Of the following nonmetals which is the least reactive A.S B. Cl C. Br D. Se

A CD spins at a constant angular velocity of 5.0 revolutions per second clockwise.

Lera25 [3.4K]

The true statement about the CD is:

<h3><em>b. No net torque acts on it at all.</em></h3>

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<h3>Further explanation</h3>

Centripetal Acceleration can be formulated as follows:

$\large {\boxed {a = \frac{ v^2 } { R } }$

<em>a = Centripetal Acceleration ( m/s² )</em>

<em>v = Tangential Speed of Particle ( m/s )</em>

<em>R = Radius of Circular Motion ( m )</em>

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Centripetal Force can be formulated as follows:

$\large {\boxed {F = m \frac{ v^2 } { R } }$

<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 !

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<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

$\Sigma \tau = I \alpha$

$\Sigma \tau = I (0)$

$\Sigma \tau = 0 \texttt{ Nm}$

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<h3>Conclusion:</h3>

The true statement about the CD is:

<em>b. No net torque acts on it at all.</em>

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<h3>Learn more</h3>

Impacts of Gravity : brainly.com/question/5330244
Effect of Earth’s Gravity on Objects : brainly.com/question/8844454
The Acceleration Due To Gravity : brainly.com/question/4189441

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Circular Motion

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Keywords: Gravity , Unit , Magnitude , Attraction , Distance , Mass , Newton , Law , Gravitational , Constant

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Pie

The answer is C, a peer group.

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A child does 350J of work while pulling a box from the ground up to his tree house with a rope. The tree house is 5.2 m above th

Advocard [28]

I think the answer to that questions is B.

3 0

3 years ago

Does the horizontal distance d travelled by the ball depend on the height of release? If it does depend on the height, what is t

elena-s [515]

Answer:

Explanation:

Yes , the horizontal distance travelled by the ball will depend upon the height of release .

When a ball is thrown at some angle from a height , it has two components , the vertical component and horizontal component . The ball goes in horizontal direction due to its horizontal component . Its vertical component has no role to play . But the horizontal range covered by the body thrown

depends upon the duration of time in which it remains in air . The longer it remains in air , the greater distance it can cover horizontally .

Horizontal distance covered = t x horizontal velocity

If V be the velocity of throw and Vx be its horizontal component

Horizontal distance covered = t x Vx

Now t depends upon the height . If height rises , time of fall will increase so horizontal distance covered will increase .

If h be the height from which the body is thrown , Vy be the vertical upward component of initial velocity

from the relation

s = ut + 1/2 at²

h = - Vy t + 1/2 at²

As h increases , t will increase and therefore horizontal distance covered will increase. If the ball has only horizontal velocity initially , Vy = 0

h = 1/2 gt²

$t = \sqrt{\frac{2h}{g} }$

Horizontal distance covered = t x Vx

= $\sqrt{\frac{2h}{g} } \times V_x$

From this expression also

Horizontal distance covered is proportional to $\sqrt{h}$ .

7 0

3 years ago

A ball is thrown down vertically with an initial speed of 31 ft/s from a height of 40 ft. (a) What is its speed just before it s

ICE Princess25 [194]

Answer:

a. 41.96ft/s

b. 1.096s

Explanation:

a. v²=u²+2gs

v²=31²+2×10×40

V=41.96ft/s

b. t=(v-u) /g

t=(41.96-31)/10

t=1.096s

5 0

3 years ago