PHYSICS COLLEGE

A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed

Answers

Answer 1

Answer:

The angular speed should be 17.18 rad / s

Calculation of the angular speed:

Since

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

Now the work done by force should be converted into the rotational kinetic energy

F x d = 1/2 I ω²

here,

F is the force applied,

d is displacement,

I is moment of inertia of disc

and ω is angular velocity of disc

So,

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18 rad / s

Learn more about speed here: brainly.com/question/18742396

Answer 2

Answer:

Answer:

Explanation:

moment of inertia of the disc I = 1/2 m r²

= .5 x 8 x .25²

= .25 kg m²

The wok done by force will be converted into rotational kinetic energy

F x d = 1/2 I ω²

F is force applied , d is displacement , I is moment of inertia of disc and ω

is angular velocity of disc

41 x .9 = 1/2 x .25 ω²

ω² = .25

ω = 17.18 rad / s

Related Questions

The minimum charge on any object cannot be less than
What sound frequency could a human detect
Use the worked example above to help you solve this problem. A ball is thrown upward from the top of a building at an angle of 30.0° to the horizontal and with an initial speed of 15.0 m/s. The point of release is h = 46.0 m above the ground.(a) How long does it take for the ball to hit the ground?(b) Find the ball's speed at impact.(c) Find the horizontal range of the ball.
A twirlers baton is 0.76 m long and spins around its center. The end of the baton has a centripetal acceleration of 47.8 m/s2?A 0.31 sB 0.56 sC 4.3sD 70s
What is the difference between V(peak voltage) and Vrms (root-mean-square) of AC voltage source?

Two children, each with a mass of 25.4 kg, are at fixed locations on a merry-go-round (a disk that spins about an axis perpendicular to the disk and through its center). One child is 0.78 m from the center of the merry-go-round, and the other is near the outer edge, 3.14 m from the center. With the merry-go-round rotating at a constant angular speed, the child near the edge is moving with translational speed of 11.5 m/s.a. What is the angular speed of each child?
b. Through what angular distance does each child move in 5.0 s?
c. Through what distance in meters does each child move in 5.0 s?
d. What is the centripetal force experienced by each child as he or she holds on?
e. Which child has a more difficult time holding on?

Answers

Answer:

a) ω₁ = ω₂ = 3.7 rad/sec

b) Δθ₁ = Δθ₂ = 18.5 rad

c) d₁ = 14.5 m d₂ = 57.5 m

d) Fc1 = 273.9 N Fc2 = 1069.8 N

e) The boy near the outer edge.

Explanation:

Since the merry-go-round is a rigid body, any point on it rotates at the same angular speed.
However, linear speeds of points at different distances from the center, are different.
Applying the definition of angular velocity, and the definition of angle, we can write the following relationship between the angular and linear speeds:

$v = \omega*r (1)$

Since we know the value of v for the child near the outer edge, and the value of r for this point, we can find the value of the angular speed, as follows:

$\omega = (v_(out) )/(r_(out) ) = (11.5m/s)/(3.14m) = 3.7 rad/sec (2)$

As we have already said, ωout = ωin = 3.7 rad/sec

Since the angular speed is the same for both childs, the angle rotated in the same time, will be the same for both also.
Applying the definition of angular speed, as the rate of change of the angle rotated with respect to time, we can find the angle rotated (in radians) as follows:
$\Delta \theta = \omega * t = 3.7 rad/sec* 5.0 sec = 18.5 rad (3)$

⇒ Δθ₁ = Δθ₂ = 18.5 rad.

The linear distance traveled by each child, will be related with the linear speed of them.
Knowing the value of the angular speed, and the distance from each boy to the center, we can apply (1) in order to get the linear speeds, as follows:

$v_(inn) = \omega * r_(inn) = 3.7 rad/sec * 0.78 m = 2.9 m/s (4)$

vout is a given of the problem ⇒ vout = 11. 5 m/s

Applying the definition of linear velocity, we can find the distance traveled by each child, as follows:

$d_(inn) = v_(inn) * t = 2.9m/s* 5.0 s = 14.5 m (5)$

$d_(out) = v_(out) * t = 11.5 m/s* 5.0 s = 57.5 m (6)$

The centripetal force experienced by each child is the force that keeps them on a circular movement, and can be written as follows:

$F_(c) = m*(v^(2))/(r) (7)$

Replacing by the values of vin and rin, since m is a given, we can find Fcin (the force on the boy closer to the center) as follows:

$F_(cin) = m*(v_(in)^(2))/(r_(in)) = 25.4 kg* ((2.9m/s)^(2) )/(0.78m) = 273.9 N (8)$

In the same way, we get Fcout (the force on the boy near the outer edge):

$F_(cout) = m*(v_(out)^(2))/(r_(out)) = 25.4 kg* ((11.5m/s)^(2) )/(3.14m) = 1069.8 N (9)$

The centripetal force that keeps the boys in a circular movement, is not a different type of force, and in this case, is given by the static friction force.
The maximum friction force is given by the product of the coefficient of static friction times the normal force.
Since the boys are not accelerated in the vertical direction, the normal force is equal and opposite to the force due to gravity, which is the weight.
As both boys have the same mass, the normal force is also equal.
This means that for both childs, the maximum possible static friction force, is the same, and given by the following expression:
$F_(frs) = \mu_(s) * m* g (10)$

If this force is greater than the centripetal force, the boy will be able to hold on.
So, as the centripetal force is greater for the boy close to the outer edge, he will have a more difficult time holding on.

