Suppose you are hiking along a trail. Make a comparison between the magnitude of your displacement and your distance traveled. C
1 answer:
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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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Answer:
Explanation:
Given that, the range covered by the sphere, , when released by the robot from the height,
, with the horizontal speed
is
as shown in the figure.
The initial velocity in the vertical direction is .
Let be the acceleration due to gravity, which always acts vertically downwards, so, it will not change the horizontal direction of the speed, i.e.
will remain constant throughout the projectile motion.
So, if the time of flight is , then
Now, from the equation of motion
Where is the displacement is the direction of force,
is the initial velocity,
is the constant acceleration and
is time.
Here, and
(negative sign is for taking the sigh convention positive in
direction as shown in the figure.)
So, from equation (ii),
Similarly, for the launched height , the new time of flight,
, is
From equation (iii), we have
Now, the spheres may be launched at speed or
.
Let, the distance covered in the direction be
for
and
for
, we have
[from equation (iv)]
[from equation (i)]
(approximately)
This is in the points range as given in the figure.
Similarly,
[from equation (iv)]
[from equation (i)]
(approximately)
This is out of range, so there is no point for .
Hence, students must choose the speed to launch the sphere to get the maximum number of points.
