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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Related Questions
2) A ray of light in air is approaching the boundary with water at an
angle of 52 degrees. Determine the angle of refraction of the light
ray. (Refractive index of air = 1, water = 1.33)
=
Answers
Answer:
Explanation:
ASSUMING the 52° is the angle of incidence measured from the perpendicular to the surface
n₁sinθ₁ = n₂sinθ₂
1 sin52 = 1.33sinθ₂
θ₂ = arcsin(sin52 / 1.33)
θ₂ = 36°
as measured from the perpendicular to the surface
HELPP
When two forces are in opposite directions, and they are the exact same magnitude, the forces will _______.
a. subtract from each other
b. cancel out
c. go on infinitely
d. eventually reach equilibrium
Answers
Answer:
i think the correct answer is B. cancel out
During a picnic, you and two of your friends decide to have a three-way-tug-of-war, with three ropes in the middle tied into a knot. Roberta pulls to the west with 277 N of force; Michael pulls to the south with 603 N. a) With what force should you pull to keep the knot from moving
Answers
Answer:
Explanation:
I'm not 10% sure but i think that if you were to do:
603 + 277 = 880
1000 - 880 = 120
So to keep it from moving, i think you would need to pull with a force of 120N
The force with which it should be pulled to keep the knot from moving is 663.58 N.
What is Newton's third law ?Newton's third law states that, for every action, there is an equal and opposite reaction.
Here,
Force exerted by Roberta, F₁ = 277 N
Force exerted by Michael, F₂ = 603 N
Since, Roberta is pulling to the west and Michael is pulling towards the south, the angle between the forces applied by them is 90°.
The force needed to keep the knot from moving is the resultant force of these two forces.
Therefore, the resultant force,
Fₙ = √F₁² + F₂²
Fₙ = √(277)² + (603)²
Fₙ = 663.58 N
Hence,
The force with which it should be pulled to keep the knot from moving is 663.58 N.
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Some amount of ideal gas with internal energy U was heated from 100^0C to 200^0C. We can predict that internal energy after heating in terms of U is:
Answers
The internal energy after heating in terms of U is 100U.
The given parameters;
initial temperature of the gas, T₁ = 100 ⁰Cfinal temperature of the gas, T₂ = 200 ⁰CAssuming a constant pressure, the internal energy of the ideal gas is equal to the change in the enthalpy of the ideal gas.
[tex]\Delta H = U \times \Delta T\\\\\Delta H = U (200 - 100)\\\\\Delta H = 100 U[/tex]
Thus, we can conclude that the internal energy after heating in terms of U is 100U.
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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:
I need to choose a theme for my physics assignment My experiment is finding g
Answers
How to find g (acceleration due to gravity)
Solution:We know,
Acceleration due to gravity (g)
[tex] = \frac{GM}{ {R}^{2} } [/tex]
where, G = Gravitational constant
[tex] = 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\ [/tex]
M = Mass of the earth
[tex] = 6 \times {10}^{24} \: kg[/tex]
R = Radius of the earth
[tex] = 6.4 \times {10}^{6} m[/tex]
Putting these values of G, M and R in the above formula, we get
[tex]g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2} [/tex]
So, the value of acceleration due to gravity is
[tex]9.8m/s ^{2} [/tex]
Hope it helps.
Do comment if you have any query.
tính các mô men quán tính chính trung tâm của tiết diện, cho biết a = 10cm
Answers
Answer:
i dont know need points
Explanation:
In an earlier chapter you calculated the stiffness of the interatomic "spring" (chemical bond) between atoms in a block of lead to be 5 N/m. Since in our model each atom is connected to two springs, each half the length of the interatomic bond, the effective "interatomic spring stiffness" for an oscillator is 4*5 N/m = 20 N/m. The mass of one mole of lead is 207 grams (0.207 kilograms).
What is the energy, in joules, of one quantum of energy for an atomic oscillator in a block of lead?
