A 0.015 kg marble sliding to the right at 0.225 m/s on a frictionless surface makes an elastic head-on collision with a 0.015 kg marble moving to the left at 0.180 m/s. After the collision, the first marble moves to the left at 0.180 m/s. Find the velocity of the second marble after the collision.

Answers

Answer 1

Explanation:

[tex]if \: the \: masses \: of \: the \:two \: marbles\: equal \\ then \: each \: velocity \: sharing \: with \: other \: velocity[/tex]

A 0.015 Kg Marble Sliding To The Right At 0.225 M/s On A Frictionless Surface Makes An Elastic Head-on

Related Questions

About the pulmonary surfactant it is true that: I. It reduces the surface tension of water. II. It is important for generating a pressure gradient between small and large alveoli.
C, I = False, Il = True
B, I = True, Il False
D, I = False, Il = False
A, I = True, II = True​

Answers

Answer:

a

Explanation:

Check all true statements about nonmetals. Group of answer choices
All nonmetals are gaseous at room temperature.
Nonmetals have low melting and boiling points. ✅
Nonmetals are brittle and have a relatively low density. ✅
Nonmetals are good at thermal and electrical conduction.

Using the colored periodic table below, match the group or area to its color.
alkali metals red alkaline ✅
earth metals orange ✅
transition metals white ✅
halogens purple ✅
Nobel gasses teal✅

How many of each subatomic particle are in a neutral atom of Potassium-39? What is its mass number? Potassium is in group 1 and has the symbol "K".
protons 19 ✅
neutrons 20 ✅
electrons 19 ✅
mass number 39✅

Match the ion with its charge.
11 protons, 12 neutrons, 10 electrons = +1 ✅
Groups 17, period 2 = -1 ✅
31 protons, 39 neutrons, 28 electrons = +3✅
Group 2, period 5 = +2

Which best describes a metal such as Silver (Ag)?
Answer: Lustrous, malleable, and forms cations.

Determine if the "quoted" word(s) makes each statement True or False.
Elements that share the most characteristics are found on the periodic table in the same "horizontal period". False ✅
If matter has a higher temperature, that means there will be "more" molecular motion. True✅
If you put a balloon in a freezer, its volume would "increase". False ✅
If you put the balloon into a chamber where there is half the pressure, its volume will "double". True ✅
If you cool down a propane tank, there will be "less" pressure in it. True ✅
The average atomic mass of Aluminum (Al) is "16.00 amu".
False✅
Oxygen's atomic number is "8" therefore it has "8" protons in its nucleus. True✅
An "electron" has the same mass as a neutron. False✅
The nucleus contains "protons and neutrons", virtually all the mass of an atom.✅

Answers

Metals are found towards the left hand side of the periodic table while Nonmetals are found towards the right hand side of the periodic table.

Non metals are found towards the right hand side of the periodic table. They are not all gaseous at room temperature some nonmetals such as iodine are solid at room temperature.

Nonmetals usually have low melting and boiling points, are brittle, have relatively low density and are not good at thermal and electrical conduction.

Potassium - 39 contains 19 protons, 20 neutrons and 19 electrons. Recall that the number of protons and electrons are equal in a neutral atom.

The following is an accurate matching of the ions;

11 protons, 12 neutrons, 10 electrons = +1 - Na^+

31 protons, 39 neutrons, 28 electrons = +3 - Ga^2+

Group 2, period 5 = +2 - Sr^2+

Silver is a lustrous metal.

Elements that share the most characteristics are found on the periodic table in the same "horizontal period". FalseIf matter has a higher temperature, that means there will be "more" molecular motion. TrueIf you cool down a propane tank, there will be "less" pressure in it. TrueIf you put a balloon in a freezer, its volume would "increase". FalseIf you put the balloon into a chamber where there is half the pressure, its volume will "double". TrueThe average atomic mass of Aluminum (Al) is "16.00 amu".  FalseOxygen's atomic number is "8" therefore it has "8" protons in its nucleus. TrueAn "electron" has the same mass as a neutron. FalseThe nucleus contains "protons and neutrons", virtually all the mass of an atom - True

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Answer:

If your answers are the green check marks all of them are correct

Explanation: I took the test

Sam wants to get a refrigerator into a moving truck. He chooses to use a ramp to accomplish the task that is 25 meters long. If the fridge weighs 250 N, what force will Sam need to use on the ramp to get the fridge into the 3 m high truck bed?​

Answers

Answer:

Explanation:

Ignoring friction and assuming one ramp end rests on level ground.

gravity acceleration acting parallel to the ramp is gsinθ.

