A whole set of birdfeeders are designed using conservation of Angular Momentum to spin when a squirrel jumps on them. This can throw the squirrel off (though not all squirrels give up that easily - see this video for an example). A bird, landing, doesn't cause the same problem. A squirrel, with a mass of 3.00 kg launches itself at the bird feeder with a velocity of 3.40 m/s. The bird feeder has a radius of 6.30 cm and a Moment of Inertia of 2.00 kg m2. Initially the bird feeder is not rotating at all, but starts rotating when the squirrel lands on the outer edge (at the same radius as described above). You can assume that the squirrel is small compared to the size of the bird feeder radius (not true in the video, but it does make this a bit easier for out calculations). What is the angular velocity of the bird feeder - squirrel system after the squirrel lands on it

Answers

Answer 1

Answer:

 w = 0.319 rad / s

Explanation:

This is an angular momentum problem, let's form a system composed of the feeder and the squirrel, therefore the forces during the collision are internal and the angular momentum is conserved.

         

initial instant. Before the squirrel jumps

           L₀ = m v r

final instant. After the trough and the squirrel are together

          L_f = (I_fetter + I_ardilla) w

angular momentum is conserved

          L₀ = L_f

          m v r = (I_fetter + I_ardilla) w

          w = [tex]\frac{mvr}{I_{fetter} + I_{ardilla} }[/tex]

the moment inercial ofbody is

         I_thed = 2.00 kg m²

We approach the squirrel to a specific mass

          I_ardilla = m r²

we substitute

            w = m v r / ( I_[feefer  + m r²)

             

           

let's calculate

              w = 3 3.40 6.30 10⁻² / (2.00 + 3.00 (6.30 10-2)² )

              w = 0.6426 / 2.0119

               w = 0.319 rad / s


Related Questions

Objects A and B, of mass M and 2M respectively, are each pushed a distance d straight up an inclined plane by a force F parallel to the plane. The coefficient of kinetic friction between each mass and the plane has the same value μ.k At the highest point is:______

a. KEA > KEB
b. KEA = KEB
c. KEA < KEB
d. The work done by F on A is greater than the work done by F on B.
e. The work done by F on A is less than the work done by F on B.

Answers

Answer:

The correct answer is option (A) that is KEA > KEB .

Explanation:

Let us calculate -

If the object is straighten up and inclined plane , the work done is

[tex]W=F_d- F_f_r_id-F_gh[/tex]

[tex]W=F_d-\mu_kmgdcos\theta-mgdsin\theta[/tex]

The change in kinetic energy is ,

   [tex]\Delta K=\frac{1}{2}mv^2-\frac{1}{2}m\nu_0^2[/tex]

At the top of the inclined plane , the velocity is zero

So,

[tex]\Delta K=\frac{1}{2} m(0)^2-\frac{1}{2}m\nu_0^2[/tex]

[tex]\Delta KE=-\frac{1}{2}m\nu_0^2[/tex]

From the work energy theorem , we have [tex]W=-\Delta K[/tex] in case of friction , so

[tex]\frac{1}{2}m\nu_0^2=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

[tex]KE=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object A-

[tex]KE_A=Fd-\mu_kmgdcos\theta-mgdsin\theta[/tex]

For object B

[tex]KE_B= Fd -2\mu_kMgdcos\theta-2Mgdsin\theta[/tex]

[tex]KE_B= Fd -2(\mu_kMgdcos\theta-Mgdsin\theta)[/tex]

Thus , larger mass is going to mean less total work and a lower kinetic energy .

From the above results , we get

[tex]KE_A >KE_B[/tex]

Therefore , option A is correct .

A negative charge of 2 C and a positive charge of 3 C are separated by 80 m. What is the force between the two charges?

Answers

Answer:

6

Explanation: i looked this up.

According to the Coloumb's law, the force is obtained from the charges and the distance between them is 8.4375×10⁶ N. This force results in the attractive force.

What is Coloumb's law?

Coloumb's law states the relation between the charges and the distance between the charges. Coloumb's law states that the Force is directly proportional to the product of charges and inversely proprotional to the square of distance between them.

The Coloumb's law gives the force of attraction or repulsion between the charged bodies. Two charges with positive or negative charges repels each other. One positive and one negative charge attract each other.

The force increases with the product of charges increases as product of charges and force are directly proprotional. The force decreases with the increase in distance of seperation and vice-versa. The SI unit of force is newton (N).

