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3 years ago

6

A bird flies 2.8 due west and then 2.7 due north. Another bird flies 2.7 due west and 2.8 due north. What is the angle between t

he net displacement vectors for the two birds?

1 answer:

Elenna [48]3 years ago

7 0

Assume a 2D plane and both the birds start at origin. +X is east, -X is west, +Y is North, -Y is South.
So, after travel position of bird1 is (-2.8, 2.7) and bird2 is (-2.7,2.8);
slope of bird1 is 2.8/2.7
slope of bird2 is 2.7/2/8
Now we have two solpes b1 and b2
formula = (b2-b1)/(1+b1*b2);

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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

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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>

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

Or

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

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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Anit [1.1K]

Answer:

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Explanation:

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