F = mgsinθ

mg = 250 N,  sinθ = 3/25

F = 250(3/25)

F = 30 N

A 25 meter long ramp strong enough to hold a person plus a 250 N refrigerator would weigh much more than 250 N.

Which is NOT a function of the
cell wall?
A. Protects cell from bursting
B. Provides support for plant cells
C. Protects cell from harsh internal
environments
D. Absorbs sunlight to give energy to the cell

Answers

Answer:

The answer is D

Explanation:

The chloroplast absorbs sunlight for energy not the cell wall

D, the sunlight is absorbed in the chloroplasts

Kinesha and her friend were watching a solar eclipse. Kinesha explains to her friend that a solar eclipse means that Earth is located between the Sun and the Moon. Her friend tells Kinesha that her explanation is incorrect. Why?

Answers

this is where the sun and moon line up where you asleep only a tiny bit of the sun it's pretty cool to see

Explanation:

a solar eclipse means when the moon goes infront of the sun and the earth turns dark


E5. A ball is thrown downward with an initial velocity of 12 m/s.
Using the approximate value of g 10 m/s2, what is the
velocity of the ball 1.0 seconds after it is released?

Answers

Hi there!

We know the following kinematic equation:

vf = vi + at

Where:

vf = final velocity

vi = initial velocity

a = acceleration

t = time

In this instance, the ball is experiencing a constant acceleration of that of gravity, thus:

vf = 12 + 10(1) = 22 m/s (if downward is considered positive in this instance)

trình bày những hiểu biết cơ bản về năng lượng

Answers

Answer:

I don't understand your language

A magnet is located above circular current. What is the direction of the magnetic force on the magnet

Answers

The magnet is attracted to the ring since the north pole of the current loop is above the ring and the south pole is below the ring.

A child at the top of a slide de has a gravitational store of 1800J. What is the child's maximum kinetic store as he slides down? Explain why

Answers

Hi there!

We know that:

Initial Total Mechanical Energy = Final Total Mechanical Energy

(Ei = Ef)

In this instance:

Ei = Gravitational Potential Energy

Ef = Kinetic Energy

In the absence of friction, ALL of the initial potential energy will be changed into kinetic energy at the bottom of the slide. Thus, the maximum kinetic energy of the child will be 1800 J.

Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer​

Answers

Answer:

thx for the points

Explanation:

no need brainliest

A power plant running at 31 % efficiency generates 270 MW of electric power. Part A At what rate (in MW) is heat energy exhausted to the river that cools the plant

Answers

The rate of heat energy exhausted to the river is 600.96 MW.

What is efficiency?

The ratio of usable output to total input can be used to objectively measure efficiency. The efficiency of the device is defined as the ratio of energy converted to a useable form to the original amount of energy supplied.

Given parameters:

Efficiency of the power plant; η = 31 %

Output  electric power; O = 270 MW.

We know that, Efficiency of the power plant;

η  = (Output  electric power/ input power)× 100%

⇒ input power = (Output  electric power × 100)/η

⇒ input power = (270 × 100)/31 MW

= 870.96 MW.

So, the rate of heat energy exhausted to the river that cools the plant =  Input power- output power

= (870.96 - 270) MW

= 600.96 MW.

Hence, heat energy exhausted to the river that cools the plant  is 600.96 MW.

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can you classify matter based on chemical properties

Answers

Answer:

Explanation:

Matter can be broken down into two categories: pure substances and mixtures. Pure substances are further broken down into elements and compounds. Mixtures are physically combined structures that can be separated into their original components. A chemical substance is composed of one type of atom or molecule.

Una turbina de vapor
recibe vapor con un flujo másico de 30 kg/s a 6205 kPa, 811 K, con una velocidad a la
entrara de 10 m/s. El vapor a la entrada tiene una energía interna específica de 3150.3
kJ/kg y un volumen específico de 0.05789 m3
/kg. El vapor sale de la turbina a 9.859 kPa,
318.8 K. El vapor sale a 200 m/s con una energía interna específica de 2211.8 kJ/kg y
un volumen específico de 13.36 m3
/kg. Encuentre la potencia producida por la turbina
si ésta pierde calor a una tasa de 30 kW.

