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uysha [10]
4 years ago
11

is dimensionally correct relation necessarily to be a correct physical relation? explain with example.​

Physics
1 answer:
Andreas93 [3]4 years ago
4 0

Answer: hope it helps you...❤❤❤❤

Explanation: If your values have dimensions like time, length, temperature, etc, then if the dimensions are not the same then the values are not the same. So a “dimensionally wrong equation” is always false and cannot represent a correct physical relation.

No, not necessarily.

For instance, Newton’s 2nd law is  F=p˙ , or the sum of the applied forces on a body is equal to its time rate of change of its momentum. This is dimensionally correct, and a correct physical relation. It’s fine.

But take a look at this (incorrect) equation for the force of gravity:

F=−G(m+M)Mm√|r|3r  

It has all the nice properties you’d expect: It’s dimensionally correct (assuming the standard traditional value for  G ), it’s attractive, it’s symmetric in the masses, it’s inverse-square, etc. But it doesn’t correspond to a real, physical force.

It’s a counter-example to the claim that a dimensionally correct equation is necessarily a correct physical relation.

A simpler counter example is  1=2 . It is stating the equality of two dimensionless numbers. It is trivially dimensionally correct. But it is false.

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3 years ago
A thin block of soft wood with a mass of 0.072 kg rests on a horizontal frictionless surface. A bullet with a mass of 4.67 g is
Olegator [25]

Answer:

366.90149 m/s

923.821735 J

324.734 J

Initial Kinetic energy > Final kinetic energy

Explanation:

m_1 = Mass of block = 0.072 kg

m_2 = Mass of bullet = 4.67 g

u_1 = Initial Velocity of block = 0

u_2 = Initial Velocity of bullet = 629 m/s

v_1 = Final Velocity of block = 17 m/s

v_2 = Final Velocity of bullet

In this system the linear momentum is conserved

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow v_2=\frac{m_{1}u_{1}+m_{2}u_{2}-m_1v_1}{m_2}\\\Rightarrow v_2=\frac{0.072\times 0+4.67\times 10^{-3}\times 629-0.072\times 17}{4.67\times 10^{-3}}\\\Rightarrow v_2=366.90149\ m/s

Final Velocity of bullet is 366.90149 m/s

The initial kinetic energy

K_i=\frac{1}{2}m_2u_2^2\\\Rightarrow K_i=\frac{1}{2}4.67\times 10^{-3}\times 629^2\\\Rightarrow K_i=923.821735\ J

The final kinetic energy

K_f=\frac{1}{2}m_2v_2^2+\frac{1}{2}m_1v_1^2\\\Rightarrow K_f=\frac{1}{2}4.67\times 10^{-3}\times 366.90149^2+\frac{1}{2}0.072\times 17^2\\\Rightarrow K_f=324.734\ J

Initial Kinetic energy > Final kinetic energy

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3 years ago
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Answer:

true for some places

Explanation:

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Question 2 (1 point)
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3 years ago
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1 point<br> How much force is needed to accelerate an 84-kg boulder at a rate of 6.4<br> m/s/s? *
Katen [24]

Answer:

<h2>537.6 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 84 × 6.4

We have the final answer as

<h3>537.6 N</h3>

Hope this helps you

5 0
3 years ago
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