At the equilibrium position of the spring, the mass has kinetic energy
1/2 (1.00 kg) (0.0400 m/s)² = 0.000800 J = 0.800 mJ = 800 µJ
and when the spring is maximally compressed, this kinetic energy is converted into potential energy.
The maximum spring potential energy of the simple harmonic motion is approximately 7.48 millijoules.
To find the maximum spring potential energy (Us) in simple harmonic motion, we need to determine the maximum displacement (A) of the mass from its equilibrium position.
In simple harmonic motion, the maximum displacement is equal to the amplitude of the oscillation.
The formula for maximum displacement (A) is given by:
A = v0 / ω
where v0 is the initial velocity (4.00 cm/s) and ω is the angular frequency of the oscillation.
The angular frequency (ω) is given by:
ω =[tex]\sqrt{(k / m)[/tex]
where k is the spring constant (9.00 N/m) and m is the mass (1.00 kg).
Substitute the values:
ω = [tex]\sqrt{(9.00 N/m / 1.00 kg)} = \sqrt{9.00 rad/s[/tex] ≈ 3.00 rad/s
Now, calculate the maximum displacement:
A = 4.00 cm/s / 3.00 rad/s ≈ 1.33 cm
Next, calculate the maximum spring potential energy:
Us =[tex](1/2) * k * A^2[/tex]
Us = (1/2) * 9.00 N/m * [tex](1.33 cm)^2[/tex]≈ 7.48 mJ
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--The compleete Question is, A 1.00 kg mass is connected to a spring with a spring constant of 9.00 N/m. If the initial velocity is 4.00 cm/s and the initial displacement is 0.00 cm, then what is the maximum spring potential energy of the simple harmonic motion--
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg. If this force serves as the action force, what is the reaction force in the action–reaction pair?
Answer:
Equal reaction from the pair in every action there's an equal and opposite reaction
The item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
What is gravitational force?All mass-bearing objects are attracted by gravitational force. Because it consistently attempts to bring masses together rather than push them apart, the gravitational force is referred to as attractive.
As we know, the gravitational force is given by:
[tex]\rm F = \dfrac{Gm_1m_2}{r^2}[/tex]
Where G is the gravitational constant.
m1 and m2 are masses.
r is the distance between the masses.
It is given that:
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
As we know,
An object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg and the item will keep moving at a consistent speed.
Thus, the item will keep moving at a consistent speed if the object is at rest on a tabletop. Earth pulls downward on this object with a force equal in magnitude to mg.
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Please help
A man stands on a freely rotating platform with his arms extended, his rotational frequency is 0.25rev/s. But when he draws them in, his frequency is 0.80revs/s. Find the ratio of his moment of inertia in the first case to that in the second.
Answer:
sorry for you
The ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
The relationship between the rotational frequency [tex](\(\omega\))[/tex] and moment of inertia (I) is given by the equation:
[tex]\[I_1\omega_1 = I_2\omega_2\][/tex]
where [tex]\(I_1\)[/tex]and [tex]\(I_2\)[/tex] are the moments of inertia in the two cases, and [tex]\(\omega_1\) and \(\omega_2\)[/tex] are the corresponding rotational frequencies.
Let's denote the moment of inertia in the first case (arms extended) as [tex]\(I_1\)[/tex] and in the second case (arms drawn in) as [tex]\(I_2\)[/tex]. The given rotational frequencies are [tex]\(\omega_1 = 0.25 \, \text{rev/s}\) and \(\omega_2 = 0.80 \, \text{rev/s}\)[/tex].
Using the equation [tex]\(I_1\omega_1 = I_2\omega_2\)[/tex], we can rearrange it to solve for the ratio of moments of inertia:
[tex]\[\frac{I_1}{I_2} = \frac{\omega_2}{\omega_1}\][/tex]
Substituting the given values, we have:
[tex]\[\frac{I_1}{I_2} = \frac{0.80 \, \text{rev/s}}{0.25 \, \text{rev/s}}\][/tex]
Simplifying the expression, we get:
[tex]\[\frac{I_1}{I_2} = 3.2\][/tex]
Therefore, the ratio of the man's moment of inertia in the first case (arms extended) to that in the second case (arms drawn in) is 3.2.