Answers
Answer:
8.01e-22
Explanation:
The energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
Extension produced by one mole of leadThe extension produced by one mole of lead atom is calculated by applying Hooke's law;
F = kx
mg = kx
x = mg/k
x = (0.207 x 9.8) / (20)
x = 0.101 m
Energy stored in the lead blockThe Energy of one quantum of energy for an atomic oscillator in a block of lead is calculated as follows;
E = ¹/₂kx²
E = ¹/₂ (20)(0.101)²
E = 0.102 J
Thus, the energy of one quantum of energy for an atomic oscillator in a block of lead is 0.102 J.
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An applied force accelerates a 4.00 kg block to an initial velocity of 11 m/s across a rough horizontal surface, in the positive x direction. As the block reaches 11 m/s, the applied force is removed. The block then slows to 1.5 m/s at a distance of 4.00 m beyond where the applied force was removed. Determine the magnitude and the direction of the non-conservative force acting on the box as it slides.
Answers
The only force opposing the block's sliding as it slows down is friction with magnitude f . By Newton's second law, the net force in this direction is
∑ F = -f = ma = (4.00 kg) a
Assuming constant acceleration a , the acceleration applied by friction is such that
(1.5 m/s)² - (11 m/s)² = 2a (4.00 m)
Solve for the acceleration :
a = ((1.5 m/s)² - (11 m/s)²) / (8.00 m) ≈ -14.8 m/s²
Then the frictional force exerted a magnitude of
-f = (4.00 kg) (-14.8 m/s²)
f ≈ 59.4 N
and was directed opposite the block's motion.
Consider 5.00 mol of liquid water. (a) What volume is occupied by this amount of water? The molar mass of water is 18.0 g/mol. (b) Imagine the molecules to be, on average, uniformly spaced, with each molecule at the center of a sma11 cube. What is the length of an edge of each sma11 cube if adjacent cubes touch but don't overlap? (c) How does this distance compare with the diameter of a molecule?
Answers
This question involves the concepts of density, volume, and mass.
(a) The volume occupied by this amount of water is "90 m³".
(b) The length of an edge of each small cube is "83 μm".
(c) The length of the edge of each small cube is "equal" to the diameter of a molecule.
(a)
The volume can be found using the following formula:
[tex]V = \frac{nM}{\rho}[/tex]
where,
V = volume occoupied = ?
n = no. of moles = 5 mol
M = molar mass = 18 g/mol
[tex]\rho[/tex] = density of water = 1 g/m³
Therefore,
[tex]V=\frac{(5\ mol)(18\ g/mol)}{1\ g/m^3}\\\\[/tex]
V = 90 m³
(b)
First, we will find the volume of an individual molecule:
[tex]V_i =\frac{V}{nN_A}[/tex]
where,
[tex]N_A[/tex] = Avogadro's number = 6.02 x 10²³ molecules/mol
Therefore,
[tex]V_i=\frac{90\ m^3}{5\ mol(6.02\ x\ 10^{23}\ molecules/mol)}[/tex]
Vi = 3 x 10⁻²³ m³
This volume can be given as a volume of the sphere:
[tex]V_i=\frac{4}{3}\pi r^3\\\\r=\sqrt[3]{\frac{3(3\ x\ 10^{-23}\ m^3)}{4\pi}}[/tex]
r = 4.15 x 10⁻⁷ m
Diameter = d = 2r = 2(4.15 x 10⁻⁷ m)
d = 8.3 x 10⁻⁷ m = 83 μm
since the cubes are adjacent to each other. Therefore, the diameter will be equal to the edge length.
Edge Length = L = d
L = 8.3 x 10⁻⁷ m = 83 μm
(c)
The edge length is equal to the diameter of the molecule.
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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 = 0.35The 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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Question 2 (4 points)
Listen
A 14,000kg freight train, coasting at 8.1m/s, strikes another car, and the two move
forward together at 3.6m/s.
What is the mass of the second car?