The Coloumb's law is:

          F = k (q₁×q₂ / r²)

k is the constant of proportionality and is equal to 9×10⁹ N.m²/C².

q₁, q₂ = charges

r² =  distance of seperation of charges.

From the given,

q₁ = -2C ( negative sign represents the negative charge)

q₂ = +3C ( + sign represents the positive charge)

r² = 80 m ( distance between the charges)

F = k (q₁×q₂ / r²)

  = 9×10⁹×2×3 / 80×80

 = 54×10⁹ / 6400

 = 8437500 N

F = 8.4375×10⁶ N

There is the presesnce of both positive and negative charges, hence it results in the attractive force. Hence, the force F between two charges separated by 80m is  8.4375×10⁶ N.

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the pygmy shrew has an average mass of 2.0 g if 49 of these shrew are placed on a spring scale with a spring constant of 24 N/m , what is the springs displacement

Answers

Answer:

Spring's displacement, x = -0.04 meters.

Explanation:

Let the spring's displacement be x.

Given the following data;

Mass of each shrew, m = 2.0 g to kilograms = 2/1000 = 0.002 kg

Number of shrews, n = 49

Spring constant, k = 24 N/m

We know that acceleration due to gravity, g is equal to 9.8 m/s².

To find the spring's displacement;

At equilibrium position:

Fnet = Felastic + Fg = 0

But, Felastic = -kx

Total mass, Mt = nm

Fg = -Mt = -nmg

-kx -nmg = 0

Rearranging, we have;

kx = -nmg

Making x the subject of formula, we have;

[tex] x = \frac {-nmg}{k} [/tex]

Substituting into the formula, we have;

[tex] x = \frac {-49*0.002*9.8}{24} [/tex]

[tex] x = \frac {-0.9604}{24} [/tex]

x = -0.04 m

Therefore, the spring's displacement is -0.04 meters.

A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.An object is made from a uniform piece of sheet metal. The object has dimensions of a

Answers

This question is not complete, the complete question is;

A centroid is an object's geometric center. For an object of uniform composition, its centroid is also its center of mass. Often the centroid of a complex composite body is found by, first, cutting the body into regular shaped segments, and then by calculating the weighted average of the segments' centroids.

An object is made from a uniform piece of sheet metal. The object has dimensions of α = 1.50 ft, where α is the diameter the semi-circle, b= 3.51 ft, and c = 2.20 ft. A hole with diameter d = 0.500 ft is centered at ( 1.21, 0.750 ).

Find x", y", the coordinates of the body's centroid.

Answer:

x" = 1.4857 ft

y" = 0.668 ft

Explanation:

Given the data in the question and as illustrated in the second image below;

from the image;

BC² = DC² - BD²

BC² = 2.2² - 1.5² = 4.84 - 2.25 = 2.59

BC = √2.59 = 1.61 ft

AB = 3.51 ft - 0.75 ft - 1.61 ft = 1.15 ft

so;

A₁ = [tex]\frac{1}{2}[/tex] × 1.51 ft × 1.61 ft  = 1.2075 ft²

x₁ = 0.75 + 1.15 + [tex]\frac{1}{3}[/tex](1.61 ft) = 2.44 ft

y₁ = [tex]\frac{1}{3}[/tex](1.5 ft) = 0.5 ft

A₂ = 1.15 ft × 1.5 ft = 1.725 ft²

x₂ = 0.75 ft + ( 1.15/2 )ft = 1.325 ft

y₂ = ( 1.5/2 ) ft = 0.75 ft

A₃ = [tex]\frac{\pi }{2}[/tex](0.75 ft)² = 0.88 ft²

x₃ = 0.75 - ([tex]\frac{4 }{3\pi }[/tex](0.75 ft)) = 0.43 ft

y₃ = 0.75 ft

diameter d = 0.5 ft and centered at ( 1.21, 0.750 )

A₄ = [tex]\frac{\pi }{4}[/tex]( d )² =  

x₄ = 1.21 ft

y₄ = 0.75 ft

Thus;

x" = [tex]\frac{A_1 x_1 + A_2 x_2 + A_3 x_3 - A_4x_4 }{A_1+A_2+A_3-A_4}[/tex]

so we substitute

x" = [tex]\frac{(1.2075X2.44) + (1.725 X 1.325) + (0.88X0.43) - (0.196 X 1.21 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]

x" = [tex]\frac{ (2.9463 + 2.285625 + 0.3784 - 0.23716)}{ 3.6165 }[/tex]