Answers

Este problema está describiendo una turbina de vapor a la que entra vapor a 30 kg/s, 6.205 kPa y 811 K con una velocidad de 10 m/s y sale a 9.859 kPa, 318.8 K y con una velocidad de 200 m/s. Adicionalmente, tanto el volumen específico como la energía interna son dados para ambas corrientes.

Con lo anterior, resulta posible escribir un balance de energía para esta turbina, despreciando todo efecto por energía potencial ya que no hay diferencia significativa entre la altura de la entrada (1) y la salida (2), pues están practicamente al mismo nivel:

[tex]mh_1+\frac{1}{2} mv^2_1=mh_2+\frac{1}{2} mv^2_2+Q_2+W_2[/tex]

Aquí vemos que la incógnita es [tex]W_2[/tex] como la potencia que produce la turbina. Ahora, el primer cáculo a realizar es el de las entalpías de las corrientes de entrada y salida, dada la energía interna, presión y volumen específico:

[tex]h_1=3150.3\frac{kJ}{kg}+6205kPa*0.05789\frac{m^3}{kg} =3509.51\frac{kJ}{kg}\\\\h_2=2211.8\frac{kJ}{kg}+9.859kPa*13.36\frac{m^3}{kg} =2342.72\frac{kJ}{kg}[/tex]

Ahora, podemos reacomodar el balance de energía con el fin de resolver [tex]W_2[/tex]:

[tex]W_2=m(h_1-h_2)+\frac{1}{2} m(v^2_1-v^2_2)-Q_2[/tex]

Finalmente, reemplazamos los valores para obtener:

[tex]W_2=10\frac{kg}{s} (3509.51-2342.72)\frac{kJ}{kg} +\frac{1}{2} *10\frac{kg}{s} [(10\frac{m}{s}) ^2-(200\frac{m}{s} )^2]*\frac{1kJ}{1000J} -30\frac{kJ}{s}\\\\W_2=11438.4 kJ/s=11438.4kW[/tex]

Es de precisar que la energía cinética como 1/2 m*v² resulta en Joules, por lo que hay que convertir a kilojoules para tener unidades consistentes de kilowatts al final.

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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the universe be in that case?

Answers

Given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Given the data in the question;

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

Age of the universe; [tex]t = \ ?[/tex]

We know that, the reciprocal of the Hubble's constant ( [tex]H_0[/tex] ) gives an estimate of the age of the universe ( [tex]t[/tex] ). It is expressed as:

[tex]Age\ of\ Universe; t = \frac{1}{H_0}[/tex]

Now,

Hubble's constant; [tex]H_0 = 51km/s/Mly[/tex]

We know that;

[tex]1\ light\ years = 9.46*10^{15}m[/tex]

so

[tex]1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m[/tex]

Therefore;

[tex]H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly} \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 = 5.39 *10^{-18}s^{-1}\\[/tex]

Now, we input this Hubble's constant value into our equation;

[tex]Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t = \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years[/tex]

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

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a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.the applied force moves through a distance of 25cm.calculate the maximum mass of a load that can be lifted by the jack and the distance through which the load is lifted.(take g=9.81ms^-2)​

Answers

The maximum mass of a load that can be lifted by the jack and the distance covered are:

m = 160.2 Kg

h = 25 cm

Given that a certain hydraulic jack is used for lifting load. a force of 250N is applied on a small piston with a diameter of 80cm while the diameter of the larger piston is 2.0m.

The parameters given are

[tex]F_{1}[/tex] = 250

[tex]A_{1}[/tex] = Area of the small piston = π[tex]r^{2}[/tex]

[tex]A_{1}[/tex] = 22/7 x [tex]0.4^{2}[/tex]

[tex]A_{1}[/tex] = 0.5 [tex]m^{2}[/tex]

[tex]F_{2}[/tex] = ?