Paragraph
B
1
U
A
+
Answers
Answer:
Explanation:
I will ASSUME that the struck car was initially at rest. Would not have to be, but a different mass will result if it was
Conservation of momentum
14000(8.1) + m(0) = (14000 + m)(3.6)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
The mass of the second car is 17500 kg.
What is law of conservation of momentum?The law of conservation of momentum asserts that, unless an external force is applied, the total momentum of two or more bodies operating upon one another in an isolated system remains constant. As a result, momentum cannot be gained or lost.
Newton's third law of motion has a direct impact on the idea of momentum conservation.
mass of the first car = 14000 kg.
Initial speed of first car = 8.1 m/s.
Final speed of the two cars = 3.6 m/s.
From law of conservation of momentum; we can write:
Total initial momentum = total final momentum
14000 kg ×8.1 m/s + m×0 = (14000 + m) kg × (3.6 m/s)
14000(8.1) = 14000(3.6) + m(3.6)
m = 14000(8.1 - 3.6) / 3.6
m = 17500 kg
Hence, the mass of the second car is 17500 kg.
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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
3. What can you infer about the position of the galaxies 100 million years before this telescope photo was taken
Answers
The galaxies are in constant motion so it can be inferred that their position will not be the same as a hundred million years ago.
A galaxy is a term to refer to the set of stars, gas clouds, planets, cosmic dust, dark matter, and energy gravitationally united in a more or less defined structure in the universe.
Galaxies are in constant motion because the elements that make them up are not static at any time; The galaxies rotate around the center of the galaxy. If they were still, the gravitational attraction would cause them to immediately fall towards the center of the galaxy: it is the same that would happen to the Earth and the other planets if they stopped rotating around the Sun, they would fall towards the Sun.
According to the above, it can be inferred that the galaxies after a while have moved from the place where they were last seen.
Note: This question is incomplete because there is some missing information. However, I can answer based on my general prior knowledge.
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49. A particle starts from rest at time t=0 and movies along the x axis. if the net force on is proportional to t its kinetic energy is proportional to?
Answers
Answer:
F net ∞ [tex]\frac{1}{\sqrt{t} }[/tex]
Explanation:
In pic
_________________
(hopet his helps can I pls have brainlist (crown)☺️)
A country is deciding what to do about pollution glven off by power plants.
Which of the following is an example of how science can be used to make
this decision?
O A. Scientists own most of the power plants in the United States.
O B. Scientists can measure how much pollution is given off
O C. Scientists know what kind of power plants people want.
OD. Scientists are allowed to pass laws about pollution
Answers
Answer:
option B is the correct answer
Explanation:
please follow me and Mark me brainliest please
Explain different layers of atmosphere and the pressure in each layer. Draw diagram
Answers
Answer:
Our atmosphere has five different layers. They are:
1. Troposphere: This is the most important layer of the atmosphere with an average height of 13 km from the earth. It is in this layer that we find the air that we breathe. Almost all the weather phenomena such as rainfall, fog and hailstorm occur here.
2. Stratosphere: This layer extends up to a height of 50 km. It presents the most ideal condition for flying airplanes. It contains a layer of ozone gas which protects us from the harmful effect of the sun rays.
3. Mesosphere: This layer extends up to a height of 80 km. Meteorites bum up in this layer on entering from the space.
4. Thermosphere: In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It extends between 80-400 km. This layer helps in radio transmission. Radio waves transmitted from the earth the reflected back to the earth by this layer.
5. Exosphere: It is the uppermost layer where there is very thin air. Light gases such as helium and hydrogen float into space from here.
calculate the change in gravitational potential energy when a 10kg object falls from 6m on earth
Answers
Answer:
600
Explanation:
GPE=mgh (mass*gravitational force*height)
m=10 kg
g=10
h=6m
10*10*6= 600 Joules
We have seen in earlier readings how to determine the speed of a wave on a string. What will happen to the wavelength of a sinusoidal wave on a string if the tension in the string is increased (assuming we keep the frequency of the wave the same)
Answers
This question involves the concepts of tension in a string and the wavelength of the wave in a string.
