Can anyone help me with physics problem?
1) check which of the following equations are dimesionally correct.
d) KE=1/2m x^2/t, where KE has units of (mass)(velocity)^2
Is the answer A?
If not correct me.
The question asks to check and does that mean i need to simplify the equation?
2) The frequency of a stretched string w (1/sec) depends on the following properties of the string:
Tension in the string, T (Newtons=kg m/s^2)
Length of string, L (meters)
Linear Density, λ (kg/m)
Find a,b,c so that w=KT^aL^bλ^c makes sense dimensionally.
im not sure what question 2 is asking me to do. Plz give me an explanation with it when ur doing the problem :)
- Ask_meLv 41 decade agoFavorite Answer
Q.1: both a) and c) are dimensionally correct.
working out a), on the LHS, you have X, which has the dimension of L (length). On the RHS, you have 3 terms-- 1st being x itself, 2nd is velocity (LT^(-1)) multiplied by T, effectively making it L, and the 3rd term is acceleration (LT^ -2) multiplied by T^2 which also works out to be L. Thus the equation is dimensionally correct.
working out b), on LHS, it is (LT^ -2)^2, i.e. L^2 T^ -4. RHS has two terms, the first one is same as the LHS dimensionally. The second one has dimensions of (LT^ -2)* L^2, becoming L^3 T^ -2 which is not the same as the LHS. The equation is not correct dimensionally.
working out c), LHS is L, while RHS has two terms (1/2 vot and 1/2 vt) of same dimension. Its dimension is LT^ -1*T, which is L again. The equation is dimensionally correct.
Working out d), LHS is M (LT^ -1)^2 i.e. ML^2T^ -2. Now RHS is M L^2 T^ -1. It is dimensionally not correct.
Regarding Q2, it can be solved very easily dimensionally as follows:
LHS is w which is T^ -1 (T being time)
RHS is (MLT^ -2)^aL^b(ML^ -1)^c
or M^(a+c)L^(a+b-c)T^ -2a
thus, -2a = -1 or a=1/2
a+c = 0. giving c= -1/2
a+b-c = 0, or 1/2 +b - (-1/2) = 0
or 1 + b = 0, giving b= -1
there you are
- David DodecaLv 51 decade ago
For Question 1 a) you would write it out like this...
m=m + m/s*s + m/s/s x s*s
and see if things cancel out right.
(by the way you are right)
in 2 he is asking you to what powers you raise T, L and lamda to.
All the units have to cancel out so that you are left with inverse seconds.
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