[tex]A_{2}[/tex] = Area of the large piston = π[tex]r^{2}[/tex]

[tex]A_{2}[/tex] = π x 1

[tex]A_{2}[/tex] = 3.14 [tex]m^{2}[/tex]

To calculate the force on the large piston, we will use the below formula

[tex]F_{1}[/tex]/ [tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]

Substitute all the parameters into the equation

250/0.5 =  [tex]F_{2}[/tex]/3.14

[tex]F_{2}[/tex] = 1570 N

To calculate the maximum mass of a load that can be lifted by the jack, let us apply Newton second law

F = mg

1570 = 9.8m

m = 1570/9.8

m = 160.2 Kg

.(take g=9.81ms^-2)​

If the applied force moves through a distance of 25cm, the distance through which the load is lifted will be

[tex]F_{1}[/tex]/ 0.25[tex]A_{1}[/tex] = [tex]F_{2}[/tex] / [tex]A_{2}[/tex]h

250/0.125 = 1570/3.14h

make h the subject of the formula

6280h = 1570

h = 1570/6280

h = 0.25 m

Therefore, the distance through which the load is lifted is 25 cm

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 A wooden box with a mass of 10.0 kg rest on a ramp that is incline at an angle of 25° to the horizontal. A rope attached to the box runs parallel to the ramp and then passes over a frictionless bully. A bucket with a mass of M hangs at the end of the rope. The coefficient of static friction between the ramp in the box is 0.50. The coefficient of Connecticut friction between the ramp in the box is 0.35.

Suppose the box remains at rest relative to the ramp. What is the maximum magnitude of the friction force exerted on the box by the ramp?

Answers

The maximum magnitude of the friction force exerted on the box by the ramp is 44.41 N.

The given parameters;

Mass of the box, m = 10 kgInclination of the ramp, θ = 25⁰Coefficient of static friction, μ = 0.5 Coefficient of kinetic friction, μk = 0.35

The normal force on the wooden box is calculated as follows;

[tex]F_n = mg \times cos(\theta)\\\\F_n = 10 \times 9.8 \times cos(25)\\\\F_n = 88.8 2 \ N[/tex]

The maximum magnitude of the friction force exerted on the box by the ramp is calculated as follows;

[tex]F_f = \mu \times F_n\\\\F_f = 0.5 \times 88.82 \\\\F_f = 44.41 \ N[/tex]

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Which quantity or quantities is/are increasing for the object represented by line B?

Answers

Answer:

C. Velocity and Position

Explanation:

The quantities that are increasing for the object represented by line B are velocity and position. The correct option is b.

What is velocity?

The direction of a body or object's movement is defined by its velocity. In its basic form, speed is a scalar quantity. In essence, velocity is a vector quantity. It is the speed at which distance changes. It is the displacement change rate.

Velocity is the pace and direction of an object's movement, whereas speed is the time rate at which an object is traveling along a path. In other words, velocity is a vector, whereas speed is a scalar value.

The graph is given which represents the velocity and time with terms A, B, and C. As opposed to the position-time graph, which describes an object's motion over time, the velocity-time graph reveals an object's speed.

Therefore, the correct option is b. velocity and position.

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The question is incomplete. Your most probably complete question is given below:

Velocity and acceleration

velocity and position

velocity only

velocity, position, and acceleration

Be able to list the three compounds that are formed as products of highly exothermic
reactions such as detonating nitrogen-based explosives?

Answers

Answer:

Ammonium perchlorate NH4ClO4

Ammonium Nitrate

Calcium Cyanamide

When detonated, the reaction products are all gases, such as water vapor, nitrogen gas, and oxides of nitrogen.

Hopefully this helps :)

Ignoring any effects of dc resistance, what is the total reactance of a 250 mH coil in series with a 4.7 microfarad capacitor at a signal frequency of 60 Hz

Answers

The total reactance of the inductor and the capacitor is 470.1 ohms.

