A 21.1-N force is applied to a cord wrapped around a pulley of mass M = 4.49-kg and radius R = 25.0-cm The pulley accelerates uniformly from rest to an angular speed of 25.5 rad/s in 2.31-s. If there is a frictional torque \tau = 2.80-mN at the axle,

(a) determine the moment of inertia of the pulley,

(b) determine the rough estimate of the moment of inertia.

(The pulley rotates about its center)

What is the difference be (a) and (b)?

Answers

Answer 1

Answer:

The difference between (a) and (b) is the deviation caused by the actual pulley not being a perfect solid disk. In (a), we took into account the additional frictional torque and calculated the more accurate moment of inertia. In (b), we made a rough estimate assuming the pulley to be a solid disk, which disregards factors such as the mass distribution and the presence of the axle. The difference between the two values is the deviation caused by these factors.


Related Questions

The left field fence in Boston's Fenway park is known as the green monster. The wall is 11.34m high, and is located 94.5m from home plate. The average exit velocity of a ball being struck by a major-league batter is 41.0m/s.


If a batter launches the ball in a direction 30° above the horizontal, will he hit a home run? (Hint: use horizontal motion to solve for t first)

Imagine that a batter hit a ball so that it was at the peak of its trajectory when it brushed against the top of the green monster.
Find the initial y velocity (Hint: need equation that does not involve time). Find the travel time.
Use that to to find the horizontal velocity.
Use the pythagorean theorem to find the total initial velocity.

Answers

To determine whether the batter will hit a home run, we need to analyze the ball's trajectory and determine if it will clear the 11.34m high Green Monster wall.

Let's break down the problem into steps:

Step 1: Calculate the time of flight (t) for the ball's horizontal motion.

We can use the horizontal distance and the average exit velocity to find the time it takes for the ball to reach the Green Monster wall. The horizontal distance (range) can be determined using the formula:

range = horizontal velocity * time

In this case, the range is given as 94.5m, and the average exit velocity is 41.0m/s. Let's solve for time:

94.5m = (41.0m/s) * t

Simplifying the equation, we have:

t = 94.5m / 41.0m/s

t ≈ 2.31s

Step 2: Find the initial vertical velocity (Viy) at the peak of the trajectory.

Since the ball brushes against the top of the Green Monster, we can assume it reaches its peak at half of the total time of flight (t/2). The vertical motion is influenced by gravity, so the equation to determine the initial vertical velocity is:

Viy = (displacement) / (time)

In this case, the displacement is half the height of the Green Monster, which is 11.34m/2 = 5.67m. The time is half of the total time of flight:

Viy = (5.67m) / (t/2)

Viy = (5.67m) / (2.31s/2)

Viy ≈ 2.46m/s

Step 3: Calculate the horizontal velocity (Vix).

Since the horizontal motion is unaffected by gravity, the horizontal velocity remains constant throughout the ball's trajectory. We can use the horizontal distance and time of flight calculated earlier to find the horizontal velocity:

Vix = (horizontal distance) / (time)

Vix = 94.5m / 2.31s

Vix ≈ 40.95m/s

Step 4: Determine the total initial velocity (Vi) using the Pythagorean theorem.

The total initial velocity of the ball can be calculated using the horizontal and vertical velocities:

Vi = √(Vix^2 + Viy^2)

Vi = √((40.95m/s)^2 + (2.46m/s)^2)

Vi ≈ √(1676.9025m^2/s^2 + 6.0516m^2/s^2)

Vi ≈ √(1682.9541m^2/s^2)

Vi ≈ 41.02m/s

Now we have found the total initial velocity of the ball, which is approximately 41.02m/s.

To determine whether it's a home run, we need to consider the ball's trajectory and the height of the Green Monster. Since the height of the wall is 11.34m and the ball's vertical velocity is 2.46m/s, the ball will not clear the Green Monster and will not result in a home run.

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Seeing the light from these distant words and watching the life cycle of the universe unfold is breathtaking reminder that light is the ultimate

Answers

From the distant stars to the smallest particles, light allows us to perceive the world and unravel its mysteries. It is through light that we gather information about our surroundings, explore the cosmos, and make scientific discoveries.

Light not only illuminates our physical environment, but it also carries the stories of the past. When we look at distant objects in space, we are actually observing light that has traveled vast distances over millions or even billions of years. By analyzing the light emitted or reflected by celestial bodies, astronomers can study their composition, temperature, and movement. This information provides invaluable insights into the nature of our universe and its evolution.

Moreover, light plays a crucial role in many areas of scientific research. In fields such as optics, photonics, and quantum mechanics, scientists harness the properties of light to develop advanced technologies. From lasers to fiber optics, these innovations have revolutionized communication, medicine, and countless other industries.

Light is not only a carrier of information, but it also embodies the electromagnetic spectrum, which encompasses various types of radiation, each with its own characteristics and applications. For instance, visible light allows us to see the world around us, while infrared light reveals heat signatures and ultraviolet light exposes hidden details. X-rays and gamma rays, on the other hand, help us explore the microscopic realm and unravel the secrets of atomic and subatomic particles.

Beyond its scientific significance, light has metaphorical and symbolic meanings as well. It is often associated with knowledge, enlightenment, and wisdom. The phrase "seeing the light" is used to describe moments of realization or understanding. Light is a universal symbol of hope, guidance, and truth.

