How fast would a 2000kg car need to be traveling to have the same momentum has a 3000kg truck driving at 18m/s?

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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8 0

3 years ago

) A 100-g ball falls from a window that is 12 m above ground level and experiences no significant air resistance as it falls. Wh

yawa3891 [41]

Answer:

Momentum = 1.534 kgm/s

Explanation:

Using the equations of motion, we can obtain the velocity of the ball as it hits the ground.

g = 9.8 m/s²

y = 12 m

u = initial velocity = 0 m/s, since the ball was released from rest

v = final velocity befor the ball hits the ground.

v² = u² + 2ay

v² = 0 + 2×9.8×12 = 235.2

v = 15.34 m/s

The momentum at any point is given as mass × velocity at that point

Mass = 100 g = 0.1 kg, velocity = 15.34 m/s

Momentum = 0.1 × 15.34 = 1.534 kgm/s

3 0

4 years ago

Explain why atomic radius decreases as you move to the right across a period for main-group elements but not for transition elem

Aleksandr-060686 [28]

Answer:

Explained.

Explanation:

Only the first question has been answered

In a period from left to right the nuclear charge increases and hence nucleus size is compressed. Thus, atomic radius decreases.

In transition elements, electrons in ns^2 orbital remain same which is the outer most orbital having 2 electrons and the electrons are added to (n-1) d orbital. So, outer orbital electron experience almost same nuclear attraction and thus size remains constant.

7 0

3 years ago

If a projectile is launched vertically from the Earth with a speed equal to the escape speed, how high above the earth's surface

Alecsey [184]

Answer: h = 3R

Explanation:

Using the law of conservation of energy,

Total energy at the beginning of the launch would be equal to total energy at any point.

kinetic energy + gravitational potential energy = constant

Initial energy of the projectile = $\frac{1}{2}mv_e^2-\frac{GMm}{R}$ ... (1)