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If work stays the same and the distance is increased, then less force is needed to do the work.
True
False
Answer:
False
Explanation:
This is because the work is the same so the force won't change
What is friction ???
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 process generates 250 watts of useful energy and 600 watts of waste energy. WhT is the efficiency of the process?
O29percent
O2.4percent
O42 percent
O71percent
The efficiency of the process is 42%
From the question given above, the following data were obtained:
Useful energy = 250 watt
Wasteful energy = 600 watt
Efficiency =?The efficiency of the process can be obtained as illustrated below:
[tex]efficiency \: = \frac{useful \: energy}{wasteful \: energy} \times 100 \\ \\ efficiency \: = \frac{250}{600} \times 100 \\ [/tex]
Efficiency = 42%Therefore, the efficiency of the process is 42%
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Answer:
Its actually 29%
Explanation:
Took the Test
A less than youthful 82.6 kg physics professor decides to run the 26.2 mile (42.195 km) Los Angeles Marathon. During his months of training, he realizes that one important component in running a successful marathon is carbo-loading, the consumption of a sufficient quantity of carbohydrates prior to the race that the body can store as glycogen to burn during the race. The typical energy requirement for runners is 1 kcal/km per kilogram of body weight, and each mole of oxygen intake allows for the release of 120 kcal of energy by oxidizing (burning) glycogen.
(a) If the professor finishes the marathon in 4:45:00 h, what is the professor's oxygen intake rate, in liters per minute, during the race if he metabolizes all of the carbo-loaded glycogen during the race and the ambient temperature is 21.5°C? 2.28 Read the problem statement again carefully. Is the air at standard temperature and pressure during the marathon? How would this affect the volume of 1 mol of oxygen? L/min
(b) The human body has an efficiency of 25.0%. Only 25.0% of the energy released from oxidizing glycogen is used as macroscopic mechanical energy, and the remaining 75.0% is used for body processes such as pumping blood and respiration, and then leaves the body through the skin via radiation, evaporative cooling, and other processes. What is the average mechanical power (in W) generated by the professor during the run? 197.561 What is the total energy required by the professor during the run? How efficient is the human body, and how long did the race last? W
(c) What is the change in entropy (in J/K) of the professor's body if his core temperature has risen to 38.3°C during the run and his skin temperature is at 36.0°C during the marathon? J/K
(d) What is the change of the entropy (in J/K) of the air surrounding the professor during the race if the ambient temperature remains constant at 21.5°C? J/K
When an alpha particle is emitted from an unstable nucleus, what happens to the atomic number of the nucleus
Answer:
The answer to your question is the number decreases by 4.
Explanation:
I hope this helps and have a great day!
A grinding wheel is a uniform cylinder with a radius of 8.50 cm and a mass of 0.380 kg. Calculate (a) its moment of inertia about its center, and (b) the applied torque needed to accelerate it from rest to 1750 rpm in 5.00 s. Take into account a frictional torque that has been measured to slow down the wheel from 1500 rpm to rest in 55.0 s.
Hi there!
(A)
A grinding wheel is the same as a disk, having moment of inertia of:
[tex]I = \frac{1}{2}MR^2[/tex]
Plug in the given mass and radius (REMEMBER TO CONVERT) to find the moment of inertia:
[tex]I = \frac{1}{2}(0.380)(0.085)^2 = 0.00137 kgm^2[/tex]
(B)
We can use the rotational equivalent of Newton's Second Law to calculate the needed torque:
Στ = Iα = τ₁ - τ₂
Begin by solving for the angular acceleration. Convert rpm to rad/sec:
[tex]\frac{1750r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 183.26 rad/sec[/tex]
Now, we can use the following equation:
ωf = wi + αt (wi = 0 rad/sec, from rest)
183.26/5 = α = 36.65 rad/sec²
τ = Iα = 0.0503 Nm
Since there is a counter-acting torque on the system, we must begin by finding that acceleration:
[tex]\frac{1500r}{min} * \frac{1 min}{60 s} * \frac{2\pi rad}{1 r} = 157.08 rad/sec[/tex]
ωf = wi + αt
-157.08/55 = α = -2.856 rad/sec²
τ₂ = Iα = 0.0039 Nm
Now, calculate the appropriate torque using the above equation:
[tex]\Sigma\tau = \tau_1 - \tau_2[/tex]
[tex]\Sigma\tau + \tau_2 = \tau_1[/tex]
[tex]0.0503 + 0.0039 = \large\boxed{0.054 Nm}[/tex]
calculate 18% of 2758 correct to 4 significant figure
Answer:
......the answer is 496.4
A Vector that starts from
Origin is called what?