In summary, light is indeed the ultimate source of knowledge. Its ability to illuminate, reveal, and transmit information has profound implications for our understanding of the universe and our place within it. Whether we contemplate the wonders of the cosmos or appreciate the metaphorical significance of light, it remains an awe-inspiring force that continues to inspire and expand our horizons.

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Particles q₁ = -29.6 μC, q2 = +37.7 μC, and
93-10.8 μC are in a line. Particles q₁ and q2 are
separated by 0.630 m and particles q2 and q3 are
separated by 0.315 m. What is the net force on
particle q₁?
Remember: Negative forces (-F) will point Left
Positive forces (+F) will point Right
-29.6 AC
91
0.630 m
+37.7 μC
+92
0.315 m
-10.8 μC
93

Answers

On particle q1, there is a net force of about +25.6 N, directed to the right.

We must take into account the electrostatic forces between particle q1 and the other two particles, q2 and q3, in order to calculate the net force on particle q1. Coulomb's Law describes the electrostatic force between two charged particles:

F = k * |q₁ * q₂| / r²

F is the force, k is the electrostatic constant (9 x 109 N m2/C2), q1 and q2 are the charges' magnitudes, and r is the distance separating them.

Let's first determine the force between q1 and q2:

F₁₂ = k * |q₁ * q₂| / r₁₂²

F₁₂ = (9 x 10^9 N m²/C²) * |(-29.6 μC) * (+37.7 μC)| / (0.630 m)²

F₁₂ = (9 x 10^9 N m²/C²) * (29.6 x 10^-6 C) * (37.7 x 10^-6 C) / (0.630 m)²

F₁₂ ≈ -7.45 N

The minus symbol denotes an attracting force between q1 and q2, pointing to the left.

Let's next determine the force between q2 and q3:

F₂₃ = k * |q₂ * q₃| / r₂₃²

F₂₃ = (9 x 10^9 N m²/C²) * |(+37.7 μC) * (-10.8 μC)| / (0.315 m)²

F₂₃ = (9 x 10^9 N m²/C²) * (37.7 x 10^-6 C) * (10.8 x 10^-6 C) / (0.315 m)²

F₂₃ ≈ +33.05 N

Positively directed to the right, the force between q2 and q3 is shown by the positive sign.

We must now add all the forces in order to determine the net force on q1:

Net force = F₁₂ + F₂₃

Net force ≈ -7.45 N + 33.05 N

Net force ≈ +25.6 N

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when the trigger is pulled on a cordless drill it takes 0.36s for the drill bit to reach 5200rpm. If the drill spins counterclockwise then, what is the angular acceleration of the drill bit?

Answers

The angular acceleration of the drill is 1512.5 rad/s².

Time taken for the drill, t = 0.36 s

Angular velocity of the drill, ω = 5200 rpm = 544.5 rad/s

The change in angular velocity that a spinning object experiences per unit of time is expressed quantitatively as angular acceleration, also known as its rotational acceleration.

It is a vector quantity that has two distinct directions or senses as well as a component of magnitude. The unit of angular acceleration is rad/s².

So,

The expression for the angular acceleration is given by,

α = ω/t

α = 544.5/0.36

α = 1512.5 rad/s²

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A 5.00-ohm resistor, a 10.0-ohm resistor, and a 15.0-ohm resistor are connected in parallel with a battery. The current through the 5.00-ohm resistor is 2.4 amperes.

Calculate the potential difference.
Group of answer choices

12.0 V
12.5 V
6.55 V
2.08 V

Answers

I’m not 100% sure but my math says it’s A. 12.0 V



Explanation: 2.4 (amperes) * 5.00 (ohm) = 12.0V

Find the position and velocity of a particle at t = 1.98 s if the particle is initially moving east at a speed of 20.4 m/s and experiences an acceleration of magnitude 4.40 m/s2, directed west.

Magnitude and direction of the position.
magnitude ---blank---
direction ---Select---

Magnitude and direction of the velocity.
magnitude ---blank---
direction ---select---

Answers

The position of the particle at t = 1.98 s is 31.7006 meters east of its initial position.The velocity of the particle at t = 1.98 s is 11.688 m/s to the east.

Position and velocity of particles

To find the position and velocity of a particle at a specific time, we can use the equations of motion.

Given:

Initial velocity (u) = 20.4 m/s (east)

Acceleration (a) = -4.40 m/s² (west)

Time (t) = 1.98 s

To find the position (displacement) of the particle at time t, we can use the equation:

s = ut + (1/2)at²

s = (20.4 m/s)(1.98 s) + (1/2)(-4.40 m/s²)(1.98 s)²

s = 40.392 m + (1/2)(-4.40 m/s²)(3.9204 s²)

s = 40.392 m - 8.6914 m

s ≈ 31.7006 m

To find the velocity of the particle at time t, we can use the equation:

v = u + at

v = (20.4 m/s) + (-4.40 m/s²)(1.98 s)

v = 20.4 m/s - 8.712 m/s

v ≈ 11.688 m/s

Therefore, the velocity of the particle at t = 1.98 s is approximately 11.688 m/s to the east.

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