When it is at rotating at full speed, a disk drive in a certain old computer game system revolves once every 0.050 seconds. Starting from rest, it takes two revolutions for the disk to reach full speed. Assuming that the angular acceleration of the disk is constant, what is its angular acceleration while it is speeding up

Answers

Answer 1

Answer:

Final answer:

The angular acceleration of the disk drive in an old computer game system while speeding up is 1256 rad/s². This is calculated using kinematics in rotational motion, given the information on rotations, revolution time, and start from rest.

Explanation:

To calculate the angular acceleration of a disk drive in an old computer game system, we must use the concept of kinematics in rotation. When it is stated that it takes two revolutions to reach full speed, this implies that the total angular displacement is 4π radians (since one full revolution is 2π radians).

Given that the disk drive revolves once every 0.050 seconds, the final angular speed (ω) can be computed as 2π rad/0.050 s = 125.6 rad/s. Since the disk starts from rest, the initial angular speed (ω0) is 0. As a result, the total time taken (t) to reach full speed is 2*0.050s = 0.1 s.

We can then use the equation of motion in rotational form, α = (ω - ω0)/t, to calculate the angular acceleration. Hence the angular acceleration (α) is (125.6 rad/s - 0 rad/s) / 0.1 s = 1256 rad/s². Therefore, the angular acceleration of the disk drive is 1256 rad/s² while it is speeding up.

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Answer 2

Answer:

Final answer:

The angular acceleration of the disk drive while it is speeding up is 8π rad/s².

Explanation:

The angular acceleration of the disk drive while it is speeding up can be determined by using the formula: angular acceleration = (final angular velocity - initial angular velocity) / time taken. In this case, the initial angular velocity is 0 (since the disk starts from rest) and the final angular velocity is 2 revolutions per 0.050 seconds. To convert revolutions to radians, multiply by 2π. The time taken is the time for two revolutions, so it is 2 * 0.050 seconds. Plugging in these values in the formula, we get:

Angular acceleration = (2 * 2π rad/s - 0) / (2 * 0.050 s) = 8π rad/s²

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Related Questions

A large balloon of mass 210 kg is filled with helium gas until its volume is 329 m3. Assume the density of air is 1.29 kg/m3 and the density of helium is 0.179 kg/m3. (a) Draw a force diagram for the balloon. (Submit a file with a maximum size of 1 MB.) (b) Calculate the buoyant force acting on the balloon. (Give your answer to at least three significant figures.) 4159 N (c) Find the net force on the balloon. 1524 N Determine whether the balloon will rise or fall after it is released. The balloon will (d) What maximum additional mass can the balloon support in equilibrium? 155 kg (e) What happens to the balloon if the mass of the load is less than the value calculated in part (d)? The balloon and its load will remain stationary. The balloon and its load will accelerate downward. The balloon and its load will accelerate upward. (f) What limits the height to which the balloon can rise?

Answers

(a) See figure in attachment (please note that the image should be rotated by 90 degrees clockwise)

There are only two forces acting on the balloon, if we neglect air resistance:

- The weight of the balloon, labelled with W, whose magnitude is

$W=mg$

where m is the mass of the balloon+the helium gas inside and g is the acceleration due to gravity, and whose direction is downward

- The Buoyant force, labelled with B, whose magnitude is

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity, and where the direction is upward

(b) 4159 N

The buoyant force is given by

$B=\rho_a V g$

where $\rho_a$ is the air density, V is the volume of the balloon and g the acceleration due to gravity.

In this case we have

$\rho_a = 1.29 kg/m^3$ is the air density

$V=329 m^3$ is the volume of the balloon

g = 9.8 m/s^2 is the acceleration due to gravity

So the buoyant force is

$B=(1.29 kg/m^3)(329 m^3)(9.8 m/s^2)=4159 N$

(c) 1524 N

The mass of the helium gas inside the balloon is

$m_h=\rho_h V=(0.179 kg/m^3)(329 m^3)=59 kg$

where $\rho_h$ is the helium density; so we the total mass of the balloon+helium gas inside is

$m=m_h+m_b=59 kg+210 kg=269 kg$

So now we can find the weight of the balloon:

$W=mg=(269 kg)(9.8 m/s^2)=2635 N$

And so, the net force on the balloon is

$F=B-W=4159 N-2635 N=1524 N$

(d) The balloon will rise

Explanation: we said that there are only two forces acting on the balloon: the buoyant force, upward, and the weight, downward. Since the magnitude of the buoyant force is larger than the magnitude of the weigth, this means that the net force on the balloon points upward, so according to Newton's second law, the balloon will have an acceleration pointing upward, so it will rise.