Calculate the following
Earth's mass: Man's Mass
Earth's Radiusi man Radius
Earth's Density: mon Density
Farth's granty: man granty
If you represented Earth's history by a line that is 1 m long, how long a segment (in cm) would represent the 400 million years since life moved onto the land
The length of the segment that will represent the 400 million years is 8.8 cm.
The given parameters:
length of age of the Earth = 1 meter longage of the Earth = 4.54 billion yearstime to be represented in the given segment = 400 million yearsThe length of the segment that will represent the 400 million years is calculated as follows;
4.54 billion years ------------- 1 m
400 million years ------------- ?
[tex]\frac{1 \ m}{4.54 \times 10^9 \ years} \times 400 \times 10^6 \ years = 0.088 \ m[/tex]
The length of the segment in centimeters (cm) is calculated as follows;
[tex]= 0.088 \ m \times 100 \ cm\\\\= 8.8 \ cm[/tex]
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a wave travels at a constant speed. how does the wavelength change if the frequency is reduced by a factor of 3? assume the speed of the wave remains unchanged.
A. the wavelength decreases by a factor of 3
B. the wavelength does not change
C. the wavelength increases by a factor of 3
D. the wavelength increases by a factor of 9
Answer:
b
Explanation:
when the wavelength increase it doesnt affect the frequency of a wave.
Answer: The wavelength increases by a factor of 3
Explanation:
4. Your group used an object with fixed mass (heavy or light). How will the results be different
for an object with a different mass?
Pls help, I really need it and plz show steps I will give brainliest..
Answer:
21.31 meters
Explanation:
Since we're working with gravitational potential energy (GPE):
GPE (Joules) = mass (kg) * gravity (m/s^2) * height (meters)
1. Figure out what we have:
GPE = mass * gravity * height
We're looking for height, and we have the other three, so we're set to move on.
2. Isolate the unknown variable (height):
(GPE) / (mass * gravity) = height
3. Plug in your numbers:
(3.78 * 10^7 J) / ((1.81 * 10^5 kg) * (9.8 m/s^2)) = 21.31 meters
I will mark brainlist
If a wave’s amplitude is 2cm, then its height is equal to:
5 cm.
0 cm
4 cm.
2 cm
Answer:
4 cm
Explanation:
Amplitude is the measure from the MIDLINE to the peak of the wave....so the wave HEIGHT is twice the ampltude
2 x 2 cm = 4 cm
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)
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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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?
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)
8. An airplane of mass 8500 kg dives has an altitude of 15,000 m. It then dives steeply to an
altitude of 11,000 m. What was the change in potential energy?
Answer:
333.5 MJ
Explanation:
ΔV = m·g·Δh
= 8500 · 9.81 · (15000-11000)
= 8500 · 9.81 · 4000
= 333 540 000
≈ 333.5 ·10⁶J = 333.5 MJ
The change in potential energy is 333.5 MJ
What is Potential Energy?
Potential energy is defined as the energy possessed by an object because of its position relative to other objects, tension within itself, its electric charge, or other factors. Any object that is raised above its rest position has energy stored in it, so it is called potential energy because it has the potential to do work when released.
