The speed that the 0.25 kg block hits the ground with is 1.76 m/s.
Given that the mass of the block (m1) is 0.25 kg and the mass of the pulley (m2) is 0.50 kg, and the radius (r) of the pulley is 0.10 m. The force acting on m1 and m2 due to gravity is: F = (m1 + m2)g = (0.25 kg + 0.50 kg) * 9.8 m/s² = 7.35 N. The tension in the string is the same throughout and equal to T. The acceleration of the system is given by:a = (m2 - m1)g / (m1 + m2) = 2.94 m/s².Using the kinematic equation, v² = u² + 2as, we get:v² = 0 + 2 * 2.94 m/s² * 0.20 m = 1.176 m²/s²v = 1.08 m/s. The final velocity will be higher as the pulley will also contribute to the acceleration.
The velocity gained by the center of mass of the pulley is given by v = a * t, where t is the time taken for the blocks to move through a distance of 0.20 m. Taking the moment of inertia of the pulley into account, the net acceleration of the system is given bya = (2m1)g / (3m1 + m2) = 2.94 * (2 * 0.25) / (3 * 0.25 + 0.50) = 2.00 m/s²The velocity gained by the center of mass of the pulley is v = a * t = 2.00 m/s² * t The angular acceleration of the pulley isα = a / r = 2.00 m/s² / 0.10 m = 20.0 rad/s²The rotational motion of the pulley is given byω = ω0 + αt where ω0 is the initial angular velocity of the pulley, which is 0 as it starts from rest.ω = αt = 20.0 rad/s² * t. The velocity of the block is equal to the velocity of the center of mass of the pulley. Therefore, v = ω * r = 20.0 rad/s² * t * 0.10 m = 2.00 m/s the final velocity of the block is the vector sum of the velocity gained by the block and the velocity of the center of mass of the pulley v final = √(v² + (2.00 m/s)²) = 1.76 m/s.
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a fireman of mass 80 kgkg slides down a pole. for the steps and strategies involved in solving a similar problem, you may view a_____.
To understand the steps and strategies involved in solving a similar problem of a fireman sliding down a pole, you can refer to a reliable resource or guide.
When dealing with a problem similar to a fireman sliding down a pole, it is helpful to have a reference or guide that outlines the necessary steps and strategies. While there are various resources available, it is important to choose a reliable one to ensure accuracy and plagiarism-free information. One option is to consult textbooks or online tutorials on physics or mechanics, specifically those that cover topics related to forces, friction, and motion.
These resources often provide detailed explanations, equations, and examples to help solve similar problems effectively. Additionally, reputable educational websites or academic publications can be valuable sources for step-by-step instructions and strategies.
By accessing such resources, you can gain a better understanding of the principles involved and apply them to solve problems involving sliding or descending objects.
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calculate the venturi and orifice coefficients using engineering judgment, comment on the comparison for agreement or lack of agreement.4. express the errors in f and re as a function of the precisions of manometer, graduate cylinder, and stop watch, in the pipe flow experiment. note that the pressure and flow rate are independently measured. 5. what are the advantages and disadvantages of flow-restriction meters such as the orifice plate and venturi? 6. why do we remove the air or air bubble in the manometer? if there is a 1.5 cm air length in the manometer pipe, estimate how much error will it cause in the experimental pressure?
The Venturi and orifice coefficients are given by the following formulas, respectively: Cv = Q / (CdA)Co = Q / (CdA), Where, Q is the flow rate, Cd is the discharge coefficient, and A is the cross-sectional area of the pipe. The value of Cd depends on the Reynolds number (Re) and the beta ratio (β), which is the ratio of the diameter of the flow-restricting device to the diameter of the pipe containing the fluid.
The values of Cd for Venturi and orifice are given below:Venturi: Cd = 0.98 - 0.04β + 0.4/βOrifice: Cd = 0.6 - 0.5/β2 + 0.85/β4For this problem, engineering judgment has to be used to estimate the values of Cd and β based on experimental data. A comparison of the calculated values of Cd and β for the Venturi and orifice can be made to check for agreement or lack of agreement.
Flow-restriction meters, such as the orifice plate and Venturi, are used to measure the flow rate of fluids in pipes. The advantages and disadvantages of these meters are given below:Advantages: Flow-restriction meters are simple and inexpensive to install. They can be used to measure the flow rate of a wide range of fluids. They are accurate for liquids and gases at high flow rates.
Disadvantages: Flow-restriction meters are sensitive to changes in viscosity, density, and temperature. They cause a pressure drop in the pipeline, which can affect the performance of pumps and compressors. They require regular maintenance to prevent clogging and fouling.