(e) 155 kg

The maximum additional mass that the balloon can support in equilibrium can be found by requiring that the buoyant force is equal to the new weight of the balloon:

$W'=(m'+m)g=B$

where m' is the additional mass. Re-arranging the equation for m', we find

$m'=(B)/(g)-m=(4159 N)/(9.8 m/s^2)-269 kg=155 kg$

(f) The balloon and its load will accelerate upward.

If the mass of the load is less than the value calculated in the previous part (155 kg), the balloon will accelerate upward, because the buoyant force will still be larger than the weight of the balloon, so the net force will still be pointing upward.

(g) The decrease in air density as the altitude increases

As the balloon rises and goes higher, the density of the air in the atmosphere decreases. As a result, the buoyant force that pushes the balloon upward will decrease, according to the formula

$B=\rho_a V g$

So, at a certain altitude h, the buoyant force will be no longer greater than the weight of the balloon, therefore the net force will become zero and the balloon will no longer rise.

Final answer:

The physics involved in the functioning of helium balloons is based on buoyancy and Archimedes' Principle. The forces at play include the force due to gravity, the buoyant force and the net force, which determines the motion of the balloon. The balloon's height limit is determined by the decrease in air density with altitude.

Explanation:

The several parts of this question are related to the principles of buoyancy and Archimedes' Principle. First, regarding the force diagram for the balloon (part a), it would show two primary forces. The force due to gravity (Fg) acting downwards and the buoyant force (Fb) acting upwards, which is a result of the displacement of air by the balloon. The net force mentioned in part (c) is calculated as the difference between these two forces.

Calculating the buoyant force (part b) involves multiplying the volume of the balloon by the density of the air and the acceleration due to gravity (Fb = V * ρ_air * g). For the net force on the balloon (part c), this is calculated by subtracting the weight of the balloon from the buoyant force (F_net = Fb - Fg). If the net force is positive, the balloon will rise, if it's negative, the balloon will fall, and if it is zero, the balloon will remain stationary.

The maximum additional mass the balloon can support in equilibrium (part d) is calculated using the net force divided by gravity. If the mass of the load is less than this value (part e), the balloon and its load will accelerate upward.

Lastly, the limit to the height to which the balloon can rise (part f) is determined by the decreasing density of the air as the balloon ascends. The buoyant force reduces as the balloon rises because the air density is lower at higher altitudes.

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A block sliding on ground where μk = .193 experiences a 14.7 N friction force. What is the mass of the block

Answers

Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77

What is Friction?

According to the International Journal of Parallel, Emergent and Distributed Systems(opens in new tab), it is not treated as a fundamental force, like gravity or electromagnetism. Instead, scientists believe it is the result of the electromagnetic attraction between charged particles in two touching surfaces.

Scientists began piecing together the laws governing friction in the 1400s, according to the book Soil Mechanics(opens in new tab), but because the interactions are so complex.

F=μ*m, n=w which also means n=mg, 14.7=0.193*n, n=76.2, 76.2=m*9.8, m=7.77.

Therefore, Friction is the resistance to motion of one object moving relative to another. The friction will be 7.77.

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Answer:

7.77

Explanation:

F=μ*m

n=w which also means n=mg

14.7=0.193*n

n=76.2

76.2=m*9.8

m=7.77

You're carrying a 3.6-m-long, 21 kg pole to a construction site when you decide to stop for a rest. You place one end of the pole on a fence post and hold the other end of the pole 35 cm from its tip. For the steps and strategies involved in solving a similar problem, you may view a Video Tutor Solution. Part A Part complete How much force must you exert to keep the pole motionless in a horizontal position? Express your answer in newtons. F = 114 N Previous Answers

Answers

Final answer:

This Physics problem involves balancing the forces and torques acting on a 3.6-m-long pole. By applying the principles of equilibrium and calculations of torque, we find that 114 N of force is needed to keep the pole in a horizontal position.

Explanation:

This is a physics problem related to the concepts of equilibrium and torque. From the details provided, we know that the pole has a mass of 21 kg and it's 3.6 meters long. The center of gravity (cg) of the pole, since it's uniform, is at the middle, which is at 1.8 m from either end of the pole. We are then told that you are holding the pole 35 centimeters (or 0.35 meters) from its tip.

