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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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worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
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In other words, if I  were standing on the surface of that neutron star,
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3 0
3 years ago
A bag of sand has a density of 45 g/cm3 and a mass of 15 kg. How much space does the sand take up?
Aleks [24]

Volume = mass/density

 

Volume = 15000 g/45 g/cm3 ≈ 333.3 cm<span>3</span>

3 0
3 years ago
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meriva

Answer:

Permanent magnetism (of the steel)

make me brainliestt :))

6 0
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A classmate is attempting to draw a free body diagram for a box being pushed across the floor at a constant speed. The image bel
aivan3 [116]

Answer:

A. The applied force should be the same size as the friction force

Explanation:

Whenever we apply a force to an object it moves if the force applied to that object is unbalanced and there is no force or a lesser force to counter it. According to Newton's Second Law of motion, when an unbalanced force is applied to an object it produces an acceleration in the object in its own direction. So, the two forces acting on this box are the frictional force and the applied force in horizontal direction. In order to move the box at constant speed, the applied force must first, overcome the frictional force, so the object can start its motion. Since, the motion has constant velocity, it means no acceleration. So, the force must be balanced in order to avoid acceleration as a consequence of Newton's Second Law of motion. Therefore, the correction in this case will be:

<u>A. The applied force should be the same size as the friction force</u>

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