A flywheel having constant angular acceleration requires 4.70 s to rotate through 164 rad . Its angular velocity at the end of this time is 101 rad/s . Find the angular velocity at the beginning of the 4.70 s interval. Find the angular acceleration of the flywheel.

Answers

Answer 1

Answer:

A) -31.2 rad/s

B) 28.1 rad/s^2

Explanation:

A Flywheel Having Constant Angular Acceleration Requires 4.70 S To Rotate Through 164 Rad . Its Angular
A Flywheel Having Constant Angular Acceleration Requires 4.70 S To Rotate Through 164 Rad . Its Angular
A Flywheel Having Constant Angular Acceleration Requires 4.70 S To Rotate Through 164 Rad . Its Angular

Related Questions

What do alcohol, drugs, and tobacco all have in common?
All have some medicinal value.

All are harmful to the body.

All are depressants.

All are stimulants.

Answers

Answer:

all are harmful to the body

This is two or more elements chemically combined in a fixed ratio.


Example: water, carbon dioxide, sodium chloride

Answers

A compound is a substance that contains two or more elements chemically combined in a fixed proportion. The elements carbon and hydrogen combine to form many different compounds.

electron and proton are projected with same velocity normal to the magnetic field which one will suffer greater deflection​? why

Answers

Answer:

Explanation:

The deflection of a charged particle by a magnetic field is proportional to its electric charge and to its velocity.   The deflection is also inversely proportional to its mass.   So given a proton and an electron going at the same velocity in a magnetic field and having equal (but opposite) electric charge the electron will deflect much more since the ratio of the masses is 1836.

An empty cylindrical barrel is open at one end and rolls without slipping straight down a hill. The barrel has a mass of 15.0 kg, a radius of 0.400 m, and a length of 0.800 m. The mass of the end of the barrel equals a fourth of the mass of its side, and the thickness of the barrel is negligible. The acceleration due to gravity is =9.80 m/s2.
What is the translational speed f of the barrel at the bottom of the hill if released from rest at a height of 33.0 m above the bottom?

Answers

Hi there!

We can use work and energy to solve this problem.

We know that:

Ei = Ef

Ei = Potential energy = mgh

Ef = Rotational kinetic + Translational kinetic = 1/2Iω² + 1/2mv²

The barrel is comprised of a hollow cylinder and disk-shaped bottom, so:

I (hollow cylinder) = mr²

I (disk) = 1/2mr²

Calculate the moment of inertias of each.

Since the mass on the base is one-fourth of its side:

x = mass of side

x + x/4 = 15

4x + x = 60

5x = 60

x = 12 kg

end mass = 3 kg

Solve for each moment of inertia:

Side: (12)(0.4²) = 1.92 Kgm²

Bottom: 1/2(3)(0.4²) = 0.24 Kgm²

Side + bottom = 2.16 Kgm²

We can now solve:

mgh = 1/2mv² + 1/2(2.16)v²/r²

(15)(9.8)(33) = 1/2(15)v² + 1/2(13.5)v²

4851 = 14.25v²

v = 18.45 m/s

A thin piece of semiconducting silicon will be used to fabricate an electrical device. This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. Electrical contacts are placed at opposite ends of its length. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

Required:
a. If the application of 1.0 volt to the contacts results in a current of 0.019 amps, what is the resistivity in (ohm-cm) of the material?
b. If the material's conductivity is due to doping with aluminum to a level of [Al]= 1x10^17 atoms/cm^3, what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material (assuming that the aluminum is fully ionized - i.e. all Al atoms donated electrons).

Answers

We have that for the Question "a)what is the resistivity in (ohm-cm) of the material? b) what is the resulting conductivity "type" and what is the mobility of these "majority" carriers in this material"

Answer:

Resistivity = [tex]1.754 ohm-cm[/tex]Conductivity = [tex]6.25*10^{25} cm^3/V-s[/tex]

From the question we are told

This layer is 0.10 cm thick and cut into a strip 0.50 cm wide by 1.50 cm long. The intrinsic carrier concentration of the silicon at room temperature (300K) is 1.0x1010/cm3 and the bandgap energy is 1.12 eV.