x" = 5.373165 / 3.6165

x" = 1.4857 ft

y" = [tex]\frac{A_1 y_1 + A_2 y_2 + A_3 y_3 - A_4y_4 }{A_1+A_2+A_3-A_4}[/tex]

so we substitute

y" = [tex]\frac{(1.2075X0.5) + (1.725 X 0.75) + (0.88X0.75) - (0.196 X 0.75 )}{ ( 1.2075 + 1.725 + 0.88 - 0.196 )}[/tex]

y" = [tex]\frac{ (0.60375 + 1.29375 + 0.66 - 0.14112)}{ 3.6165 }[/tex]

y" = 2.41638 / 3.6165

y" = 0.668 ft

Therefore,

x" = 1.4857 ft

y" = 0.668 ft

1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points

Answers

Answer:

24 J

Explanation:

[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]

How does the presence of a nucleus provide a method of basic cell
classification? *

Answers

Answer:

The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.

Explanation:

I hope this is what you were looking for?!

Hope this helps!

Have a great day!

-Hailey!

The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

What is a nucleus?

The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.

A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.

Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.

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Transverse thrusters are used to make large ships fully maneuverable at low speeds without tugboat assistance. A transverse thruster consists of a propeller mounted in a duct; the unit is then mounted below the waterline in the bow or stern of the ship. The duct runs completely across the ship. Calculate the thrust developed by a 1900 kW unit supplied to the propeller if the duct is 2.6 m in diameter and the ship is stationary.

Answers

Answer:

Thrust developed = 212.3373 kN

Explanation:

Assuming the ship is stationary

Determine the Thrust developed

power supplied to the propeller ( Punit ) = 1900 KW

Duct distance ( diameter ; D  ) = 2.6 m

first step : calculate the area of the duct

A = π/4 * D^2

   =  π/4 * ( 2.6)^2  = 5.3092 m^2

next : calculate the velocity of propeller

Punit = (A*v*β ) / 2  * V^2     ( assuming β = 999 kg/m^3 ) also given V1 = 0

∴V^3 = Punit * 2 / A*β

         = ( 1900 * 10^3 * 2 ) / ( 5.3092 * 999 )

hence V2 = 8.9480 m/s

Finally determine the thrust developed

F = Punit / V2

  = (1900 * 10^3) / ( 8.9480)

  = 212.3373 kN

An object is pushed with a force of 400 N for 20 meters; what is the work done on the object? Solve using W=F*d

Answers

Answer:

800joules

Explanation:

work is measured in joules

1. A train is moving north at 5 m/s on a straight track. The engine is causing it to accelerate northward at 2 m/s^2.
How far will it go before it is moving at 20 m/s?
A) 83
B) 43
C) 39
D) 94
E) 20

Answers

Answer:

(D) x = 93.8 m

Explanation:

v^2 = v0^2 + 2ax

(20 m/s)^2 = (5 m/s)^2 + 2(2 m/s^2)x

Solving for x,

x = 93.8 m

A ball bounces changing velocity from vi=15m/s[D] to vf=15m/s[U] in t=0.01s. The balls acceleration is *
A) 0
B) 30m/s/s
C) 3000m/s/s
D) 0.3m/s/s

Answers

Answer:

Option (A) is correct

Explanation:

a= (vf-vi)/ t

put the values

hence,

a= ( 15-15)/0.01

a=0

Consider a coaxial cable (like the kind that is used to carry a signal to your TV). In this cable, a current I runs in one direction along the inner central wire, and the same current I runs in the opposite direction in the hollow conductor. How does the magnetic field B outside the entire cable (outside both the inner wire as well as the hollow conductor) vary as a function of distance away from the cable

Answers

Answer:

0 < r < r_exterior     B_total = [tex]\frac{\mu_o I}{2\pi r}[/tex]

r > r_exterior            B_total = 0

Explanation:

The magnetic field created by the wire can be found using Ampere's law

        ∫ B. ds = μ₀ I

bold indicates vectors and the current is inside the selected path

           

outside the inner cable

          B₁ (2π r) = μ₀ I

          B₁ = [tex]\frac{\mu_o I}{2\pi r}[/tex]

the direction of this field is found by placing the thumb in the direction of the current and the other fingers closed the direction of the magnetic field which is circular in this case.