It can be expressed as:
P.E.= mgh
where, m is the mass of an object measured in grams 'g'
g is the acceleration due to gravity which is [tex]9.8 m/s^2[/tex]
h is the height measured in meter 'm'
Here,
the given information is
m= 8500kg
Δh= [tex]h_2 - h_1[/tex] = 15000-11000 = 4,000m
So, P.E. = 850 x 9.8 x 4000 = 333 540 000= 333.5 ·10⁶J = 333.5 MJ
Thus, the change in potential energy is 333.5 MJ
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A phone cord is 2.28 m long. The cord has a mass of 0.2 kg. A transverse wave pulse is produced by plucking one end of the taut cord. The pulse makes four trips down and back along the cord in 0.849 s. What is the tension in the cord?
The characteristics of the speed of the traveling waves allows to find the result for the tension in the string is:
T = 10 N
The speed of a wave on a string is given by the relationship.
v =[tex]\sqrt{\frac{T}{\mu } }[/tex]
Where v es the velocty, t is the tension ang μ is the lineal density.
They indicate that the length of the string is L = 2.28 m and the pulse makes 4 trips in a time of t = 0.849 s, since the speed of the pulse in the string is constant, we can use the uniform motion ratio, where the distance traveled e 4 L
v = [tex]\frac{d}{t}[/tex]
v = [tex]\frac{4 L}{t}[/tex]
v = [tex]\frac{4 \ 2.28 }{0.849}[/tex]
v = 10.7 m / s
Let's find the linear density of the string, which is the length of the mass divided by its mass.
μ = [tex]\frac{m}{L}[/tex]
[tex]\mu = \frac{0.2}{2.28}[/tex]
μ = 8.77 10⁻² kg / m
The tension is:
T = v² μ
Let's calculate
T = 10.7² 8.77 10⁻²
T = 1 0 N
In conclusion using the characteristics of the velocity of the traveling waves we can find the result for the tension in the string is:
T = 10 N
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Students perform a set of experiments by placing a block of mass m against a spring, compressing the spring a distance x along a horizontal surface of negligible friction, releasing the block, and measuring the velocity v of the block as it leaves the spring, as shown in Figure 1. The experiments indicate that as x increases, so does v in a linear relationship. The surface is now lifted so that the surface is at an angle θ above the horizontal. Which of the following indicates how the relationship between v and x changes?
Increasing the angle of inclination of the plane decreases the velocity of the block as it leaves the spring.
The statement that indicates how the relationship between v and x changes is; As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of x.Reasons:
The energy given to the block by the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
According to the principle of conservation of energy, we have;
On a flat plane, energy given to the block = [tex]0.5 \cdot k \cdot x^2[/tex] = kinetic energy of
block = [tex]0.5 \cdot m \cdot v^2[/tex]
Therefore;
0.5·k·x² = 0.5·m·v²
Which gives;
x² ∝ v²
x ∝ v
On a plane inclined at an angle θ, we have;
The energy of the spring = [tex]\mathbf{0.5 \cdot k \cdot x^2}[/tex]
The force of the weight of the block on the string, [tex]F = m \cdot g \cdot sin(\theta)[/tex]
The energy given to the block = [tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta)[/tex] = The kinetic energy of block as it leaves the spring = [tex]\mathbf{0.5 \cdot m \cdot v^2}[/tex]
Which gives;
[tex]0.5 \cdot k \cdot x^2 - m \cdot g \cdot sin(\theta) = 0.5 \cdot m \cdot v^2[/tex]
Which is of the form;
a·x² - b = c·v²
a·x² + c·v² = b
Where;
a, b, and c are constants
The graph of the equation a·x² + c·v² = b is an ellipse
Therefore;
As x increases, v increases, however, the value of v obtained will be lesser than the same value of x as when the block is on a flat plane.Please find attached a drawing related to the question obtained from a similar question online
The possible question options are;
As x increases, v increases, but the relationship is no longer linear and the values of v will be less for the same value of xThe relationship is no longer linear and v will be more for the same value of xThe relationship is still linear, with lesser value of vThe relationship is still linear, with higher value of vThe relationship is still linear, but vary inversely, such that as x increases, v decreasesLearn more here:
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Which strategy should you use if your research question is too broad for the scope of your project? (1 O narrow the focus of research question o add another research question o use the very first source you find for your project O change the scope of your project
Answer:
"Narrow the focus of research question"
Explanation:
O Narrow the focus of research question
This is good! You can still use your question, but focus in on something so you have a proper research project.
