A solid, horizontal cylinder of mass 18.0 kg and radius 1.70.0 m rotates with an angular speed of 40 rad/s about a fixed vertical axis through its center. A 0.8 kg piece of putty is dropped vertically onto the cylinder at a point 0.300 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

Answers

Answer 1
Answer:

Answer:39.88 rad/s

Explanation:

Given

mass of cylinder m_1=18 kg

radius R=1.7 m

angular speed \omega =40rad/s

mass of m_2=0.8 kg dropped at r=0.3 m from center

let \omega _2 be the final angular velocity of cylinder

Conserving Angular momentum

L_1=L_2

\left ( (m_1R^2)/(2)\right )\omega =\left ( (m_1R^2)/(2)+m_2r^2\right )\omega _2

\left ( (18\cdot 1.7^2)/(2)\right )\cdot 40=\left ( (18\cdot 1.7^2)/(2)+0.8\cdot 0.3^2\right )\omega _2

26.01* 40=26.082* \omega _2

\omega _2=39.88 rad/s


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Denise is riding her bike and falls to the ground. During the collision, she hits both grass and cement.Which best explains why Denise had fewer injuries to the area of her body that hit the grass? The grass did not exert a reaction force.
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The grass shortens the time of the collision.
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Answers

Answer : Because the grass extends the time of the collision.

Explanation :

Denise is riding her bike and falls to the ground. During the collision, she hits both grass and cement.

He had fewer injuries to the area of her body that hit the grass because the grass extends the time of the collision and hence it imparts less impulse.

i.e. I=(F)/(t)

Where

F is the force

t is the time of collision

Hence, the correct option is (d) " The grass extends the time of the collision ".

The grass extends the time of the collision

How to solve for vi in vf=vi+at?

Answers

Subtract 'at' from each side of that equation.

A gas has an initial volume of 24.6 L at a pressure of 1.90 atm and a temperature of 335 K. The pressure of the gas increases to 3.50 atm, and the volume of the gas increases to 31.3 L.

Answers

The problem can be solved by using the formula associated with the gas laws. It is as follows:

PV/T

Given:

P1 = 1.9 atm
V1 = 24.6 L
T1 = 335 K 

P2 = 3.5 atm
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Substituting the given:

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a coil has n turns enclosing an area of a. in a physics laboratory experiment, the coil is rotated during the time interval δt from a position in which the plane of each turn is perpendicular to earth's magnetic field to one in which the plane of each turn is parallel to the field. the magnitude of earth's magnetic field at the lab location is b.

Answers

The correct answer for the question that is being presented above is this one:

Phi = BAsin(theta) 
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Here are the follow-up questions:

1. What is the total magnitude Phi_initial of the magnetic flux through the coil before it is rotated? 
2. What is the magnitude Phi_final of the total magnetic flux through the coil after it is rotated? 
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A glass lens, n glass = 1.52 , has a 131 nm thick antireflective film coating one side, n film = 1.25. White light, moving through the air, is perpendicularly incident on the coated side of the lens. What is the largest wavelength of the reflected light that is totally removed by the coating?

Answers

Answer:

655 nm

Explanation:

When the intereference is destructive then the thickness, d of antireflective film coating one side is given by

2d=w/2n

Where w is wavelength and n is the reflective index of the film

Making w the subject of formula then

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w=4*1.25*131=655 nm

Therefore, the wavelength is equivalent to 655 nm

The formula for calculating the wavelength in an antireflective film involves thickness (d) and refractive index (n). For n = 1.25 and d = 131 nm, the resulting wavelength is 655 nm.

When light waves encounter a thin film, some of the waves are reflected from the top surface of the film, and some pass through it. These waves can interfere with each other, leading to constructive or destructive interference. In the case of antireflective coatings, destructive interference is desired to minimize reflection.

The formula you mentioned is used to calculate the thickness (d) of an antireflective film that results in destructive interference for a specific wavelength (w) of light. The formula is:

2d = w / (2n)

Where:

d is the thickness of the film.

w is the wavelength of light.

n is the refractive index of the film.

To find the wavelength (w) when given the thickness (d) and refractive index (n), you can rearrange the formula:

w = 4 * n * d

Now, let's calculate the wavelength using the provided values:

n = 1.25 (refractive index)

d = 131 nm (thickness in nanometers)

Substitute these values into the formula:

w = 4 * 1.25 * 131 = 655 nm

Therefore, the calculated wavelength (w) is 655 nanometers (nm). This means that for a film with a refractive index of 1.25 and a thickness of 131 nm, destructive interference occurs at a wavelength of 655 nm.

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A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s

Answers

it will be traveling exactly 24 miles per hour 

Final answer:

The final velocity of the car can be found using the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time. Plugging in the given values, we find that the car will be traveling at 24 m/s at 6 seconds.

Explanation:

To find the final velocity of a car accelerating at a constant rate, we can use the equation:
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Where:
v is the final velocity, u is the initial velocity (which is zero in this case), a is the acceleration (4.0 m/s²), and t is the time (6 seconds).

Plugging in the values, we get:
v = 0 + (4.0 m/s²)(6s) = 24 m/s

Therefore, the car will be traveling at 24 m/s at 6 seconds.

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