 

A) Resistivity is given as,

[tex]p = \frac{RA}{l}[/tex]

where,

[tex]R = \frac{V}{I}[/tex]

Therefore,

[tex]p = \frac{VA}{Il}\\\\p = \frac{1*(0.1*0.5)}{0.019*1.5}\\\\p = 1.754 ohm-cm[/tex]

B) Conductivity is given as,

[tex]U = \frac{\rho}{pe}\\\\U = \frac{10^{17}}{10^{10}*1.6*10^{-19}}\\\\U = 6.25*10^{25} cm^3/V-s[/tex]

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Children in a tree house lift a small dog in a basket 3.85 m up to their house. If it takes 201 J of work to do this, what is the combined mass of the dog and basket

Answers

Answer:

Explanation:

The work will equal the increase in potential energy.

PE = mgh

m = PE/gh = W/gh = 201/(9.81(3.85)) = 5.32 kg

100 J of work was done to lift a 10-N rock and set it at Position A near the edge of a cliff.
1. If the 100 Joules of work lifted the rock to the top of the cliff, how much potential energy did the rock gain?
2. At point C, the rock's potential energy will be
3. The rock's kinetic energy at point A is
4. At point B, some of the rock's potential energy will be changed to Kinetic energy
5. What is the mass of the rock?
6. What is the rock's velocity just before it hits the ground?
The rock to the right is sitting at the top of a ramp. I wonder how much work it required to get that rock up there.

Answers

Answer:

lol

Explanation:

plz help me on this question thank you

Answers

Answer:

D

Explanation:

how do all organisms begin life

Answers

Answer:

All organisms begin their lives as single cells.Overtime,these organisms grow and take on the characteristics of their species...All organisms grow,and different parts of organisms may grow at different rates.Organisims made out of only one cell

may change little during their lives, but they do grow

Explanation:

brainlest me please

A cyclist on a training ride records the distance she travels away from home. The data only shows the first150 minutes of the ride before her cycling computer ran out of battery.

Answers

Answer:

A) 58 km

B) 30 mins

Explanation:

In pic details

graph in pic


3) A force of magnitude Fx acting in the x-direction on a 2.00 kg particle varies in time as shown
in FIGURE 2. Find
a) The impulse of the force
b) The final velocity of the particle if it is initially at rest
c) The final velocity of the particle if it is initially moving along the x-axis with velocity
of -2.00 ms -1

Answers

Answer:

a) Impuise of force =F∗?(t) = area of F-T graph area= impulse =triangle + rectangle + triangle = 0.5*4*2 + 4*1 + 0.5*4*2 = 12 N-s (b) impulse = change in momentum \(= mExplanation:

How much distance does a car travel with a speed of 2m/s in 15 min?​

Answers

1800m because there 900 seconds in 15 minutes and 2*800=1800 so 1800m

Explanation:

1 minute is 60 seconds so you multiply 60 * 15 and then multiply that answer * 2

Velocity and Acceleration Quick Check

C

D

E

During which of the labeled time segments is the object moving forward but slowing down?

(1 point)

Ο Α

0 С

OD

ОВ

Answers

Answer:

Explanation:

1 Object C has an acceleration that is greater than the acceleration for D.

2 B

3 17M

4 The velocity is zero.

5 a straight line with negative slope

just took it

A 2457 kg car moves with initial speed of 18 ms-l. It is stopped in 62 m by its brakes.
What is the force applied by the brakes?

Answers

Answer:

Explanation:

The work of the brakes will equal the initial kinetic energy of the car

Fd = ½mv²

F = mv²/2d

F = 2457(18²) / (2(62))

F = 6,419.903...

F = 6.4 kN

The oscillation of the 2.0-kg mass on a spring is described by x = 3.0 cos (4.0 t) where x is in centimeters and t is in seconds. What is the force constant k of the spring?

Answers

X is the answer because x can be any thing

The force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

What is force?

Force is the influence of either pull or pushes in the body. Basically, gravitation forces, nuclear forces, and friction forces are the types of forces. For e.g. when the wall is hit by a hand then a force is exerted by the hand on the wall as well as the wall also exerts a force on the hand. There are different laws given to Newton to understand force.

Newton is a unit of force used by physicists that is part of the International System (SI). The force required to move a body weighing one kilogram one meter per second is known as a newton.

Given:

The mass of the block, m = 2 kg,

The oscillation of spring, x  = 3 cos 4t,

Calculate the omega  by comparing the standard equation given below,

[tex]x = A cos \omega t[/tex]

ω = 4

Calculate the spring constant by the formula given below,

[tex]\omega = \sqrt{\frac{k}{m} }[/tex]

4² = k / 2

k = 32 N / m

Therefore, the force constant k of the spring, if The oscillation of the 2 kg mass of spring is described by x = 3.0 cos (4.0 t) is 32 N / m.

