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3 years ago

Every chemical reaction demonstrate what important scientific principle

1 answer:

4 0

The answer to this question would be the Law of conservation of energy.

In every chemical reaction, 10 atoms of reacted will make 10 atoms as result. The number of atoms produced will be same as the number of ingredients. This will prove that the energy is <span>can neither be created nor destroyed; The energy is just changed into a different shape.</span>

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4 years ago

Read 2 more answers

A potter's wheel is spinning with an initial angular velocity of 11 rad/s . It rotates through an angle of 80.0 rad in the proce

Grace [21]

The angular acceleration of the wheel approximately <u>-0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s</u>.
It need approximately <u>14.474 s</u> to come to rest.

<h2>Introduction</h2>

Hi ! I will help you to discuss about Proportionally Changes in Circular Motion. The analogy of proportionally changes in circular motion is same as the analogy of proportionally changes in direct motion. Here you will hear again the terms acceleration and change in speed, only expressed in the form of a certain angle coverage. Before that, in circular motion, it is necessary to know the following conditions:

1 rotation = 2π rad
1 rps = 2π rad/s
1 rpm = $\sf{\frac{1}{60} \: rps}$ = $\sf{\frac{1}{30}\pi \: rad/s}$

<h2>Formula Used</h2>

The following equations apply to proportionally changes circular motion:

<h3>Relationship between Angular Acceleration and Change of Angular Velocity </h3>

$\boxed{\sf{\bold{\omega_t = \omega_0 + \alpha \times t}}}$

With the following conditions:

$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)

<h3>Relationship between Angular Acceleration and Change of $\sf{\theta}$ (Angle of Rotation) </h3>

$\boxed{\sf{\bold{\theta = \omega_0 \times t + \frac{1}{2} \times \alpha \times t^2}}}$

$\boxed{\sf{\bold{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}}}$

With the following condition :

$\sf{\theta}$ = change of the sudut (rad)
$\sf{\alpha}$ = angular acceleration (rad/s²)
t = interval of the time (s)
$\sf{\omega_t}$ = final angular velocity (rad/s)
$\sf{\omega_0}$ = initial angular velocity (rad/s)

<h2>Problem Solving</h2>

We know that :

$\sf{\omega_t}$ = final angular velocity = 0 rad/s >> see in the sentence "in the process of coming to rest."
$\sf{\omega_0}$ = initial angular velocity = 11 rad/s
$\sf{\theta}$ = change of the sudut = 80.0 rad

What was asked :

$\sf{\alpha}$ = angular acceleration = ... rad/s²
t = interval of the time = ... s

Step by step :

$\sf{\alpha}$ = ... rad/s²

$\sf{(\omega_t)^2= (\omega_0)^2 + 2 \times \alpha \times \theta}$

$\sf{0^2= (11)^2 + 2 \times \alpha \times 80}$

$\sf{0 = 121 + 160 \alpha}$

$\sf{-160 \alpha = 121}$

$\sf{\alpha = \frac{121}{-160}}$

$\sf{\alpha = -0.75625 \: rad/s^2 \approx \boxed{-0.76 \: rad/s^2}}$

t = ... s

$\sf{\alpha = \frac{\omega_0 - \omega_t}{t}}$

$\sf{-0.76 = \frac{0 - 11}{t}}$

$\sf{-0.76t = -11}$

$\sf{t = \frac{- 11}{-0.76}}$

$\boxed{\sf{t \approx 14.474 \: s}}$

<h3>Conclusion</h3>

So :

The angular acceleration of the wheel approximately -0.76 rad/s² or proportionally as deceleration approximately 0.76 rad/s.
It need approximately 14.474 s to come to rest.

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2 years ago

Which of the following is an example if harmonic motion?

german

Answer:

"A pendulum swinging back and forth" is an example of harmonic motion

X = Xo cos ω t

Explains the back and forth motion of the pendulum

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2 years ago