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--
A weather forecaster uses a computational model on a Monday to predict the weather on Friday. Why might that forecast change? (1 point)
A. The forecaster may have done the calculations a second time because he made a mistake in the calculation.
B. Someone may have reported the weather incorrectly before the first computation.
C. The forecaster may have had the computer model do the calculations a second time, and he found that the prediction changed.
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Answer:
D. An area of low pressure might move more quickly on Tuesday and Wednesday than expected.
Explanation:
I took the test and got it wrong :/
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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Identify the direction of the net force acting on the object. Explain your reasoning
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
help i am having a mental breakdown help! need this done...
*graph is below* help
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?
5. The net external force on a rock of mass 4.2 kg is 8.0 N forward. Find the acceleration of the rock.
Answer:
1.904
Explanation:
F= ma
8 = 4.2 a
a = 8/4.2
a = 1.904
Look up the specs on a c6 rocket engine. How many C6 engines would it take to launch Mr. Blazey (90kg)
from the elevation of Boulder to the elevation of longs peak? (3 points) What would the max power be? (3
points) The max acceleration? (3 points) For all 12 points, you'll need to account for the mass of the added
rockets (you can assume their mass declines linearly over their burn and is O after they finish burning).
Answer:
The Estes C6-0 engine is a booster stage engine designed for model rocket flight and has to be used with a standard engine. This engine is for flights in rockets weighing less than 4 ounces, including the engines. Each package includes 3 engines, 4 starters and 4 plugs.
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?
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.
Comparing energy resources
Answer:
Sorry, This Photo is not clear.
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?
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 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?
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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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
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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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
Answer:
25.0 cm3
Explanation:
The volume is 25.0 cm3 .
Which of the following correctly describes electromagnetic waves?
A. transverse waves
B. Longitudinal waves
C. Have a constant wavelength
D. Need a medium to transfer energy
Answer:
answer
transverse waves
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?
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
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
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.
Which quantity or quantities is/are increasing for the object represented by line B?
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
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?
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
What is the maximum height above ground a projectile of mass 0.79 kg, launched from ground level, can achieve if you are able to give it an initial speed of 80.3 m/s?
From kinematic, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.
The parameters given from the question are
mass m = 0.79kg
initial speed U = 80.3 m/s
Maximum height H = ?
The object will be going up under the influence of gravity. Acceleration due to gravity g = -9.8m/[tex]s^{2}[/tex]
At maximum height, final velocity V = 0
From kinematic formula, the best formula to use is
[tex]V^{2} = U^{2} - 2gH[/tex]
Substitute all the parameters into the equation
0 = [tex]80.3^{2}[/tex] - 2 x 9.8 x H
19.6H = 6448.09
H = 6448.09 / 19.6
H = 328.98 m
Therefore, the maximum height above ground a projectile of mass 0.79 kg, launched from ground level can achieve is 329 meters approximately.
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What was different about the molecules you needed to make protein 3 compared to the molecules you used to make protein 2?
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.
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?
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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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
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 HzThe 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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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
Answer:
its b it must be charge
Explanation:
Easiest way to Find fahrenhiet to celsius please i need necessary 20 for the fastest correct answer
Answer:
thx for the points
Explanation:
no need brainliest
What is the effect of erosion?
A. New land forms at the mouth of a river.
B. New land forms at the top of a mountain.
C. A mountain forms.
D. A fossil is created.
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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A solid sphere starts from rest and rolls down a slope that is 6.4 m long. If its speed at the bottom of the slope is 5.3 m/s, what is the angle of the slope
From the relationship between acceleration a and g on an inclined plane, the angle of the slope is 13 degrees
Given that a solid sphere starts from rest and rolls down a slope that is 6.4 m long. The speed at the bottom of the slope is 5.3 m/s, the distance travelled is 6.4 m. That is,
Initial velocity U = 0 ( since it starts from rest)
Final velocity V = 5.3 m/s
distance S = 6.4 m
Let us first calculate its acceleration by using third equation of motion.
[tex]V^{2}[/tex] = [tex]U^{2}[/tex] + 2aS
[tex]5.3^{2}[/tex] = 0 + 2 x 6.4a
28.09 = 12.8a
a = 28.09 / 12.8
a = 2.2 m / [tex]s^{2}[/tex]
To calculate the angle of the slope, let us use the relationship between acceleration a and g on an inclined plane.
acceleration a = gsin∅
substitute all the relevant parameters
2.2 = 9.8 sin∅
sin∅ = 2.2/9.8
sin∅ = 0.224
∅ = [tex]Sin^{-1}[/tex](0.224)
∅ = 12.97 degrees
∅ = 13 degrees (approximately)
Therefore, the angle of the slope is 13 degrees
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A magnet is located above circular current. What is the direction of the magnetic force on the magnet
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.
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
Answer:
a
Explanation:
trình bày những hiểu biết cơ bản về năng lượng
Answer:
I don't understand your language
A block slides down a smooth ramp, starting from rest at a height h. When it reaches the bottom it's moving at speed v. It then continues to slide up a second smooth ramp. At what height is its speed equal to v/2
Answer:
3h/4
Explanation:
At speed v/2 height will be 3/4 h
What is equation of motion in kinematics?Equation that describes the motion of point , bodies , and system of bodies without considering the force that cause them to move is called equation of motion in kinematics
When block is at top of first ramp
u=0 ( block was at rest )
a = g ( acceleration due to gravity
using equation of motion
2as = v^2 - u^2
2gh = v^2
Then the block continued and reached a speed of v1 = v/2 on second ramp
now , final velocity = v= v1 =[tex]\sqrt{2gh}[/tex] / 2
u= [tex]\sqrt{2gh\\}[/tex]
s= h1
using equation of motion , we get
2as = v^2 - u^2
2(-g)h1 =( [tex]\sqrt{2gh}[/tex]/2)^2 - [tex]\sqrt{2gh}[/tex]
2(-g)h1 = (g h - 4 g h) / 2
h1 = 3/4 h
At speed v/2 height will be 3/4 h
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