Gravitational
Acceleration H Is The Height.
(a) x = x0 +ut + (1/2) at2
Answers
Answered by
3
As LHS=RHS, formula is dimensionally correct.
Writing the dimensions of either side of the given equation.
LHS=8=displacement=[M
0
LT
0
]
RHS=ut=velocity×time=[M
0
LT
−1
][T]=[M
0
LT
0
]
and
2
1
at
2
=(acceleration)×(time)
2
=[M
0
LT
−2
][T]
2
=[M
0
LT
0
]
As LHS=RHS, formula is dimensionally correct.
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