The wavelength of a sinusoidal wave will "increase by square power" on a string if the tension in the string is increased when the frequency is kept constant.
The speed of a wave on a string is given by the following formula:
[tex]v=\sqrt{\frac{T}{\mu}}[/tex]
where,
v = speed of wave = fλ
f = frequency of the wave
λ = wavelength of the wave
T = tension force
μ = linear mass density of the string
Therefore,
[tex]f\lambda=\sqrt{\frac{T}{\mu}}\\\\T = f^2\lambda^2\mu[/tex]
It is given that the frequency is kept constant. The linear mass density is also constant for a string. Therefore,
[tex]T=(constant)\lambda^2\\T\ \alpha\ \lambda^2[/tex]
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Another box of samples is hoisted up by the same rope. If the rope is shaken with the same frequency as before, and the wavelength is found to be 7.9 m , what is the mass of this box of samples
Answers
The motion of the rope which is perpendicular to the direction of the
propagation of the wave is a transverse wave motion.
The mass of the box is approximately 9.93 kgReasons:
The given function for the wave speed is presented as follows;
[tex]\displaystyle v = \sqrt{\frac{T}{\mu} } \\[/tex]
Where;
[tex]\displaystyle \mu = \frac{Mass \ of \ rope }{Length \ of \ rope}[/tex]
Taking the mass of the rope as, m = 2.00 kg
The length of the rope, L = 80.0 m
The mass hanging on the rope, M = 20.0 kg
We have;
T = 20.0 kg × 9.81 m/s² = 196.2 N
[tex]\displaystyle \mu = \frac{2.0 }{80.0} = \frac{2.0 }{80.0} = 0.025[/tex]
Therefore;
Taking the wavelength as, λ = 7.9 m, and the frequency as 20 Hz, we have;
v = f × λ
Therefore;
v = 7.9 Hz × 7.9 m = 62.41 m/s
Which gives;
[tex]\displaystyle 62.41 = \sqrt{\frac{T}{0.025} }[/tex]
T = 62.41² × 0.025 = 97.3752025
[tex]\displaystyle Mass, \ m = \mathbf{\frac{T}{g}}[/tex]
Where;
g = The acceleration due to gravity which is approximately 9.81 m/s²
[tex]\displaystyle Mass, \ m = \frac{97.3752025}{9.81} \approx 9.93[/tex]
Therefore;
The mass of the box, m ≈ 9.93 kg
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The parameters obtained from a similar question online are;
[tex]\displaystyle The \ equation \ applied, \, v = \sqrt{\frac{T}{\mu} } \\[/tex]
Length of the rope, L = 80.0 m
Mass of the rope, m = 2.0 kg
Frequency of a point on the rope, f = 20 Hz
the wheel of bicycle has a radius of 25cm. what will be the magnitudes of the angular displacement in radian and revolution respectively, when the wheel has rolled a distance of 350cm on straight level road?
Answers
Answer:
Explanation:
350 cm / 25 cm = 14 radians
14 rad / 2π rad/rev = 2.23 revolutions
I need the answer to this question
Answers
Answer:
I don't see anything so you should repost
What is the volume of 150g of a substance that has a density of 150g of a substance that has a density of 11.3g/cm3
Answers
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
How does climate change?
Answers
Answer:
Con el cielo del agua
Explanation:
espero que te ayude
hope it helps
An astronaut is doing thee simple pendulum experiment on a different planet to measure the acceleration due to gravity. If the length of the pendulum is 45 cm and the period of oscillations is equal to 1.428 s, what is the acceleration due to the gravity of the planet
Answers
The acceleration due to gravity on the given planet is 8.71 m/s².