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An object of mass m is hanging by a string from the ceiling of an elevator. The elevator is moving down at constant speed. What is the tension of the string?

A. Zero
B. Equal to mg
C. Less than mg
D. Greater than mg​

Answers

Answer:

D. Greater than mg​

Explanation:

According to Newton’s second law of motion, the net force equals mass times acceleration. We are going to use a free body diagram (force diagram) to show that the equation of the motion is given by

T – mg = – ma

Thereby,

T = mg – ma

and the answer is: (d)

D. Greater than mg​

_________________________________

(hopet his helps can I pls have brainlist  (crown)☺️)

Acceleration of a Car A car traveling along a straight road at accelerated to a speed of over a distance of ft. What was the acceleration of the car, assuming that it was constant

Answers

Answer:

how many feet?

Explanation:

What is the direction of the torque produced on the crankset by the 2-kg mass attached to the pedal bar

Answers

A Torque is a twisting force, or turning moment, it is a vector quantity with both magnitude and direction e.g Turning the handle of a cork-screw clockwise and then counterclockwise will advance the screw first inward and then outward By convention, counterclockwise torques are positive and clockwise torques are negative.

The direction is perpendicular to both the radius from the axis and to the force. It is conventional to choose it in the right hand rule direction along the axis of rotation.

Counterclockwise is the positive rotation direction and clockwise is the negative direction.

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Which is the most famous book of the philosopher Alexis karpouzos? I think it is the ''Cosmology, philosophy and physics''.

Answers

Answer: Yes, the " Cosmology, philosophy and physics" is the most famous book of the philosopher, Alexis karpouzos. But and the other books are important. For example, the " The self-criticism of science", the "Universal conscilusness" and the "Non-duality".

Explanation:

PLEASE HELP ON THIS QUESTION ​

Answers

[tex]r = 1.29×10^8\:\text{m}[/tex]

Explanation:

According to Newton's law of universal gravitation, the gravitational force between Uranus and Miranda is

[tex]F_G = G\dfrac{M_UM_M}{r^2}[/tex]

where [tex]M_U[/tex] is the mass of planet Uranus, [tex]M_M[/tex] is the mass of its satellite Miranda, r is the distance between their centers and G is the universal gravitational constant. Moving the variable r to the left side, we get

[tex]r^2 = G\dfrac{M_UM_M}{F_G}[/tex]

Taking the square root of the equation above, we get

[tex]r = \sqrt{G\dfrac{M_UM_M}{F_G}}[/tex]

Plugging in the values, we get

[tex]r = \sqrt{(6.67×10^{-11}\:\text{N-m}^2{\text{/kg}}^2)\dfrac{(8.68×10^{25}\:\text{kg})(6.59×10^{19}\:\text{kg})}{2.28×10^{19}\:\text{N}}}[/tex]

[tex]\:\:\:\:\:=1.29×10^8\:\text{m}[/tex]

Uranus (mass = 8.68 x 1025 kg) and its moon Miranda (mass = 6.59 x 1019 kg) exert a gravitational force of 2.28 x 1019 N on each other. How far apart are they? cs [?] x 10?'m Coefficient (green) Exponent (yellow) Enter​

Answers

Answer:

Explanation:

F = GMm/d²

d = √(GMm/F)

d = √(6.674e-11(8.68e25)(6.59e19) / 2.28e19)

d = 1.29398e8 = 1.29 x 10^8 m  center to center

Answer:

1.29 x 10^8 m  apart

Explanation:

Works in Acellus!

what memory are you using to remember who the president of the united states is

Answers

Answer:

The First 8 Presidents

For this exercise, we're going to use a silly story made of silly sentences. The letters that represent the last names of these presidents are W, A, J, M, M, A, J, V. One silly sentence to help you remember this sequence is: Wilma and John made merry and just vanished

working memory.

sensory memory.

short-term memory.

long-term memory.

c. Boat travels north then west

A boat travels 76.0 km due north in 8.0 hours then 56.0 km due west in 5.0 hours.

Determine the direction (as a bearing) of the average velocity (to 1 decimal places) of the boat in the 8 + 5 hour period.


PLEASE HELP!!

