A propeller is modeled as five identical uniform rods extending radially from its axis. The length and mass of each rod are 0.777 m and 2.67 kg, respectively. When the propellor rotates at 573 rpm (revolutions per minute), what is its rotational kinetic energy?

Answers

Answer 1

Answer: The formula for the rotational kinetic energy is

$KE_(rot) = (1)/(2)(number \ of\ propellers)( I)( omega)^(2)$

where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.

$I=m L^(2)=(2.67 \ kg) (0.777 \ m)^(2) =2.07459 \ kgm^(2)$

ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s

Then,

$KE_(rot) =( (1)/(2) )(5)(2.07459 \ kgm^(2)) (60\ rad/s)^(2)$

$KE_(rot) =18,671.31 \ J$

Answer 2

Answer:

Answer:

4833J

Explanation:

Length=0.777

mass=2.67

# rods= 5

ω=573 rpm--> $573*2\pi *(1)/(60) =60$ rad/s

I= $(1)/(3) mL^2=(1)/(3) (2.67kg)(0.777m)^2=0.537$ kgm^2

K=1/2(number of rods)(I)(ω)= $(1)/(2) *(5)(0.537)(60)^2=4833$ J

I know it's very late, but hope this helps anyone else trying to find the answer.

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A Hall-effect probe to measure magnetic field strengths needs to be calibrated in a known magnetic field. Although it is not easy to do, magnetic fields can be precisely measured by measuring the cyclotron frequency of protons. A testing laboratory adjusts a magnetic field until the proton's cyclotron frequency is 9.70 MHz . At this field strength, the Hall voltage on the probe is 0.549 mV when the current through the probe is 0.146 mA . Later, when an unknown magnetic field is measured, the Hall voltage at the same current is 1.735 mV .A) What is the strength of this magnetic field?

Answers

Answer:

The value of the magnetic field is 2.01 T when Hall voltage is 1.735 mV

Explanation:

The frequency of the cyclotron can help us find the magnitude of the magnetic field, thus then we can compare the effect of increasing Hall voltage on the probe.

Magnetic field magnitude at initial Hall voltage.

The cyclotron frequency can be written in terms of the magnetic field magnitude as follows

$f = \cfrac{qB}{2\pi m}$

Solving for the magnetic field.

$B = \cfrac{2\pi mf}q$

Thus we can replace the given information but in Standard units, also remembering that the mass of a proton is $m_p=1.67 * 10^(-27) kg$ and its charge is $q_p=1.6 * 10^(-19) C$ .

So we get

$B = \cfrac{2\pi * 1.67 * 10^(-27) kg * 9.7 * 10^6 Hz}{1.6 * 10^(-19)C}$

$B =0.636 T$

We have found the initial magnetic field magnitude of 0.636 T

Magnetic field magnitude at increased Hall voltage.

The relation given by Hall voltage with the magnetic field is:

$V_H =\cfrac{R_HI}t B$

Thus if we keep the same current we can write for both cases:

$V_(H1) =\cfrac{R_HI}t B_1\nV_(H2) =\cfrac{R_HI}t B_2$

Thus we can divide the equations by each other to get

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{\cfrac{R_HI}t B_1}{\cfrac{R_HI}t B_2}$

Simplifying

$\cfrac{V_(H1) }{V_(H2)}=\cfrac{ B_1}{ B_2}$

And we can solve for $B_2$

$B_2 =B_1 \cfrac{V_(H2)}{V_(H1)}$

Replacing the given information we get

$B_2= 0.636 T * \left(\cfrac{1.735 mV}{0.549 mV} \right)$

We get

$\boxed{B=2.01\, T}$

Thus when the Hall voltage is 1.735 mV the magnetic field magnitude is 2.01 T

Answer all three parts and show work.

Answers

The distance for both Parts A and B are given in the question.

A balloon drifts 140m toward the west in 45s.

The wind suddenly changes and the balloon flies 90m toward the east in the next 25s.

