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3 years ago

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To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a

nd the other at an angel 19.0° east of north, as they pull the tanker a distance 0.710 km toward the north

1 answer:

irina1246 [14]3 years ago

6 0

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

= 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

W = F x S joules

Substituting the values in the above equation

W = 1.5 x 10⁶ x 355 · sin θ

= 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J

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The change in speed of this object is 3m/s

According to Newton's second law;

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Given the following parameters

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Required

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Substitute the given parameters into the formula

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A woman drives 200 miles west and then turns south and drives 375 miles. Her trip takes her 11.75 hrs. What is the woman's avera

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The woman's average velocity during the trip is 36.2 miles/hour.

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From the given question,

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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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