For the outer shell

for the case   r> r_exterior

         

           B₂ = \frac{\mu_o I}{2\pi  r}

This current is in the opposite direction to the current in wire 1, so the magnetic field has a rotation in the opposite direction

for the case r <r_exterior

in this case all the current is outside the point of interest, consequently not as there is no internal current, the field produced is zero

           B₂ = 0

Now we can find the field created by each part

0 < r < r_exterior

          B_total = B₁

          B_total = [tex]\frac{\mu_o I}{2\pi r}[/tex]  

r > r_exterior

          B_total = B₁ -B₂

          B_total = 0

how do stocks and bonds differ

Answers

Answer:

Explanation:

Stocks give you partial ownership in a corporation, while bonds are a loan from you to a company or government. The biggest difference between them is how they generate profit: stocks must appreciate in value and be sold later on the stock market, while most bonds pay fixed interest over time.

Answer:

Stocks give you partial ownership in a corporation, while bonds are a loan from you to a company or government. The biggest difference between them is how they generate profit: stocks must appreciate in value and be sold later on the stock market, while most bonds pay fixed interest over time.

Explanation:

None

can someone please help me answer this!!

Answers

i may be wrong, but i believe it is C, as there is no mention of sunlight being a control variable. the amount of sunlight greatly determines plant growth since they obviously use it for photosynthesis

Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.

Answers

It’s says that the temperature in the room is 23C so the air temperature in the air would be 23C

The average temperature of the air in the room at 20 min is 23°C.

What is temperature?

Temperature is the degree of hotness or coldness of the object.

A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.

The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.

Thus, the average temperature of the air in the room at 20 min is 23°C.

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Model the Earth's atmosphere as 79% N2, 19% O2, and 2% Argon, all of which are in thermal equilibrium at 280 K. At what height is the density of O half its value at sea level

Answers

Answer:

[tex]9.495 \times 10^3\ m[/tex]

Explanation:

From the given information:

Using the equation of Barometric formula as related to density, we have:

[tex]\rho (z) = \rho (0) e^{(-\dfrac{z}{H})} \ \ \ \ --- (1)[/tex]

Here;

[tex]p(z) =[/tex] the gas density at altitude z

[tex]\rho(0) =[/tex] the gas density  at sea level

H = height of the scale

[tex]H = \dfrac{RT}{M_ag } \ \ \ --- (2)[/tex]

Also;

R represent the gas constant

temperature (T) a= 280 K

g = gravity

[tex]M_a =[/tex] molaar mass of gas; here, the gas is Oxygen:

[tex]M_a =[/tex] 15.99 g/mol

= 15.99 × 10⁻³ kg/mol

[tex]H = \dfrac{8.3144 \times 280}{15.99 \times 10^{-3} \times 9.8 }[/tex]

[tex]H =14856.43 \ m[/tex]

Now we need to figure out how far above sea level the density of oxygen drops to half of what it is at sea level.

This implies that we have to calculate z;

i.e. [tex]\rho(z) =\dfrac{\rho(0) }{(2)}[/tex]

By using the value of H and [tex]\rho(z)[/tex] from (1), we have:

[tex]\dfrac{\rho(0) }{(2)} = \rho (0) e^{(-\dfrac{z}{14856.43})}[/tex]

[tex]\dfrac{1}{2} = e^{(-\dfrac{z}{14856.43})} \\ \\ e^{(-\dfrac{z}{14856.43})} =\dfrac{1}{2}[/tex]

By rearrangement and taking the logarithm of the above equation; we have:

[tex]- z = 14856.43 \times \mathtt{In}\dfrac{1}{2} \\ \\ -z = 14856.43 \times (-0.6391) \\ \\ z = 9495 \ m \\ \\ z = 9.495 \times 10^3\ m[/tex]

As a result, the oxygen density at  [tex]9.495 \times 10^3\ m[/tex] is half of what it is at sea level.

the minimum speed on the interstate highway is
1. 40 mph
2. 60 mph
3. 55 mph
4. 50 mph
I'm in Nebraska btw​

Answers

It’s 1 I’m not from Nebraska but I looked it up lol.

An Object, Start from rest w Confront Aiceleration 8m/s2 along a
Straight line. Find
A, the speed At the end Of 5 second
B, The average Speed for the 5second interval​

Answers

Answer:

A)   v = 40 m / s, B)   v_average = 20 m / s

Explanation:

For this exercise we will use the kinematics relations

         

A) the final velocity for t = 5 s and since the body starts from rest its initial velocity is zero

         v = vo + a t

         v = 0 + 8 5

         v = 40 m / s

B) the average velocity can be found with the relation

         v_average = vf + vo / 2

         v-average = 0+ 40/2

          v_average = 20 m / s

Which category best describes this group of shapes? rhombuses rectangles parallelograms squares II​

Answers

they're parallelograms

The universe cooled after the Big Bang.At some point hydrogen atoms combined to form helium.What is this process called?

