A scientific law is ____________________A. A rule enacted by the National Academy of Sciences. B. What scientists expect will al
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The true statement about the CD is:
<h3><em>b. No net torque acts on it at all.</em></h3>Centripetal Acceleration can be formulated as follows:
<em>a = Centripetal Acceleration ( m/s² )</em>
<em>v = Tangential Speed of Particle ( m/s )</em>
<em>R = Radius of Circular Motion ( m )</em>
Centripetal Force can be formulated as follows:
<em>F = Centripetal Force ( m/s² )</em>
<em>m = mass of Particle ( kg )</em>
<em>v = Tangential Speed of Particle ( m/s )</em>
<em>R = Radius of Circular Motion ( m )</em>
Let us now tackle the problem !
<em>Complete Question:</em>
<em>A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise. Which of the following statements about the CD is true?</em>
<em>a. A net torque acts on it clockwise to keep it moving</em>
<em>b. No net torque acts on it at all.</em>
<em>c. A net torque acts on it counterclockwise to keep it moving</em>
<u>Given:</u>
angular velocity = ω = 5.0 revolutions per second
<u>Asked:</u>
net torque = Στ = ?
<u>Solution:</u>
Constant angular velocity → angular acceleration = α = 0 rad/s²
The true statement about the CD is:
<em>b. No net torque acts on it at all.</em>
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Grade: High School
Subject: Physics
Chapter: Circular Motion
Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant
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To solve this problem it is necessary to apply the concepts related to acceleration due to gravity, as well as Newton's second law that describes the weight based on its mass and the acceleration of the celestial body on which it depends.
In other words the acceleration can be described as
Where
G = Gravitational Universal Constant
M = Mass of Earth
r = Radius of Earth
This equation can be differentiated with respect to the radius of change, that is
At the same time since Newton's second law we know that:
Where,
m = mass
a =Acceleration
From the previous value given for acceleration we have to
Finally to find the change in weight it is necessary to differentiate the Force with respect to the acceleration, then:
But we know that the total weight (F_W) is equivalent to 600N, and that the change during each mile in kilometers is 1.6km or 1600m therefore:
Therefore there is a weight loss of 0.3N every kilometer.