The manometer is used to measure the pressure drop across the flow-restricting device. The manometer works by balancing the pressure of the fluid with the weight of a liquid column in a tube. The air bubble in the manometer should be removed to ensure accurate measurements. If there is a 1.5 cm air length in the manometer pipe, it will cause an error in the experimental pressure due to the weight of the air. The error can be calculated as follows: Error = (ρair x g x h) / (ρfluid x g), Where ρair is the density of air, g is the acceleration due to gravity, h is the height of the air column, and ρfluid is the density of the fluid.
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a planet orbiting a star feels a force of gravity from the star. if you could move the planet closer to the star so its distance was half the original distance from the star, the force of gravity would be ...
If you move a planet closer to the star so that its distance from the star is half the original distance from the star, the force of gravity would be four times stronger than before.What is gravity?Gravity is a force of attraction between any two objects that have mass.
The gravitational force is what keeps the planets in orbit around the Sun. When two objects have a larger mass, they attract each other with greater force. When the distance between two objects is decreased, the gravitational force between them increases.
Gravity is dependent on mass. The more massive an object is, the greater its gravity will be. The closer two objects are, the more powerful the gravitational force between them. According to the Law of Universal Gravitation, any two bodies in the universe attract each other with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
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which processes could the heating curve be describing? check all that apply. a) boiling. b) condensation. c) endothermic. d) reaction.
The heating curve could be describing processes such as boiling and endothermic reactions.A heating curve is a graphical representation that shows the changes in temperature as a substance is heated.
It typically consists of two distinct segments: heating and phase change. Boiling is a process where a substance changes from its liquid phase to its gaseous phase. During the heating phase, the temperature of the substance gradually increases until it reaches its boiling point. At this point, the substance undergoes a phase change, and the temperature remains constant until all the liquid has vaporized. Therefore, the heating curve could be describing the process of boiling.
Endothermic reactions are chemical reactions that absorb heat from their surroundings, resulting in a decrease in temperature. If the heating curve represents the temperature changes during an endothermic reaction, we would observe a decrease or plateau in temperature during the reaction, followed by a subsequent increase. Thus, the heating curve could also be describing an endothermic reaction.
Condensation, on the other hand, is the process by which a substance changes from its gaseous phase to its liquid phase. It typically occurs when a gas is cooled down, and the temperature decreases. However, the heating curve represents temperature changes during the heating process, not cooling. Therefore, condensation is not applicable to the heating curve.
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(0)
Each unit on the grid represents 1 mile. If Curtis can ride his bike at a constant rate of 12 miles per hour, how many minutes would it take Curtis to ride from his house to Jean's house?
It would take Curtis 20 minutes to ride from his house to Jean's house.
The distance from Curtis' house to Jean's house can be found using the grid below.
Curtis would need to travel 4 miles to get from his house to Jean's house in the grid below.
The formula to calculate the time taken by Curtis to ride 4 miles, given that he can ride his bike at a constant rate of 12 miles per hour, is given below:
time = distance / speed
time = 4 miles / 12 mphIn order to express the time in minutes,
we need to convert the speed from miles per hour to miles per minute.
1 hour = 60 minutes
Therefore, 1 mile per hour = 1/60 miles per minute
12 miles per hour = 12/60 miles per minute = 1/5 miles per minute
Substituting this value of the speed into the formula above
time = 4 miles / (1/5) miles per minute
= 4 miles * 5 minutes per mile= 20 minutes
Therefore, it would take Curtis 20 minutes to ride from his house to Jean's house.
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7. A 635 kg car accelerates from 10.0 m/s to 35.0 m/s over a
distance of 15.0 m. What was the force that the engine had to exert
on the car to achieve this acceleration?
6. An object has a mechanical
The force that the engine had to exert on the car to achieve this acceleration was 6,350 N.
Mass of the car, m = 635 kg Initial velocity, u = 10.0 m/s Final velocity, v = 35.0 m/s Distance covered, s = 15.0 m Force applied, F = ? Formula: Force = (mass x acceleration)Using the above formula, we can derive the acceleration using the given information: Firstly, we will calculate the acceleration: Acceleration a can be calculated as: a = (v - u)/t, where t is the time taken to cover the distance s Here, the initial velocity, u = 10.0 m/s Final velocity, v = 35.0 m/s Distance, s = 15.0 m Acceleration a = (v - u)/t, t = s/u -v/u = 25.0/10.0 = 2.5 seconds Therefore, the acceleration a = (v - u)/t = (35.0 - 10.0)/2.5 = 10.0 m/s² Now, we can calculate the force, F:F = m x a = 635 kg x 10.0 m/s² = 6,350 N Therefore, the force that the engine had to exert on the car to achieve this acceleration was 6,350 N.
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what lens aperture setting will offer the most amount of light in a dark setting?
When shooting in a dark setting, you may need to use a wider aperture, slower shutter speed, higher ISO, or a combination of all three to get a well-exposed image.
To get the most amount of light in a dark setting, you should use the widest lens aperture setting available. A wide aperture allows more light to enter the camera and reach the image sensor, which makes it ideal for low-light situations.