Answers

Answer:

Explanation:

θ = arctan(56.0/76.0) = 36.4° West of North

average velocity is √(56.0² + 76.0²) / (8 + 5) = 94.4/13 = 7.26 m/s

If the voltage across a 5-F capacitor is 2*e^-3
V find the current and the power

Answers

Answer correct in this picture………….

Check Pic please, need help immediately ​

Answers

It’s d


I did this already

MCQ
A body of mass 5kg is pushed for distance x with accleration a. Then workdone against static friction is

1.ma*X cosB
2.ma*X sinB
3.zero
4.ma/X​

Answers

Answer:

ma*XsinB

option 2 is correct

I need help been struggling on this question

Answers

Answer:

440 m

Explanation:

S=(u+v) t / 2

S = (11+33) × 20/2

S= 44× 20/2

S=440 m

Tectonic plate movement is the reason why northern California has a very different landscape than southern California. Two different tectonic plates, each moving in different directions, border the western side of the North American Plate. Use the map to identify the two tectonic plates that border the North American Plate to the west.

Answers

Answer:

Remember, NORTH ^, EAST >, SOUTH v, WEST <

Explanation:

It doesn't have to be a super complex answer. All you have to do is look to the left (west) of the North American plate. What are the 2 plates that you see? The Pacific and the Juan de Fuca, yeah? To the South, there is the Cocos amongst a few others.

I am not going to share the answer for sure as I haven't completed the test yet but that's how I'm solving it. You should write the answer in your own words anyways. Hope this helps! Have a good day :)

Answer:

The Juan de Fuca Plate and the Pacific Plate both border the west side of the North American Plate.

Explanation:

Edmentum

An object is dropped from a vertical height of 1.89 m above the balcony level. What is the object’s speed when it is 2.20 m below the balcony level if 10.0% energy is lost due to the air resistance? Does it matter when to apply 10% loss before V calculations or after? [8.49m/s] [yes it does, 0.9Energy result in √0.9Velocity]

Answers

a.

The object's speed at 2.20 m below balcony level is 8.74 m/s

Let the balcony level be 0 m and the height above the balcony level be positive and height below the balcony level negative.

Using the principle of conservation of energy, the total energy at a vertical height of 1.89 m above the balcony level equals the total mechanical energy when the object is 2.20 m below the balcony level and

So, E = E'

U + K + f = U' + K' + f'

where U = initial potential energy at 1.89 m = mgh, K = initial kinetic energy at 1.89 m = 0 J(since it is released from rest), f = energy loss at 1.89 m = 0 J, U' = final potential energy at 2.20 m below balcony level = mgh', K = final kinetic energy at 2.20 m = 1/2mv², f' = energy loss at 1.89 m = 10%U = 0.10mgh(since 10% of the initial energy is lost).

So,

U + K + f = U' + K' + f'

mgh + 0 + 0 = mgh' + 1/2mv² + 0.10mgh

mgh = mgh' + 1/2mv² + 0.10mgh

Dividing through by m, we have

gh = gh' + 1/2v² + 0.10gh

So, gh -  0.10gh = gh' + 1/2v²

0.90gh = gh' + 1/2v²

1/2v² = 0.90gh - gh'

1/2v² = g(0.90h - h')

v² = 2g(0.90h - h')

Taking square-root of both sides, we have

v = √[2g(0.90h - h')]

where v = velocity of object at 2.20 m below balcony level, h = height above the balcony level = 1.89 m, h' = height below the balcony level = -2.20 m and  g = acceleration due to gravity = 9.8 m/s²

Substituting the values of the variables into the equation, we have

v = √[2g(0.90h - h')]

v = √[2 × 9.8 m/s²{0.90 × 1.89 m - (-2.20 m)}]

v = √[2 × 9.8 m/s²(1.701 m + 2.20 m)]

v = √[2 × 9.8 m/s²(3.901 m)]

v = √[76.4596 m²/s²]

v = 8.74 m/s

So, the object's speed at 2.20 m below balcony level is 8.74 m/s

b.

Yes it does matter when we apply 10% loss before V calculations

We need to apply the 10 % loss before V calculations because this would give us a proper value for V since the energy is lost before V is obtained.

So, yes it does matter when we apply 10% loss before V calculations

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Accelerations are produced by

A. Masses
B.accelerations
C. Velocities
D.unbalanced, net forces

Answers

The correct answer is D.
Acceleration is produced by a net force on an object.
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