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3 years ago

PART ONE

Physics

1 answer:

Lubov Fominskaja [6]3 years ago

4 0

Answer:

change in momentum is (delta,p))= 2*m*v*sin(23) ,parallel velocity to the window remains unchanged ignore, perpendicular velocity changes by twice the initial perpendicular velocity as the hail stone changes direction with respect to the perpendicular . force=total(delta,p)/t

hence force=450*2*m*v*sin(23) /46, pressure =force/area=450*2*m*v*sin(23) /46*1.187

This is not fluid mechanics,

Just think about splitting the velocity in to vector components use elastic collision fact is energy is same after collision hence so is the speed ,use known equations.

Explanation:

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Electronic signals that pulse current in a series of ones and zeros are known as

grandymaker [24]

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3 years ago

he fan blades on a jet engine make one thousand revolutions in a time of 54.9 ms. What is the angular frequency of the blades?

Gnesinka [82]

So, the angular frequency of the blades approximately <u>36.43π rad/s</u>.

<h3>Introduction</h3>

Hi ! Here I will discuss about the angular frequency or what is also often called the angular velocity because it has the same unit dimensions. <u>Angular frequency occurs, when an object vibrates (either moving harmoniously / oscillating or moving in a circle)</u>. Angular frequency can be roughly interpreted as the magnitude of the change in angle (in units of rad) per unit time. So, based on this understanding, the angular frequency can be calculated using the equation :

$\boxed{\sf{\bold{\omega = \frac{\theta}{t}}}}$

With the following condition :

$\sf{\omega}$ = angular frequency (rad/s)
$\sf{\theta}$ = change of angle value (rad)
t = interval of the time (s)

<h3>Problem Solving</h3>

We know that :

$\sf{\theta}$ = change of angle value = 1,000 revolution = 1,000 × 2π rad = 2,000π rad/s >> Remember 1 rev = 2π rad/s.
t = interval of the time = 54.9 s.

What was asked :

$\sf{\omega}$ = angular frequency = ... rad/s

Step by step :

$\sf{\omega = \frac{\theta}{t}}$

$\sf{\omega = \frac{2,000 \pi}{54.9}}$

$\boxed{\sf{\omega \approx 36.43 \pi \: rad/s}}$

<h3>Conclusion :</h3>

So, the angular frequency of the blades approximately 36.43π rad/s.

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erastovalidia [21]

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