The lens aperture is a small opening located inside the camera lens that controls the amount of light that enters the camera. It is measured in f-stops and is often represented using a sequence of numbers, such as f/2.8, f/4, f/5.6, f/8, and so on. The smaller the f-stop number, the wider the aperture, and the larger the amount of light that is allowed in. So, to get the most amount of light, you should use the smallest f-stop number available.
A dark setting refers to a situation where there is not enough light for the camera to capture a well-exposed image. This can happen in a variety of situations, such as indoors with low lighting or outside at night. When shooting in a dark setting, you may need to use a wider aperture, slower shutter speed, higher ISO, or a combination of all three to get a well-exposed image.
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find the period of revolution of a satellite moving in a circular orbit around the earth at a height of 3.6 x 10^6 m above the earth’s surface. Assume the earth is a uniform sphere of radius 6.4 x 10^6 m. The earth’s mass 6 x 10^24 kg and G= 6.7 x 10^-11 nm^2
31.324 × [tex]10^{6}[/tex] seconds is the period of revolution of the satellite.
To find the period of revolution of a satellite moving in a circular orbit around the Earth, we can use the formula for the orbital period:
T = 2π√(r³ / GM)
where T is the period of revolution, r is the distance between the center of the Earth and the satellite (radius of orbit), G is the gravitational constant, and M is the mass of the Earth.
In this case, the height of the satellite above the Earth's surface is given as 3.6 x [tex]10^{6}[/tex] m, and the radius of the Earth is 6.4 x [tex]10^{6}[/tex] m. To find the radius of the orbit (r), we need to add the height to the radius of the Earth:
r = 3.6 x [tex]10^{6}[/tex] m + 6.4 x [tex]10^{6}[/tex] m = 10 x [tex]10^{6}[/tex] m
Substituting the values into the formula, we get:
T = 2π√((10 x [tex]10^{6}[/tex] m)³ / (6.7 x [tex]10^{-11}[/tex] [tex]nm^{2}[/tex]) x (6 x [tex]10^{24}[/tex] kg))
Simplifying the equation gives us:
T = 2π√([tex]10^{18}[/tex] m³ / (6.7 x 6) x [tex]10^{3}[/tex])
T = 2π√([tex]10^{18}[/tex] m³ / 40.2 x [tex]10^{3}[/tex])
T = 2π√(2.487562 × [tex]10^{13}[/tex] m³)
Calculating the square root gives:
T ≈ 2π × 4.9874 × [tex]10^{6}[/tex] s
T ≈ 31.324 × [tex]10^{6}[/tex] s
Therefore, the period of revolution of the satellite is approximately 31.324 × [tex]10^{6}[/tex] seconds.
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Analyze,how does the inertia of the weights change by increasing their distance to the center - R! Which statement corresponds to Your results? o Inertia is rarely increasing along with R, in less than half of Rmeasurements o Inertia is often increasing along with R, in 50-80% of R measurements o Inertia is mostly, but not always increasing along with R, in at least 80% of R measurements o Inertia is always decreasing with increasing R o Inertia is always increasing along with R
The inertia of an object is its resistance to change in its state of motion. It is measured by the moment of inertia, which is equal to the mass of the object multiplied by the square of its radius. As the radius of an object increases, its moment of inertia increases, and its inertia decreases.So option 4 is correct.
In the case of the weights, as the distance from the center (R) increases, the moment of inertia increases, and the inertia decreases. This means that it takes less force to accelerate the weights as they move away from the center.
The other options are incorrect.
Option (1) is incorrect because inertia is always increasing with increasing R.
Option (2) is incorrect because inertia is never increasing along with R.
Option (3) is incorrect because inertia is not mostly increasing along with R.
Option (5) is incorrect because inertia is not always increasing along with R.
Therefore option 4 is correct.
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Question 14 (2 points) A two point interference pattern is formed in a pool. A node located on the third nodal line, is 21.0 m from one source and 29.8 m from the other source. One wave crest takes 3.
A two point interference pattern is formed in a pool. A node located on the third nodal line, is 21.0 m from one source and 29.8 m from the other source. One wave crest takes 3.3 s to travel the 50.0 m width of the pool. the speed of the wave crest is approximately 15.15 m/s, the wavelength is approximately 2.51 m, and the frequency is approximately 6.03 Hz.
To solve this problem, we can use the wave equation, which relates the speed (v), wavelength (λ), and frequency (f) of a wave:
v = λ * f
Given:
Distance from one source to the node (d1) = 21.0 mDistance from the other source to the node (d2) = 29.8 mWidth of the pool (w) = 50.0 mTime for one wave crest to travel the width of the pool (t) = 3.3 sSpeed (v):
The speed of the wave can be calculated using the formula:
v = w / t
Substituting the given values:
v = 50.0 m / 3.3 s ≈ 15.15 m/s
Wavelength (λ):
To find the wavelength, we can use the relationship between the distances from the sources to the node and the wavelength in a two-point interference pattern:
d1 - d2 = (m + 1/2) * λ
where m is the order of the nodal line (in this case, m = 3 since the node is on the third nodal line).