O Add another research question
Would adding another question to an already broad question help? No.
O Use the very first source you find for your project
If your question is too broad, you should not use whatever you see first as it may be incorrect or does not answer the question
O Change the scope of your project
You could, but if you have a set scope for your project (a) you might not be able to change it (b) you don't need to restart
Have a nice day!
I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly. (ノ^∇^)
- Heather
*graph is below*
1. What is Peter’s total distance traveled? What is Peter's displacement?
2. Is there a time when Peter is not moving? If so, when?
The total distance covered is 24 Km and Peter was not moving between the points marked 10 mins and 30 mins on the graph. His displacement according to the graph is zero.
The distance time graph shows the distance covered plotted on the vertical axis against the time taken plotted on the horizontal axis. Using this graph, the total distance covered can easily be obtained.
The total distance covered is 12 km + 12 km since equal distance was covered to and fro. Hence the total distance covered is 24 km. Perter was not moving between the points marked 10 mins and 30 mins on the graph.
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Balance the following chemical equation:
CO2 + VH20 - CH1206+ voz
[
~
CO₂+
Answer:
Explanation:
Start with Carbon and assume we only get 1 sugar molecule from the process.
you have 6 carbons in the sugar on the right, so you need 6 carbons on the left which only come from CO₂
6 CO₂
you have 12 hydrogen atoms in the sugar on the right, so you need 12 hydrogen atoms on the left which only come from H₂O. At 2 hydrogen atoms per water molecule means you need 6 waters.
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆
you are supplied with 12 oxygen from the CO₂ and 6 oxygen from the H₂O, but you only need 6 oxygen for the sugar. That means there are 12 oxygen remaining which will become 6 O₂ molecules
6 CO₂ + 6 H₂O → 1 C₆H₁₂O₆ + 6 O₂
HELP ASAP!!!!! Choose all the answers that apply. Technology A)influences science
B)helps scientists observe fast phenomena
C)is the same as science
D) influences history
E)helps scientists observe slow phenomena
Astone has a mass of 200 grams. When it is immersed in a measuring cylinder of water,the water rises 100 ml.What is the density of the stone
Answer:
2 g/mLExplanation:
The density of a substance can be found by using the formula
[tex]d = \frac{m}{v} \\ [/tex]
m is the mass
v is the volume
From the question
m = 200g
v = 100 mL
We have
[tex]d = \frac{200}{100} = 2 \\ [/tex]
We have the final answer as
2 g/mLHope this helps you
what is translator disadvantage?
Machine translators:
Level of accuracy can be very low.
Accuracy is also very inconsistent across different languages.
Machines can't translate context.
Mistakes are sometimes costly.
Sometimes translation simply doesn't work.
Human translators:
Turnaround time is longer.
Translators rarely work for free.
Unless you use a translation agency, with access to thousands of translators, you're limited to the languages any one translator can work with.
what were your preparetion before going the different physical fitness test?
Answer:
Avoid heavy strenuous exercise for the 24 hours prior to testing. Do not exercise at all on the day of testing to ensure you are well rested. Wear appropriate clothing for the conditions (e.g. shorts/track pants and t-shirt/singlet/sports top) and non-slip athletic footwear with laces securely fastened
Please write a paragraph explaining the bible verse below in your own words.
Exodus 16:19-20
Answer:
Moses had told them to not keep the food till morning but some kept some anyways because they probably thought they were going to starve or not have food the next morning but what I think it means is that you have to trust in God that he will provide for you and so when the people kept the food cuz he thought they were probably going to start of the next day it got maggots
10. What is Newton's 3rd Law?
O A. For every action there is an equal and opposite reaction.
B. Acceleration depends on two variables, the mass of the object and the amount of
force.
C. An object at rest will stay at rest, an object in motion will stay in motion, unless an
unbalanced force acts upon it.
D. The amount of matter in an object.
7. What is Newton's 2nd Law? *
Answer:
A. For every action there is an equal and opposite reaction.