Answers

Answer:

Nuclear fusion

Explanation:

Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?

Answers

Answer:

i dont know but i should know try g o o g l e

                               

Explanation:

7)
Which wave carries the most energy?
-)
A)

B)
C)
D)
Hover over answer image to enlarge

Answers

C.

Explanation:

C. is a gamma ray and they carry the most energy.

The most energy is carried by the gamma rays. The correct option is C.

What is a gamma ray?

Gamma radiation, also known as gamma rays, is a type of electromagnetic radiation produced by the radioactive decay of atomic nuclei. It is made up of electromagnetic waves with the shortest wavelengths, which are typically shorter than those of X-rays.

Waves with a higher frequency have more energy. Gamma rays have the shortest wavelengths and highest frequencies of any electromagnetic wave. Gamma rays have more energy than any other electromagnetic wave due to their extremely high frequencies.

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8. What is the mass and charge of a proton?
A. 1 amu, no charge
B. 1 amu, charge of +1
C. Mass of O, charge of -1
D. None of the above

Answers


B.

Step By Step Explanation:

HURRY IM TIMED!

The age and gender of an audience are important to consider when deciding on a subject.
True
False

Answers

Answer:

True.

Explanation:

true

knowing your audience there General age gender education level religion language culture and gave group members is the single most important aspect of developing your speech this means the speaker dock smart the audience wasn't often without asking question or spending with any feedback

Average running speed is 8km/h. Convert this to m/s :
a)8m/s
b)2.2m/s
c)28.8m/s
d)8km/h

Answers

Answer:

Im not sure if you mean miles per hour or meters per hour but if it is miles it is 4.97 miles per hour and if it is meters then it is 8000 meters per hour

Explanation:

Sorry if I'm wrong

Which electromagnetic wave has the shortest wavelength?



Which electromagnetic wave would have a wavelength approximately the size of the tip of a pencil?



Which electromagnetic wave has the longest wavelength?

Answers

Answer:

shortest - gamma Ray's

size of a pencil tip : most likely microwaves

longest - radio waves

Gamma rays have the shortest wavelength whereas Radio waves have the longest wavelength. The wave whose wavelength is about the size of the tip of a pencil is infrared ray.

What is an Electromagnetic spectrum?

The whole electromagnetic waves are divided into several classes based on their frequencies or energies as follows, in the order of increasing frequency:

Radio waves

Microwaves

Infrared

Visible

Ultraviolet

X-rays

Gamma rays

The Gamma rays are the shortest wavelength group with frequencies greater than 3 × 10²⁰ Hz.

The Radio waves are the longest wavelength group. Their frequency is lower than 3 × 10⁷ Hz.

The tip of a pencil is about 0.5mm in width.

Calculate the frequency of the wave whose wavelength is about the size of the tip of a pencil i.e. λ = 0.5 mm,

ν = 3×10⁸/(0.5×10⁻³) = 6×10¹¹ Hz.

The infrared rays have frequencies in the range of 3×10¹¹ Hz to 3×10¹³ Hz. So, the wave is infrared.

Infrared waves have a wavelength about the tip of a pencil.

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In an elastic collision between a moving 10-kg mass and a stationary 10-kg mass half the momentum is transferred to the stationary mass. In this situation the total kinetic energy after the collision is less than it was before the collision. Where did the kinetic energy go?

A) The kinetic energy was destroyed during the collision.

B) Some of the kinetic energy was turned into momentum during the collision.

C) Some of the kinetic energy was turned into heat or used to deform the masses.

D) Some of the kinetic energy was turned into potential energy during the collision.

Answers

Answer: C

Explanation:

USAtestprep

Two risks of exposure to High levels of UV radiation

Answers

Answer:

uv radiation cause cancer

uv radiation effect our eyes

Answer:

you can get

1:skin cancer

2:eye damage

3:skin damage

4:immune system suppression

choose which two u want

hope this helped

:)

Explanation:

You are riding on a carousel that is rotating at a constant 24 rpm. It has an inside radius of 4 ftand outside radius of 12 ft. You begin to run from the inside to the outside along a radius. Your peak velocity with respect to the carousel is 6 mph and occurs at a radius of 8 ft.What are your maximum Coriolis acceleration magnitude and its directionwith respect to the carousel

Answers

Answer:

magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

Explanation:

Given the data in the question;

Speed of carousel N = 24 rpm

From the diagram below, selected path direction defines the Axis of slip.