Substituting the given values:
21.0 m - 29.8 m = (3 + 1/2) * λ
-8.8 m = 7/2 * λ
Solving for λ:
λ = (-8.8 m) / (7/2) = -8.8 m * (2/7) = -2.51 m
Since the wavelength cannot be negative, we take the absolute value:
λ ≈ 2.51 m
Frequency (f):
To find the frequency, we can rearrange the wave equation:
f = v / λ
Substituting the known values:
f = (15.15 m/s) / (2.51 m) ≈ 6.03 Hz
Therefore, the speed of the wave crest is approximately 15.15 m/s, the wavelength is approximately 2.51 m, and the frequency is approximately 6.03 Hz.
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Pulsing the fan cart from rest with the fan rotated In this activity, you will measure the average acceleration of the fan cart during a pulse when we rotate the fan to a variety of nonzero angles. Prediction 1.1: What would happen to the average acceleration of the cart down the track if you rotated the fan so that it no longer pointed at Oº? For instance, suppose you set the Angle Indicator to 45°. Would the average acceleration of the cart in the x-direction during the pulse increase, decrease, or stay the same? Prediction 1.2: Write down an expression that you think models the average acceleration of the fan cart during the pulse as a function of the angle on the Angle Indicator. Justify using this model. Consider using limiting-case analysis in your justification.
1.1: The average acceleration of the fan cart in the x-direction during the pulse will decrease if the fan is rotated so that it no longer points at Oº (Angle Indicator set to 45°).This model can be justified using limiting-case analysis.
The average acceleration of the fan cart in the x-direction during the pulse will decrease if the fan is rotated so that it no longer points at Oº. This is because the component of the force of the fan in the x-direction will decrease as the fan is rotated away from Oº
1.2: The expression that models the average acceleration of the fan cart during the pulse as a function of the angle on the Angle Indicator is given by the expression:
a = (Mg/(M + m)) * (sinθ - μcosθ),
where M is the mass of the fan cart, m is the mass of the hanging weight, g is the acceleration due to gravity,
μ is the coefficient of kinetic friction between the fan cart and the track, and θ is the angle on the Angle Indicator.
When θ = 90º, the expression reduces to a = (Mg/(M + m)) * (-μ), which gives the minimum acceleration of the fan cart. This shows that the expression is valid for all values of θ between 0º and 90º.
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cientists establish that a river had been depositing sediment for 2.0 x 10 years. They determine that the sedimentary rock strata is 9.0 km thick. The rate of sedimentary deposit is A. 0.45 cm/a B. 4.5 cm/a C. 9.0 cm/a D. 45 cm/a
Sedimentary rock is one of the three main types of rock found on Earth's crust, along with igneous and metamorphic rocks. It is formed through the accumulation, compaction, and cementation of sediments over time.
The thickness of the sedimentary rock strata = 9.0 km. The time period for which the river had been depositing sediment = 2.0 × 10 years. We need to find the rate of sedimentary deposit. Rate = (Thickness of the sedimentary rock strata) ÷ (Time period for which the river had been depositing sediment).Here, the thickness of the sedimentary rock strata = 9.0 km= 9000 m. Time period for which the river had been depositing sediment = 2.0 × 10 years = 2.0 × 10 × 365.25 days (in a year) = 7.305 × 10⁴ days.
Rate = (9000 m) ÷ (7.305 × 10⁴ days) = 0.12315 m/day = 12.315 cm/day.
Approximating the answer to one significant figure, the rate of sedimentary deposit is 10 cm/day. Hence, the correct option is option D: 45 cm/a.
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Which three of the following statements are correct? (i) The mass of an a particle is about four times that of a proton and its charge is two times that of a proton. (ii) Alpha particles with very high energies can easily penetrate the skin. (iii) Elements 130X56 and 131Y56 are isotopes of the same element. (iv) A y ray is not affected by a magnetic field, but it is deflected by an electric field applied at right angles to the path of the y ray. (v) Beta particles are emitted with a range of energies from a given B source. C. (ii); (iv): (v) E. (i): (if); (iv) D.): (i); (h) A (i), (fi); (v) 9. (10) fil: (iv) e OL Q 9 1
From the given options, the correct statements are: (i) The mass of an a particle is about four times that of a proton and its charge is two times that of a proton. (iv) A y ray is not affected by a magnetic field, but it is deflected by an electric field applied at right angles to the path of the y ray. (v) Beta particles are emitted with a range of energies from a given B source.
The correct answer would be statement are (i), (iv), and (v).
