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Serhud [2]
3 years ago
14

A disk is rotating around an axis located at its center. The angular velocity is 0.5 rad/s. The radius of the disk is 0.4 m. Wha

t is the magnitude of the velocity at a point located on the outer edge of the disk, in units of m/s?
Engineering
2 answers:
dimaraw [331]3 years ago
4 0

Answer:

0.2 m/s

Explanation:

The velocity of a point on the edge of a disk rotating disk can be calculated as:

v=\omega*r

Where \omega is the angular velocity and r the radius of the disk. This leads to:

v=0.5\,rad/s\,*\,0.4\,m=0.2\,m/s

Anna007 [38]3 years ago
4 0

Answer:

0.2 m/s

Explanation:

<u>Step 1: </u>identify the given parameters

angular velocity, ω = 0.5 rad/s

radius of the disk, r = 0.4m

<u>Note</u>: the inner part of the disk and outer edge spin at the same rate. This means that the velocity at inner part of the disk is the same as the outer part.

<u>Step 2:</u> calculate the velocity of the disk at the outer edge in m/s

Velocity = Angular velocity (rad/s) X radius (m)

Velocity = 0.5 rad/s X 0.4 m

Velocity = 0.2 m/s

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3 years ago
W10L1-Show It: Pythagorean Theorem<br> Calculate the total material in the picture.<br> 4<br> 3
Fantom [35]

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7 0
3 years ago
A train consists of a 50 Mg engine and three cars, each having a mass of 30 Mg . If it takes 75 s for the train to increase its
ohaa [14]

Answer:

T = 15 kN

F = 23.33 kN

Explanation:

Given the data in the question,

We apply the impulse momentum principle on the total system,

mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

we substitute

[50 + 3(30)]×10³ × 0 + FΔt = [50 + 3(30)]×10³ ×  ( 45 × 1000 / 3600 )  

F( 75 - 0 ) =  1.75 × 10⁶

The resultant frictional tractive force F is will then be;

F =  1.75 × 10⁶ / 75

F = 23333.33 N

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mv₁ + ∑\int\limits^{t2}_{t1} {Fx} \, dt = mv₂

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