The given parameters;

inductance of the coil, L = 250 mHcapacitance, C = 4.7 μfrequency of the circuit, f = 60 Hz

The inductive reactance of the coil is calculated as follows;

[tex]X_l = \omega L\\\\X_l = 2\pi f L\\\\X_l = 2\pi \times 60 \times 250 \times 10^{-3}\\\\X_l = 94.26 \ ohms[/tex]

The capacitive reactance of the capacitor is calculated as follows;

[tex]X_c = \frac{1}{\omega C} \\\\X_c = \frac{1}{2\pi f C} \\\\X_c = \frac{1}{2\pi \times 60 \times 4.7 \times 10^{-6}} \\\\X_c = 564.31 \ ohms[/tex]

The impedance of the circuit is calculated as follows;

[tex]Z = \sqrt{(X_c - X_l)^2} \\\\Z = X_c - X_l\\\\Z = 564.31 - 94.26\\\\Z = 470.1 \ ohms[/tex]

Thus, the total reactance of the inductor and the capacitor is 470.1 ohms.

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Identify the direction of the net force acting on the object. Explain your reasoning

Answers

Answer:

See Below

Explanation:

I am not sure what you exact question is to know what direction it is but here is how you do it.

The direction of the net force is the direction of the largest force.

for example if you were to push a box forward with 100 newton's of force and someone pushed at the same time 50 newton's backwards on the box, the box would move forwards because the was a greater force on the box in a forward direction. hope this helps

Một khối khí hidro bị nén đến thể tích bằng 1/2 lúc đầu khi nhiệt độ không đổi. Nếu vận tốc trung bình của phân tử hidro lúc đầu là V thì vận tốc trung bình sau khi nén là bao nhiêu ?

Answers

answer: what language is this???

5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.

Answers

Answer:

1.904

Explanation:

F= ma

8 = 4.2 a

a = 8/4.2

a = 1.904

what is a non-economic benefit to international trade

Answers

Answer:

non economic benefits to international trade means not being too much profit not selling too much things /items in international trade

An object with mass 1.2 kg is moving at a constant speed in a circle with radius 2.5 m. If the
object makes exactly 12 revolutions in a 1 minute, what is the acceleration of the object?

Answers

Answer:

3.9m/s² is the acceleration

The acceleration of the object is 3.9m/s².

How do you find the acceleration of an object?

Acceleration (a) is the change in velocity (Δv) over the change in time (Δt), represented by the equation a = Δv/Δt. This allows you to measure how fast velocity changes in meters per second squared (m/s^2).

What is an example of an object that has acceleration?

When you fall off a bridge. The car turning at the corner is an example of acceleration because the direction is changing. The quicker the turns, the greater the acceleration.

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If you were told an atom was an ion, you would know the atom must have a?
A neutral charge
B charge
C negative charge
D positive charge

Answers

Answer:

its b it must be charge

Explanation:

What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?

Answers

Answer:

the different about the molecules we needed to make protein 3 compared to the molecules we used to make protein 2 is that if we used 2 molecules than it will be smaller than using protein 3.

A motorcycle, which has an initial linear speed of 9.7 m/s, decelerates to a speed of 4.0 m/s in 4.4 s. Each wheel has a radius of 0.61 m and is rotating in a counterclockwise (positive) directions.
What is (a) the constant angular acceleration (in rad/s2) and (b) the angular displacement (in rad) of each wheel?

Answers

Hi there!

We can begin by solving for the linear acceleration as we are given sufficient values to do so.

We can use the following equation:

vf = vi + at

Plug in given values:

4 = 9.7 + 4.4a

Solve for a:

a = -1.295 m/s²

We can use the following equation to convert from linear to angular acceleration:

a = αr

a/r = α

Thus:

-1.295/0.61 = -2.124 rad/sec² ⇒ 2.124 rad/sec² since counterclockwise is positive.

Now, we can find the angular displacement using the following:

θ = ωit + 1/2αt²

We must convert the initial velocity of the tire (9.7 m/s) to angular velocity:

v = ωr

v/r = ω

9.7/0.61 = 15.9 rad/sec

Plug into the equation:

θ = 15.9(4.4) + 1/2(2.124)(4.4²) = 20.56 rad

An electron is held up against the force of gravity by the attraction of a fixed proton some distance above it. How far above the electron is the proton

Answers

5.08 m

Explanation:

The weight of the electron is being counteracted by the attractive electrostatic force exerted by the proton above it. We can write the force equation as follows:

[tex]m_eg = k_e\dfrac{Q_pQ_e}{r^2}[/tex]

where the Q's are the charges of the proton and electron, r is the distance between the particles, g is the acceleration due to gravity, [tex]m_e[/tex] is the mass of the electrons and [tex]k_e[/tex] is the Coulomb constant. So solving for r, we get