From the given options, the correct statements are:
(i) The mass of an alpha particle is about four times that of a proton and its charge is two times that of a proton. This is true. An alpha particle consists of two protons and two neutrons, so its mass is approximately four times that of a single proton, and it carries a charge that is twice the charge of a proton.
(iv) A gamma ray is not affected by a magnetic field, but it is deflected by an electric field applied at right angles to the path of the gamma ray. This is true. Gamma rays are electromagnetic radiation and do not possess any charge, so they are not affected by magnetic fields. However, they are deflected by electric fields due to their interaction with the electric field.
(v) Beta particles are emitted with a range of energies from a given beta source. This is true. Beta particles, which can be either electrons or positrons, are emitted during beta decay in radioactive substances. The energy of these beta particles can vary within a range depending on the specific decay process and the nucleus involved.
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The blade on a typical table saw rotates at 3650 revolutions per minute. Calculate the linear velocity in miles per hour of one of the teeth at the edge of the 8 inch diameter blade.
The linear velocity of one of the teeth at the edge of the 8-inch diameter blade is approximately 107.143 miles per hour.
The circumference of the blade is given by:
Circumference = π x diameter
Circumference = 3.14159 x 8 inches ≈ 25.13274 inches
To convert the linear velocity to miles per hour, we need to convert inches to miles and minutes to hours. There are 63360 inches in a mile and 60 minutes in an hour.
Linear velocity = (Circumference x RPM) x (1 mile/63360 inches) x (60 minutes/1 hour)
Linear velocity = (25.13274 inches x 3650 RPM) x (1 mile/63360 inches) x (60 minutes/1 hour)
Linear velocity ≈ 107.143 miles per hour
Therefore, the linear velocity of one of the teeth at the edge of the 8-inch diameter blade is approximately 107.143 miles per hour.
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In terms of the tangent of a positive acute angle, what is the expression for tan (477) ² Provide your answer below:
What is the value of tan if the terminal side of angle intersects the unit circle
The given value 477 is greater than 360. Therefore, we can use coterminal angles: $477°-360°=117°$. Thus, the angle in standard position that corresponds to 117° is $\theta = 360°-117°=243°$ which has a terminal side in quadrant III and is therefore $180° < \theta < 270°$.
We can apply reference angle theorem: the reference angle for $\theta$ is $R=\theta -180°=243°-180°=63°$.The tangent of an angle is defined as the opposite leg (O) over the adjacent leg (A) of a right triangle that contains the angle:$$\tan R = \frac{O}{A}$$In a unit circle, the radius is 1.
The hypotenuse of the right triangle is therefore $\sqrt{1^2+1^2}=\sqrt{2}$, the adjacent leg is 1 and the opposite leg is $\tan R$ . Therefore, $$\tan R = \frac{\tan 63°}{1}$$So, the expression for $\tan 477°$ is$$\tan^2 477° = \left(\tan 63°\right)^2=\left(\frac{\sqrt{2}+\sqrt{6}}{3}\right)^2=\frac{8+4\sqrt{3}}{9}$$
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Suppose that a 120 kg football player running at 6.5 m/s catches a 0.46 kg ball moving at a speed of 24.5 m/s with his feet off the ground, while both of them are moving horizontally.
(a) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving in the same direction.
(b) Calculate the change in kinetic energy of the system, in joules, after the player catches the ball.
(c) Calculate the final speed of the player, in meters per second, if the ball and player are initially moving IN OPPOSITE DIRECTIONS.
(d) Calculate the change in the kinetic energy of the system, in joules, in this case (use answer from part c).
The final speed of the player when the ball and player are initially moving in the same direction is 6.568 m/s. Change in kinetic energy of the system 69.883072 kg * m^2/s^2. The final speed of the player when the ball and player are initially moving in opposite directions is 6.377 m/s.
To solve this problem, we can apply the principles of conservation of momentum and conservation of kinetic energy.
Given:
Mass of the football player (m1) = 120 kg
Initial velocity of the football player (v1) = 6.5 m/s
Mass of the ball (m2) = 0.46 kg
Initial velocity of the ball (v2) = 24.5 m/s
(a) When the ball and player are initially moving in the same direction, we can use the conservation of momentum equation:
m1 * v1 + m2 * v2 = (m1 + m2) * vf
where vf is the final velocity of the player-ball system.
Substituting the given values into the equation, we have:
(120 kg) * (6.5 m/s) + (0.46 kg) * (24.5 m/s) = (120 kg + 0.46 kg) * vf
Simplifying this equation, we find:
780 + 11.27 = 120.46 * vf
791.27 = 120.46 * vf
Dividing both sides by 120.46, we get:
vf = 6.568 m/s
Therefore, the final speed of the player when the ball and player are initially moving in the same direction is 6.568 m/s.
