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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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A model rocket is launched straight upward with an initial speed of 52.0 m/s. It accelerates with a constant upward acceleration
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Explanation:

(a) After the engines stop, the rocket reaches a maximum height at which it will stop and begin to descend in free fall due to gravity.

(b) We must separate the motion into two parts, when the rocket's engines is on  and when the rocket's engines is off.

First we must find the rocket speed when the engines stop:

v_f^2=v_0^2+2ay_1\\v_f^2=(52\frac{m}{s})^2+2(1\frac{m}{s^2})(160m)\\v_f^2=3024\frac{m^2}{s^2}\\v_f=\sqrt{3024\frac{m^2}{s^2}}=54.99\frac{m}{s}

This final speed is the initial speed in the second part of the motion, when engines stop until reach its maximun height. Therefore, in this part the final speed its zero and the value of g its negative, since decelerates the rocket:

v_f^2=v_0^2+2gy_{2}\\y_{2}=\frac{v_f^2-v_0^2}{2g}\\y_{2}=\frac{0^2-(54.99\frac{m}{s})^2}{2(-9.8\frac{m}{s^2})}=154.28m

So, the maximum height reached by the rocket is:

h=y_1+y_2\\h=160m+154.28m=314.28m

(c) In the first part we have:

v_f=v_0+at_1\\t_1=\frac{v_f-v_0}{a}\\t_1=\frac{54.99\frac{m}{s}-52\frac{m}{s}}{1\frac{m}{s^2}}\\t_1=2.99s

And in the second part:

t_2=\frac{v_f-v_0}{g}\\t_2=\frac{0-54.99\frac{m}{s}}{-9.8\frac{m}{s^2}}\\t_2=5.61s

So,  the time it takes to reach the maximum height is:

t_3=t_1+t_2\\t_3=2.99s+5.61s=8.60s

(d) We already know the time between the liftoff and the maximum height, we must find the rocket's time between the maximum height and the ground, therefore, is a free fall motion:

v_f^2=v_0^2+2ay\\v_f^2=0^2+2(9.8\frac{m}{s^2})(314.28m)\\v_f=\sqrt{6159.888\frac{m^2}{s^2}}=78.48\frac{m}{s}

t_4=\frac{v_f-v_0}{g}\\t_4=\frac{78.48\frac{m}{s}-0}{9.8\frac{m}{s^2}}\\t_4=8.01s

So, the total time is:

t=t_3+t_4\\t=8.60s+8.01s\\t=16.61s

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