1 year ago

Is dimensionally correct equation must be physically correct . Explain your logic behind physically correct equation ?

1 answer:

8 0

An equation can be dimensionally correct but that does not mean that it is physically correct. Explanation is given in the photo.

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Coming to the bottom of a mountain, a skier moving with speed v collides with a barrier and is brought to a stop in an amount of

IRISSAK [1]

Answer:

Explanation:

Impulse of a force is measured by force x time or F X t

Impulse also equals change in momentum or

F x t = m v₂ - m v₁

The given case is as follows

in the first case

F x t = mv - o = mv

F = mv / t

in the second case

F₁ x 4 t = mv

F₁ = 1/4 x mv /t

F₁ = F / 4

option a) is correct .

iii )

In the last case

F₂ X t = m v/2 -0

F₂ = 1/2 x mv / t

= 1/2 x F

F₂ = F/2

Option e ) is correct.

8 0

3 years ago

The word parallel means ____

timurjin [86]

C. Is the correct answer for this question

7 0

4 years ago

Read 2 more answers

A sample from of water is heated with 5000 J of energy and its temperature goes up by 6 K. What is the mass of the sample?

Dima020 [189]

Answer:

mass= 0.1993 kg

Explanation:

Using the formula c = Q / (mΔT)

8 0

3 years ago

Imagine a scenario in which balanced forces are exerted on a box. What does the motion of the box look like?

zepelin [54]

It would depend on the amount it would be still or crushed<span />

3 0

3 years ago

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A particle executes simple harmonic motion with an amplitude of 2.00 cm. At what positions does its speed equal one fourth of it

White raven [17]

Answer:

The positions are 0.0194 m and - 0.0194 m.

Explanation:

Given;

amplitude of the simple harmonic motion, A = 2.0 cm = 0.02 m

speed of simple harmonic motion is given as;

$v = \omega \sqrt{A^2-x^2}$

the maximum speed of the simple harmonic motion is given as;

$v_{max} = \omega A$

when the speed equal one fourth of its maximum speed

$v =\frac{v_{max}}{4}$

$\omega\sqrt{A^2-x^2} = \frac{\omega A}{4} \\\\\sqrt{A^2-x^2}= \frac{A}{4}\\\\A^2-x^2 = \frac{A^2}{16} \\\\x^2 = A^2 - \frac{A^2}{16} \\\\x^2 = \frac{16A^2 - A^2}{16} \\\\x^2 = \frac{15A^2}{16} \\\\x= \sqrt{\frac{15A^2}{16} } \\\\x = \sqrt{\frac{15(0.02)^2}{16} }\\\\x = 0.0194 \ m \ \ or\ - 0.0194 \ m$

Thus, the positions are 0.0194 m and - 0.0194 m.

8 0

3 years ago

Is dimensionally correct equation must be physically correct . Explain your logic behind physically correct equation ?​​

Is dimensionally correct equation must be physically correct . Explain your logic behind physically correct equation ?