Some snakes have special sense organs that allow them to see the heat emitted from warm blooded animals what kind of an electromagnetic waves does the sense organs detect?A. Visible light waves
B. Ultraviolet light waves
C. Infrared Waves
D. Microwaves

Answers

The heat emitted from anything is carried in the form of infrared waves. (C)

If a microwave oven produces electromagnetic waves with a frequency of 2.30 ghz, what is their wavelength?

Answers

Answer: wavelength is $<strong> 1.30 * 10^8 nm.</strong>$

The frequency of the microwave is, f = 2.30 GHz.

To Find frequency use the formula:

$c=f$ λ

Where, c is the speed of electromagnetic wave or light. f is the frequency, and λ is the wavelength of light.

Rearranging, $\lambda = (c)/(f)$

Plug in the values,

$\lambdam = (3 * 10^8 m/s)/(2.30 GHz(10^9 Hz)/(1 GHz))=0.130 m(10^9 nm)/(1 m) = 1.30 * 10^8 nm.$

$\lambda = (c)/(\nu) = (3 \cdot 10^8)/(2.3 \cdot 10^9) = (3)/(23) \approx 0.130435 \approx 0.13 \ m.$

An airplane with a speed of 92.3 m/s is climbing upward at an angle of 51.1 ° with respect to the horizontal. When the plane's altitude is 532 m, the pilot releases a package. (a) Calculate the distance along the ground, measured from a point directly beneath the point of release, to where the package hits the earth.
(b) Relative to the ground, determine the angle of the velocity vector of the package just before impact. (a) Number Units (b) Number Units

Answers

Answer:

$D = 1162.7 \ m$

$\beta =- 65.55^o$

Explanation:

From the question we are told that

The speed of the airplane is $u = 92.3 \ m/s$

The angle is $\theta = 51.1^o$

The altitude of the plane is $d = 532 \ m$

Generally the y-component of the airplanes velocity is

$u_y = v * sin (\theta )$

=> $u_y = 92.3 * sin ( 51.1 )$

=> $u_y = 71.83 \ m/s$

Generally the displacement traveled by the package in the vertical direction is

$d = (u_y)t + (1)/(2)(-g)t^2$

=> $-532 = 71.83 t + (1)/(2)(-9.8)t^2$

Here the negative sign for the distance show that the direction is along the negative y-axis

=> $4.9t^2 - 71.83t - 532 = 0$

Solving this using quadratic formula we obtain that

$t = 20.06 \ s$

Generally the x-component of the velocity is

$u_x = u * cos (\theta)$

=> $u_x = 92.3 * cos (51.1)$

=> $u_x = 57.96 \ m/s$

Generally the distance travel in the horizontal direction is

$D = u_x * t$

=> $D = 57.96 * 20.06$

=> $D = 1162.7 \ m$

Generally the angle of the velocity vector relative to the ground is mathematically represented as

$\beta = tan ^(-1)[(v_y)/(v_x ) ]$

Here $v_y$ is the final velocity of the package along the vertical axis and this is mathematically represented as

$v_y = u_y - gt$

=> $v_y = 71.83 - 9.8 * 20.06$

=> $v_y = -130.05 \ m/s$

and v_x is the final velocity of the package which is equivalent to the initial velocity $u_x$

$\beta = tan ^(-1)[-130.05}{57.96 } ]$

$\beta =- 65.55^o$

The negative direction show that it is moving towards the south east direction

1. On a force vs. mass graph, what would be the slope of the line?2. On a Free Body Diagram, if the forces are all balanced, what do you know about the
object? Can it be moving?

Answers

1. By Newton's second law,

F = ma

so the slope of the line would represent the mass of the object.

2. If all the forces are balanced, then the object is in equilibrium with zero net force, which in turn means the object is not accelerating. So the object is either motionless or moving at a constant speed.

Final answer:

The slope on a Force vs. Mass graph represents acceleration. In a Free Body Diagram, if all the forces are balanced, the object could be either at rest or moving at a constant velocity.

Explanation:

1. On a Force vs. Mass graph, the slope of the line represents acceleration, according to Newton's second law of motion, which is force equals mass times acceleration (F=ma). The slope of the line is calculated as the change in force divided by the change in mass, which results in acceleration.

2. In a Free Body Diagram, if all the forces are balanced, it means the net force acting on the object is zero. This does not necessarily mean that the object is stationary. The object could be at rest, or it could be moving at a constant velocity. If an object is moving at a constant velocity, it is said to be in equilibrium because the forces are balanced.

Learn more about the Physics of Forces and Motion here:

brainly.com/question/36922560

#SPJ11

What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

$P=68*6894.74\ Pa$

$P=468842.32\ Pa$

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

$P=(1)/(2)\rho v^2$

$468842.32=0.5*1000* v^2$

$v=\sqrt{(468842.32)/(0.5*1000)}$

$v=30.62\ m/s$

Now, We will convert m/s to ft/s

$v =30.62*3.281$

$v=100.46\ ft/s$

Hence, The velocity at discharge is 100.46 ft/s

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

Learn more about Pressure and Velocity here:

brainly.com/question/32124707

#SPJ3

Other Questions

A 4.0-m-long, 500 kg steel beam extends horizontally from the point where it has been bolted to the framework of a new building under construction. A 70 kg construction worker stands at the far end of the beam.What is the magnitude of the torque about the bolt due to the worker and the weight of the beam?

The four tires of an automobile are inflated to a gauge pressure of 2.2 105 Pa. Each tire has an area of 0.023 m2 in contact with the ground. Determine the weight of the automobile.

a painting in an art gallery has height h and is hung so that its lower edge is a distance d above the eye of an observer. How far from the wall should the observer stand to get the best view?

Which configuration is common to the outer shells of the elements neon, argon, and krypton?sp8s2dp6s2p6s2p4