Hence, The Coriolis is acting along the axis of transmission

Now, we determine the angular speed ω of the carousel.

ω = 2πN / 60

we substitute in the value of N

ω = (2π × 24) / 60

ω = 2.5133 rad/s

Next, we convert the given velocity from mph to ft/s

we know that; 1 mph = 1.4667 ft/s

so

[tex]V_{slip[/tex] = 6 mph = ( 6 × 1.4667 ) = 8.8002 ft/s

Now, we determine the magnitude of the Coriolis acceleration

[tex]a_c[/tex] = 2( [tex]V_{slip[/tex] × ω )

we substitute

[tex]a_c[/tex] = 2( 8.8002 ft/s × 2.5133 rad/s )

[tex]a_c[/tex] = 44.235 ft/s²

Hence, magnitude of the Coriolis acceleration is 44.235 ft/s² and the direction of the acceleration is along the axis of transmission

The first extra-solar planet around a main sequence star (i.e., not a neutron star or white dwarf) was found around the star 51 Pegasi in 1995. The large planet causes a measureable motion of the star around the center of mass of the system. Pegasi 51's orbital motion had a period of 4 days indicating a very large planet very close to the star. A large planet of mass M around a small star of mass 4M. The distance between planet and star is L. Both the planet and star will orbit around the center of mass of the system (marked by the red X).

Required:
Where is the center of mass of the star-planet system?

Answers

Answer:

   [tex]r_{cm}[/tex]= 1/5 L

Explanation:

To find the center of mass of the system let's use

        [tex]r_{cm} = \frac{1}{M}[/tex] ∑ r_i x_i

where m is the total mass of the system

let's apply this expression to our case

       

Let's set the reference frame on the star

         [tex]r_{cm} = \frac{1}{M +4M} ( 4M 0 + M L)[/tex]

         r_{cm} = [tex]\frac{1}{5}[/tex] L

         [tex]r_{cm}[/tex]= 1/5 L

Hey, the breast center is 1/5 of the distance between the star and the planet.

A bucket of water of mass 14.6 kg is suspended by a rope wrapped around a windlass, that is a solid cylinder with diameter 0.320 m with mass 11.1 kg . The cylinder pivots on a frictionless axle through its center. The bucket is released from rest at the top of a well and falls a distance 11.0 m to the water. You can ignore the weight of the rope.

Required:
a. What is the tension in the rope while the bucket is falling?
b. What is the time of fall?
c. While the bucket is falling, what is the force exerted on the cylinder by the axle?

Answers

Answer:

a. T = 39.41 N

b. t = 1.76s

c. 150.78 N

Explanation:

Given:

Mass of bucket of water, Mb = 14.6 kg

Mass of cylinder, Mc = 11.1 kg

Diameter of cylinder, D =  0.320 m, or radius, r = D/2 = 0.16m

Displacement of the bucket from the top, that is the vertical displacement , y = 11.0 m

a. The tension in the rope while the bucket is falling is:

F = mg - T = ma

Where F= The force

m= mass

g= Acceleration due to gravity

T = tension in the rope

a = acceleration

T= m(g - a)

Then, calculating the angular acceleration of the pulley system in relation to its radial acceleration

T= 1/2Ma

Merging the two final equation so as to solve for a

M(g - a) = 1/2Ma

Make a the subject of the formula

Mg - Ma = 1/2Ma

1/2Ma + Ma = Mg

a (1/2 M + M) = Mg

Divide both side by (1/2 M + M)

a = Mg ÷ (1/2 M + M)

Inputing the given value in the formula above

g= 9.8m/s2

a = (14.6 kg) (9.8m/s2) ÷ 1/2 (11.1 kg) + 14.6 kg

a = 7.1007m/s2

Now it is easy to input the value into T= 1/2Ma

T = 1/2 (11.1 kg) (7.1007m/s2) = 39.41 N

B. Time of fall is:

Using one of the equation of motion

s = ut + 1/2 at^2

U = Initial velocity

t = time

a = acceleration

s= distance in this case displacement y

making t the subject of the formula

t = √(2s ÷ a)

u is 0 since the bucket starts from rest

so, t = √((2)(11.0 m) ÷ 7.1007m/s2)

t = 1.76s

c.  the force exerted on the cylinder by the axle = T + Mg

  = 42 N + (11.1 kg) (9.8m/s2)

= 150.78 N

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