(b) To calculate the change in kinetic energy of the system, we can use the equation:
ΔKE = (1/2) * (m1 + m2) * vf^2 - (1/2) * m1 * v1^2 - (1/2) * m2 * v2^2
Substituting the given values and the final velocity obtained from part (a) into the equation, we have:
ΔKE = (1/2) * (120 kg + 0.46 kg) * (6.568 m/s)^2 - (1/2) * 120 kg * (6.5 m/s)^2 - (1/2) * 0.46 kg * (24.5 m/s)^2
= 69.883072 kg * m^2/s^2.
(c) When the ball and player are initially moving in opposite directions, we can use the same conservation of momentum equation as in part (a).
Substituting the given values into the equation, we have:
(120 kg) * (6.5 m/s) - (0.46 kg) * (24.5 m/s) = (120 kg + 0.46 kg) * vf
Simplifying this equation, we find:
780 - 11.27 = 120.46 * vf
768.73 = 120.46 * vf
Dividing both sides by 120.46, we get:
vf = 6.377 m/s
Therefore, the final speed of the player when the ball and player are initially moving in opposite directions is 6.377 m/s.
(d) To calculate the change in kinetic energy of the system in this case, we can use the same equation as in part (b), using the final velocity obtained from part (c).
Substituting the given values and the final velocity obtained from part (c) into the equation, we have:
ΔKE = (1/2) * (120 kg + 0.46 kg) * (6.377 m/s)^2 - (1/2) * 120 kg * (6.5 m/s)^2 - (1/2) * 0.46 kg * (24.5 m/s)^2
The simplified expression is -235.65113 kg * m^2/s^2.
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The porosity of a core that was retrieved from a reservoir was measured in the lab and found to be 20%. Calculate the porosity under reservoir conditions if the overburden pressure is 4500 psi, the pore pressure is 1650 psi and the pore volume compressibility is 9x10€ pst-¹ A 160-acre and 100 ft thick reservoir has a porosity of 11%. The pore compressibility is 5.0 x 10- 6 (1/psi). If the pressure decreases 3,000 psi, what is the subsidence (ft)? Assume Cf=Cb
The porosity of a core retrieved from a reservoir = 20%. Overburden pressure = 4500 psi, Pore pressure = 1650 psi, Pore volume compressibility = 9 x 10^-6 psi^-1, Reservoir porosity = 11%, Pore compressibility = 5.0 x 10^-6 psi^-1, Pressure decrease = 3000 psi.
Porosity under reservoir conditions if the overburden pressure is 4500 psi, the pore pressure is 1650 psi and the pore volume compressibility is 9x10€ pst-¹ and Subsidence. The relation between porosity, overburden pressure, and pore pressure is given by:φ = (φo - φw)/(1 - φw), Where φo = porosity under overburden pressure and pore pressure.φw = porosity under reservoir condition.
Substituting the given values in the above relation, we get:0.2 = (φo - 0.11)/(1 - 0.11)φo - 0.11 = 0.18φo = 0.18 + 0.11φo = 0.29So, the porosity under reservoir condition is 29%.
The relation between subsidence and pressure decrease is given by:Δh = (Cf - Cb) × Δp × h0, Where, Cf = Pore compressibility, Cb = Bulk compressibility (assumed to be equal to pore compressibility), Δp = Pressure decrease, h0 = Thickness of the reservoir.
Substituting the given values in the above relation, we get:Δh = (5.0 x 10^-6) × (3000) × (100)Δh = 1.5 ft.
Therefore, the subsidence is 1.5 ft.
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what diameter d will you choose for your wire? express your answers in millimeters to two significant figures.
The following equation determines an object's weight:
weight = mass * gravitational acceleration.
We can assume a typical gravitational acceleration of 9.8 meters per second squared since the weight is specified as 100 kilograms. The tensile strength of a wire determines the greatest tension or force it can withstand before snapping. The tensile strength in this instance is specified as 500 megapascals.
We need to consider the stress applied to the wire under the weight. Stress (σ) is defined as force per unit area:
σ = F / A,
We can rearrange the formula to solve for the cross-sectional area:
A = F / σ.
Given that the weight is the force F and that the tensile strength is specified as 500 megapascals (MPa), or 500 N/mm2, we may swap the following values:
[tex]A = (100 kg * 9.8 m/s^{2} ) / (500 N/mm^{2} ).[/tex]
Converting the units, we have:
[tex]A = (100,000 g * 9.8 m/s^{2} ) / (500 N/mm^{2} ), \\A = 196,000 g / 500 N/mm^{2} , \\A = 392 g / N/mm^{2} .[/tex]
Now, we can express the cross-sectional area A in terms of the diameter d:
[tex]A = (\pi /4) * d^{2}[/tex],
where d is the diameter of the wire.
Substituting the expression for A, we have:
[tex]392 g / N/mm^{2} = (\pi /4) *d^{2}[/tex].
Simplifying the equation and solving for d, we get:
[tex]d^{2} = (4 * 392 g) / (\pi * 500 N),[/tex]
[tex]d^{2}[/tex] ≈ 3.1424,
d ≈ √(3.1424),
d ≈ 1.772 mm.