The given parameters;
Length of the pendulum, L = 45 cm Period of the oscillation, T = 1.428 sThe acceleration due to gravity on the planet is calculated by applying following formula as follows;
[tex]T = 2\pi \sqrt{\frac{l}{g} } \\\\\frac{T}{2 \pi } = \sqrt{\frac{l}{g}} \\\\\frac{T^2}{4\pi ^2} = \frac{l}{g} \\\\g = \frac{4\pi^2 l}{T^2} \\\\g = \frac{4 \times (\pi)^2 \times 0.45}{1.428^2} \\\\g = 8.71 \ m/s^2[/tex]
Thus, the acceleration due to gravity on the given planet is 8.71 m/s².
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A car of mass 5000 kg was initially moving at 100 km/h and stops at a distance of 55 m. Find the
magnitude of the net force (in N) acting to stop the car.
Answers
Answer:
|F| = 35 kN
Explanation:
a = F/m
100 km/hr(1000 m/km / 3600 s/hr) = 27.8 m/s
v² = u² + 2as
a = (v² - u²) / 2s
F/m = (v² - u²) / 2s
F = m(v² - u²) / 2s
F = 5000(0² - 27.8²) / 2(55)
F = - 35,072.9517...
HOW DO U FEEL WHEN U PLAY OR WATCH BADMINTON?
Answers
Answer:
I feel exited and happy I enjoy it with my friend
A 5 kg bowling ball travelling at 2 m/s hits a motionless 10 kg bowling ball. If the smaller ball bounces back at a speed of -1 m/s, what will be the speed of larger ball after the collision? Hint: Use the conservation of momentum equation to solve this problem.
Answers
Answer:
1.5 m/s
Explanation:
Conservation of momentum means the momentum of the system before the collision is the same as after.
The before, after momentum of each ball is ...
5 kg ball: (5 kg)(2 m/s), (5 kg)(-1 m/s)
10 kg ball: (10 kg)(0 m/s), (10 kg)(v)
The sum of the "before" products is the same as the sum of the "after" products:
(5 kg)(2 m/s) +0 = (5 kg)(-1 m/s) +(10 kg)v
(10 +5) kg·m/s = (10 kg)·v . . . . . add (5 kg)(1 m/s) to both sides
v = (15 kg·m/s)/(10 kg) = 1.5 m/s
The speed of the larger ball will be 1.5 m/s. Its direction of motion will be the opposite of that of the 5 kg ball after the collision.
A camera takes a properly exposed photo at f/5.6 and 1/250 s. What shutter speed should be used if the lens is changed to f/4.0?
a.1/65 s
b.1/125 s
c.1/250 s
d.1/500 s
e.1/1000 s
Answers
The shutter speed that should be used if the lens is changed to f/4.0 is; Choice D: 1/500 s.
The relationship between the shutter speed and the focal ratio is an inverse relationship.
Ratio of Areas = 5.6²/4²
A1/A2 = 1.96 = 2Since; T1A1 = T2A2
T2 = T1(A1/A2)T2 = 250 × 2 = 500 secondsIn essence; when the focal ratio is reduced as in this case; from f/5.6 to f/4.0; the shutter speed is increased.
Ultimately, the shutter speed of the camera in discuss increases to; 1/500 s.
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can someone help me answer this question? A student connected an ammeter as shown in this picture. Did the student connect the ammeter correctly? Explain
Answers
Answer:
No - It is connected in parallel instead of series
Explanation:
The answer is No, the reason behind the answer that the ammeter is connected in parallel which is wrong it must be connected in series.
What is Ammeter?An ammeter is a device for detecting electric current in amperes, either directly (DC) or alternating (AC). Due to the fact that a shunt running parallel to the meter carries the majority of the electricity at high current values, the digital multimeter can measure a wide range of current values. A circle with the capital A in it serves as the icon for an ammeter for circuit diagrams.
The moving coil of the electrodynamics ammeter rotates in the field created by the fixed coil. It measures alternating and direct current with 0.1 to 0.25 percentage points (by converting Ac power To DC power using a rectifier).
The ammeter is connected in series and, on the other hand, the voltmeter is connected in parallel.
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