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pantera1 [17]
3 years ago
11

If an equation is dimensionally correct, does this mean that the equation must be true? If an equation is not dimensionally corr

ect, does this mean that the equation cannot be true? How is dimension analysis used as a check on the plausibiity of derived equations and computations? How does this knowledge help you to understand and characterize particular phenomenon? Explain.
Physics
1 answer:
tatiyna3 years ago
3 0

Answer:

Explanation: the explanation goes as follow;

  • If an equation is dimensionally correct, does this mean that the equation must be true?

answer: A dimensionally correct equation deosn't make the equation true. It is always possible to construct an arbitrary meaningless equation that satisfies the dimensional analysis. we can observe this in different physical quantities that have the same dimensional formula, by inserting a wrong physical parameter into an equation, it may turn out as dimensionally correct, but it wont have any physical meaning.

  • If an equation is not dimensionally correct, does this mean that the equation cannot be true?

answer:  A dimensionally incorrect equation necessarily implies that the equation is wrong.

  • How is dimension analysis used as a check on the plausibiity of derived equations and computations?

answer: dimensional analysis is a very strong tool that is used to check the trustworthiness of  a derived equation. It helps to compare the dimensions of the output variable from an equation against the true dimensions of the output variable. If both of them are not same, the equation/computation/derivation is incorrect.

  • How does this knowledge help you to understand and characterize particular phenomenon?

answer:  the knowledge of dimensional analysis helps in understanding which variables are necessary for defining a process.

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They are equal

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Calculate the conductance of a conduit the cross-sectional area of which is 3.0 cm2 and the length of which is 9.0 cm, given tha
pshichka [43]
For resistance we have R=ρ l/a
 thus for conductance we have K=σ a/l
conductance,K=1/R
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6 0
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Read 2 more answers
A lawn roller in the form of a thin-walled, hollow cylinder with mass M is pulled horizontally with a constant horizontal force
olga nikolaevna [1]

Answer:

a_{cm}=\frac{F}{2M}\\\\F_{fr}=\frac{F}{2}

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v_{cm}=wR, a_{cm}=\alpha R

\sum F_x=Ma_x, -F+F_{fr}=-Ma_{cm}\\\\a_{cm}=\frac{F-Fr}{M}    \ \ \ \ \ \ \ \ eqtn1\\\\\sum \tau_{cm}=I_{cm}\alpha, F_{fr}(R)}=(MR^2)(\frac{a_{cm}}{R}), ->F_{fr}=Ma_c_m\ \ \  \ \ \ \ eqtn2\\\\a_{cm}=\frac{F-Ma_{cm}}{M}, ->F=2Ma_{cm}\\\\a_{cm}=\frac{F}{2M}\\\\\\\therefore F_{fr}=M(\frac{F}{2M})\\\\F_{fr}=\frac{F}{2}

6 0
3 years ago
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Explanation:

It is known that relation between force and acceleration is as follows.

                      F = m \times a

I is given that, mass is 1090 kg and acceleration is 21 m/s. Therefore, we will calculate force as follows.

              F = m \times a      

                 = 1090 \times (\frac{21}{16})

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Also, it is known that

      sin(\theta) = \frac{\text{Force car can exert}}{\text{Force gravity pulls car}}

      sin(\theta) = \frac{1430.625 N}{(1090 \times 9.8) N}

        \theta = 7.70 degrees

Thus, we can conclude that the maximum steepness for the car to still be able to accelerate is 7.70 degrees.

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