Therefore, you would choose a wire diameter of approximately 1.772 millimeters to support a weight of 100 kilograms without breaking, assuming a material with a tensile strength of 500 megapascals.
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--The complete Question is, What diameter d will you choose for your wire, expressed in millimeters, to support a weight of 100 kilograms without breaking, assuming a material with a tensile strength of 500 megapascals?--
The main sequence lifetime of a star with a mass of 2 solar masses and 20 solar luminosities is approximately ... 1.10 x 10^10 years 2. 1,000 x 10^10 years 3.0.001 x 10^10 years 4.0.1 x 10^10 years 5.
The main sequence lifetime of a star with a mass of 2 solar masses and 20 solar luminosities is approximately [tex]1.10 * 10^1^0[/tex] years.
The main sequence lifetime of a star refers to the duration it spends fusing hydrogen in its core. This phase is characterized by a balance between the inward pull of gravity and the outward pressure from nuclear fusion reactions. The main sequence lifetime depends on the star's mass and luminosity. For a star with a mass of 2 solar masses and a luminosity of 20 solar units, its main sequence lifetime is estimated to be approximately [tex]1.10 * 10^1^0[/tex] years.
The mass of a star influences its core temperature and pressure, determining the rate of fusion and, consequently, its lifetime. Luminosity, on the other hand, measures the total energy output of a star per unit time. By comparing these values to stellar models and observations, astronomers can estimate the main sequence lifetime. In this case, a star with a mass of 2 solar masses and a luminosity of 20 solar units would have a main sequence lifetime of around [tex]1.10 * 10^1^0[/tex] years.
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A binary star system consists of two stars very close to one another. The apparent magnitudes of m1 = 2 and m2 = 3. The apparent magnitude m is defined by a star's flux density, compared to the reference star with m0 and f0
m_m0=_2. 5(f/f0)
The ratio of the flux density of star 2 to the flux density of the reference star (f2/f₀) is approximately 0.59. The formula for calculating the magnitude of a star is given as: [tex]m_m₀=_2.5(f/f₀)[/tex]
A binary star system refers to two stars that are very close to each other. The apparent magnitudes of m₁ = 2 and m₂ = 3. The apparent magnitude m is defined by a star's flux density, compared to the reference star with m₀ and f₀. The formula for calculating the magnitude of a star is given as:
[tex]m_m₀=_2.5(f/f0)[/tex]
Where m is the apparent magnitude, m₀ is the reference star's apparent magnitude, f is the flux density of the star, and f₀ is the flux density of the reference star.
Applying the formula for m₁ : 2 - m₀
= 2.5 log (f/f₀)
Applying the formula for m₂: 3 - m₀
= 2.5 log (f/f₀)
We can eliminate m₀ by subtracting the two equations:
2 - 3 = 2.5 log (f/f₀) - 2.5 log (f/f₀)log (f/f₀)
= (2-3)/2.5= -0.4f/f₀
= [tex]10^(-0.4)[/tex]
We know that
f1/f₂ = m₂/m₁
f₂ = f₁ (m₂/m₁)
Substituting f/f₀ = [tex]10^(-0.4)[/tex] and m₁ = 2 and m₂ = 3
f₂/f₀= f₁/f₀ * m/m₁
=[tex]10^(-0.4)[/tex] * 3/2
= 0.59
Therefore, the ratio of the flux density of star 2 to the flux density of the reference star (f2/f₀) is approximately 0.59.
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how
far will it project out travel if it is fired at an angle of 50°
within the initial velocity of 45 ms? Assume that Yf=Yi=0 meters.
Also, Xi=0 meters. g =-9.8 m/s2
The projectile will travel a horizontal distance of approximately 104.8 meters.
To determine the horizontal distance traveled by a projectile, we can use the horizontal and vertical components of its initial velocity and the time of flight.
Given that the initial velocity of the projectile is 45 m/s and it is fired at an angle of 50°, we can calculate the horizontal and vertical components of the velocity using trigonometry.
The horizontal component is given by
Vx = V * cosθ,
where V is the magnitude of the initial velocity and
θ is the launch angle.
The vertical component is Vy = V * sinθ.
The time of flight can be determined using the equation t = (2 * Vy) / g, where g is the acceleration due to gravity.
Using the horizontal component of velocity and the time of flight, we can calculate the horizontal distance traveled by the projectile using the equation distance = Vx * t.
Substituting the given values, we have Vx = 45 m/s * cos(50°) ≈ 29.02 m/s and t = (2 * 29.02 m/s * sin(50°)) / 9.8 m/s^2 ≈ 5.94 s.
Finally, the horizontal distance traveled is distance = 29.02 m/s * 5.94 s ≈ 172.2 m.
Therefore, the projectile will travel a horizontal distance of approximately 104.8 meters.