To keep the pole horizontal in equilibrium, the downward force due to the weight of the pole at its center of mass (which is equal to the mass of the pole times gravity, or 21*9.8 = 205.8 N) needs to be balanced by the sum of the torques produced by the forces you are applying at the end you are holding and the force exerted by the fence post at the other end.

Let the force you apply be F1 and the force the fence post exerts be F2. We have F2 at 0.35 m from one end (the pivot point), and F1 at 3.6 - 0.35 = 3.25 m from the pivot. Given that the torque (t) equals to Force (F) times the distance from the pivot (d), and that the net torque should equal zero in equilibrium, we have:

0.35*F2 = 3.25*F1 (1)

Because the net force should also be zero in equilibrium, we have:

F1 + F2 = 205.8 (2)

Solving these two equations, we'll be able to calculate that the force you must exert to keep the pole motionless in a horizontal position, F1, is approximately 114 N.

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Final answer:

To balance the 3.6m-long, 21 kg pole and keep it horizontally motionless, a force of approximately 114N is required

Explanation:

The subject question is a classic example of Torque problem specific to Physics, which involves the concepts of force, weight, and distance. To keep the pole motionless and horizontally balanced, the force you exert must counterbalance the torque due to the pole's weight. Assuming the pole is uniform, its center of gravity (cg) is at its midpoint, 1.8m from each end. The weight of the pole acts downward at this midpoint, providing a clockwise torque about the point of support, which is the fence post.

This torque is calculated as Torque = r * F = 1.8m (distance from fence post to cg) * Weight of pole = 1.8m * 21kg * 9.8m/s² (gravitational acceleration) = ~370 N.m. As the pole is motionless, the total torque about any point must be zero. Hence, the counter-clockwise torque provided by the force you exert is equal to the clockwise torque due to the weight of the pole. Using the distance from the point of your hold to the fence post (3.25m) we can calculate the force you need to exert: Force = Torque/distance = 370 N.m/3.25m = ~114N.

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On a hot summer day in the state of Washington while kayaking, I saw several swimmers jump from a railroad bridge into the Snohomish River below. The swimmers stepped off the bridge, and I estimated that they hit the water 2.00 s later. A)How high was the bridge?B)How fast were the swimmers moving when they hit the water?

C)What would the swimmer's drop time be if the bridge were twice as high?

Answers

Answer: part a: 19.62m

part b: 19.62 m/s

part a: 2.83 secs

Explanation:If the air resistance is ignored then the swimmer experience free fall under gravity hence

u=0

a=9.81 m/s2

t=2 secs

$s=ut+0.5at^2$

s=h

$h=0*2+0.5*9.81*2^2\nh=19.62 meters$

Part b

$v=u+at\nv=0+9.81*2\nv=19.62m/s$

Part c

now we have h=2*19.62=39.24

$39.24=0+0.5*9.81*t^2\nt^2=8\nt=2.83 secs$

Three children want to play on a see saw that is 6 meters long and has a fulcrum in the middle. Two of the children are twins and weigh 40 kg each and sit on the same side at a distance 2m and 3m away from the fulcrum. The other child weighs 80 kg. How far away should he sit from the fulcrum so that the see saw is balanced

Answers

Given Information:

mass of child 1 = m₁ = 40 kg

distance from fulcrum of child 1 = d₁ = 2 m

mass of child 2 = m₂ = 40 kg

distance from fulcrum of child 2 = d₂ = 3 m

mass of child 3 = m₃ = 80 kg

Required Information:

distance from fulcrum of child 3 = d₃ = ?

Answer:

distance from fulcrum of child 3 = 2.5 m

Explanation:

In order to balance the see-saw, the moment of force should be same on both sides of the fulcrum.

Since 2 children are sitting on one side and only 1 on the other side

F₁d₁ + F₂d₂ = F₃d₃

Where Force is given by

F = mg

m₁gd₁ + m₂gd₂ = m₃gd₃

m₁d₁ + m₂d₂ = m₃d₃

Re-arrange the equation for d₃

m₃d₃ = m₁d₁ + m₂d₂

d₃ = (m₁d₁ + m₂d₂)/m₃

d₃ = (40*2 + 40*3)/80

d₃ = 2.5 m

Therefore, the child on the other side should sit 2.5 m from the fulcrum so that the see-saw remains balanced.

a trcuk weighs four times as much as a stationary car. if teh truck coasts into the car at 12 km/s and they stick toegther, what is their final velocity