[tex]r^2 = k_e\dfrac{Q_pQ_e}{m_eg}[/tex]

Taking the square root of r^2, we then get the distance as

[tex]r = \sqrt{k_e\dfrac{Q_pQ_e}{m_eg}}[/tex]

The values are given as follows:

[tex]m_e = 9.11×10^{-31}\:\text{kg}[/tex]

[tex]g = 9.8\:\text{m/s}^2[/tex]

[tex]Q_p = Q_e = 1.60×10^{-19}\:\text{C}[/tex]

[tex]k_e = 8.99×10^9\:\text{N-m}^2\text{/C}^2[/tex]

Putting in all of these values in our equation for r,

[tex]r = \sqrt{\dfrac{(8.99×10^9\:\text{N-m}^2\text{/C}^2)(1.60×10^{-19}\:\text{C})^2}{(9.11×10^{-31}\:\text{kg})(9.8\:\text{m/s}^2)}}[/tex]

[tex]\:\:\:\:\:= 5.08\:\text{m}[/tex]

What is friction ??? ​

Answers

Answer:

Frictional force is produced when two bodies are rubbed against each other. It is the force that oppose the motion and therefore it stops or slow down a moving body.It depends upon the roughness or smoothness of the surface of the body in contact.Rough surface have more friction that the smooth surface. Similarly, the heavier body produces more friction than a lighter body. Frictional force acts in the opposite direction of the motion of the body.

A very bouncy mushroom can be modeled as mass of 30g(the mushroom cap) on top of a spring(the mushroom stalk) when spring constant 20 N/m. A bird of mass 50g lands on the mushroom gently so that its velocity is zero when it lands. Questions are in the image.

Answers

The motion of the bouncy mushroom can be described as a simple

harmonic motion, SHM.

a) The equilibrium height of the mushroom is 0.024525 m below its initial heightb) The frequency of resulting oscillation is 0.5 Hzc) The maximum compression of the mushroom 0.03924 md) The equation that describes the oscillation of the mushroom as a function of time is; [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]

Reasons:

The given parameters are;

Mass of the mushroom cap, m = 30 g = 0.03 kg

Mass of the bird = 50 g = 0.05 kg

The spring constant, K = 20 N/m

a) The equilibrium height of the mass spring system, is given as follows;

F = -K·x

[tex]x = \dfrac{F}{K}[/tex]

The applied force, F = The weight of the bird

∴ F = (0.05 kg) × 9.81 m/s² = 0.4905 N

[tex]x = \dfrac{0.05 \, kg \times 9.81 \ m/s^2}{20 \, N/m} = 0.024525 \, m[/tex]

The equilibrium height of the mushroom is 0.024525 m below its initial height.

b) The frequency of oscillation of a spring, ω, is given as follows;

[tex]\omega = \sqrt{\dfrac{K}{m} }[/tex]

Therefore;

[tex]\omega = \sqrt{\dfrac{20 \, N/m }{80 \, kg} } = \dfrac{1}{2} \, Hz[/tex]

The frequency of resulting oscillation, ω = [tex]\dfrac{1}{2} \, Hz[/tex] = 0.5 Hz

c) The applied force, F = The weight of the bird and the mushroom cap

F = (0.03 kg + 0.05 kg) × 9.81 m/s² = 0.7848 N

[tex]x = \dfrac{0.7848 \, N}{20 \, N/m } = 0.03924 \, m[/tex]

The maximum compression of the mushroom = 0.03924 m

d) The motion of the mushroom is a Simple Harmonic Motion, SHM.

The equation of a SHM as a function of time is; x(t) = A·cos(ω·t + Φ)

For the mushroom, we have;

The amplitude, A = 0.03924 m - 0.024525 m = 0.014715 m

Ф = The phase angle

When t = 0, cos(ω × 0 + Φ) = 1

cos(Φ) = 1

Ф = arcos(1) = 0

The equation is therefore;

x(t) = 0.014715·cos(0.5·t)

Equation of the oscillation of the mushroom is;  [tex]\underline{x(t) = 0.014715 \cdot cos(0.5\cdot t)}[/tex]

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