To find the total distance, we can just add.

140 + 90 = 230m

Best of Luck!

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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Two wooden crates rest on top of one another. The smaller top crate has a mass of m1 = 25 kg and the larger bottom crate has a mass of m2 = 91 kg. There is NO friction between the crate and the floor, but the coefficient of static friction between the two crates is µs = 0.79 and the coefficient of kinetic friction between the two crates is µk = 0.62. A massless rope is attached to the lower crate to pull it horizontally to the right (which should be considered the positive direction for this problem).1)The rope is pulled with a tension T = 234 N (which is small enough that the top crate will not slide). What is the acceleration of the small crate?2)In the previous situation, what is the magnitude of the frictional force the lower crate exerts on the upper crate?3)What is the maximum tension that the lower crate can be pulled at before the upper crate begins to slide?4)The tension is increased in the rope to 1187 N causing the boxes to accelerate faster and the top box to begin sliding. What is the acceleration of the upper crate?m/s25)As the upper crate slides, what is the acceleration of the lower crate?

Answers

Answer:

Explanation:

Let the bigger crate be in touch with the ground which is friction less. In the first case both m₁ and m₂ will move with common acceleration because m₁ is not sliding over m₂.

1 ) Common acceleration a = force / total mass

= 234 / ( 25 +91 )

= 2.017 m s⁻².

2 ) Force on m₁ accelerating it , which is nothing but friction force on it by m₂

= mass x acceleration

= 25 x 2.017

= 50.425 N

The same force will be applied by m₁ on m₂ as friction force which will act in opposite direction.

3 ) Maximum friction force that is possible between m₁ and m₂

= μ_s m₁g

= .79 x 25 x 9.8

= 193.55 N

Acceleration of m₁

= 193 .55 / 25

= 7.742 m s⁻²

This is the common acceleration in case of maximum tension required

So tension in rope

= ( 25 +91 ) x 7.742

= 898 N

4 ) In case of upper crate sliding on m₂ , maximum friction force on m₁

= μ_k m₁g

= .62 x 25 x 9.8

= 151.9 N

Acceleration of m₁

= 151.9 / 25

= 6.076 m s⁻².

John, who has a mass of 65kg stands at rest on the ice. He catches a 10kg ball that is thrown to him at 5m/s.

Answers

The momentum of John after catching the ball is 50 kg.m/s.

"Your question is not complete, it seems to be missing the following information";

find John's momentum

The given parameters;

mass of John, m = 65 kg
mass of the ball caught by John, m' = 10 kg
initial velocity of John, u = 0
initial velocity of the ball, v = 5 m/s

Apply the principles of conservation of linear momentum to determine the momentum of John.

The momentum of John is calculated as follows;

P = mu + mv

P = (65 x 0) + (10 x 5)

P = 0 + 50

P = 50 kg.m/s

Thus, the momentum of John after catching the ball is 50 kg.m/s.

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Why might a scientist want to use a model to study the solar system? O A. Its extreme simplicity makes it difficult to see patterns in observations. B. Its extremely slow movement makes it difficult to see the motions of different planets. C. Its extremely large size makes it difficult to see all of its parts at the same time. D. Its extremely small size makes it difficult to see planets that are far away

Answers

A scientist wants to use a model to study the solar model because its extremely large size makes it difficult to see all of its parts at the same time. Hence, option C is correct.

What is a Solar System?

The Sun and all the smaller movable objects that orbit it make up the Solar System. The eight main planets are the largest objects in the Solar System, excluding the Sun. Mercury, Venus, Earth, and Mars are the four relatively tiny, rocky planets closest to the Sun.

The asteroid belt, which is home to millions of stony objects, lies beyond Mars. These are remains from the planets' creation 4.5 billion years ago.

Jupiter, Saturn, Uranus, and Neptune are the four gas giants that can be found on the opposite side of the asteroid belt. Despite being much larger than Earth, these planets are rather light. Their main components are hydrogen and helium.

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