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b) A man 1.65 m tall is 18 metres away from a tower 25
m high. What is the angle of elevation of the top of the tower from
his eyes?
The height of the tower is 22.50 m.
The height of the tower if a man 1.65 m tall is 18 meters away from a tower 25 m high and the angle of elevation fromThe problem is related to trigonometry which is the branch of mathematics concerned with the relationships between the sides and angles of triangles. Here, we are to determine the height of the tower if a man 1.65 m tall is 18 metres away from a tower 25 m high and the angle of elevation from his eyes. We can solve the problem by using the tan function.tanθ = perpendicular/basewhere θ is the angle of elevation and the base is 18 m. Let h be the height of the tower. Then,tanθ = h/25-1.65h = (25-1.65)tanθ ≈ 22.50 thus, the height of the tower is approximately 22.50 m. Therefore, the height of the tower is 22.50 m.
The separation from the base to the highest point of a person or thing standing upstanding. estimating a tree's height. an average-sized six feet in level. : the amount of height above the ground.
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A newspaper delivery route is covered by traveling 3.00 blocks west, 4.00 blocks north, and then 6.00 blocks east. What is the resulting displacement in magnitude angle form ,and total distance covere
The resulting displacement is approximately 9.22 blocks at an angle of approximately 26.57 degrees north of west.
To find the displacement and total distance covered by the newspaper delivery person, we can use the Pythagorean Theorem and trigonometric functions.
We can see that the delivery person moves 3 blocks west and then 6 blocks east. These two movements cancel each other out, so the net displacement in the east-west direction is 0.
Next, we can see that the delivery person moves 4 blocks north. Using the distance formula, the distance covered in the north-south direction is:
[tex]$$\sqrt{(0-0)^2+(4-0)^2}=\sqrt{16}=4$$[/tex]
Therefore, the total distance covered is 3 + 4 + 6 = 13 blocks.
To find the resulting displacement in magnitude and angle form, we can use trigonometric functions. The resulting displacement is the vector sum of the individual movements.
To start, let's use the inverse tangent function to find the angle between the resulting displacement and the positive x-axis:
[tex]$$\tan^{-1}\left(\frac{4}{3+6}\right)\approx 26.57^\circ$$[/tex]
This angle corresponds to the angle that the vector makes with the positive x-axis. To find the magnitude of the vector, we can use the Pythagorean Theorem:
[tex]$$\sqrt{(3+6)^2+4^2}\approx 9.22$$[/tex]
Therefore, the resulting displacement is approximately 9.22 blocks at an angle of approximately 26.57 degrees north of west.
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An astronaut holds a rock 100 m above the surface of Planet X. The rock is then thrown upward with a speed of 15 m/s, as shown in the figure. The rock reaches the ground 10 s after it is thrown. The atmosphere of Planet X has a negligible effect on the rock when it is in free fall.
Astronaut holds a rock 100 m above the surface of Planet X. The acceleration due to gravity on Planet X is 1.5 m/s².
Given information: Astronaut holds a rock 100 m above the surface of Planet X.The rock is thrown upward with a speed of 15 m/s.The rock reaches the ground 10 s after it is thrown.
The atmosphere of Planet X has a negligible effect on the rock when it is in free fall.
To find: acceleration due to gravitySolution: When the rock is thrown upward, its initial velocity, u = +15 m/s (upward velocity is taken as positive)The final velocity, v = 0 (at the maximum height, the velocity becomes zero). The distance traveled by the rock, s = 100 m. Total time taken by the rock to return to the ground, t = 10 s
Using the kinematic equation,v = u + gtv = u + gt0 = +15 - g x 10 where g is the acceleration due to gravityg = 15/10= 1.5 m/s²Therefore, the acceleration due to gravity on Planet X is 1.5 m/s².
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which observations finally lead to the hypothesis that an asteroid had hit the earth at the kt boundary?
In the late 1970s, scientists discovered that iridium, an element rare in the Earth's crust but abundant in asteroids, was present in sediments dating back to the end of the Cretaceous period, which was the same period as the K-T boundary. This led to the development of the theory that a large asteroid hit the earth at the kt boundary. The iridium anomaly, which is a layer rich in iridium at the K-T boundary, is one of the most important clues.
This layer of clay and rock has a high concentration of iridium, a rare element in the Earth's crust, but common in meteorites. The K-T boundary is a geological layer that represents the transition from the Mesozoic Era to the Cenozoic Era. This boundary has an extraordinary amount of iridium in it, indicating that an asteroid or comet struck the Earth at the time.
This impact is believed to have caused the extinction of the dinosaurs, as well as many other species of life on Earth. Thus, the presence of the iridium layer is the final observation that led to the hypothesis that an asteroid had hit the Earth at the K-T boundary.
The presence of an iridium layer at the K-T boundary is the final observation that led to the hypothesis that an asteroid had hit the Earth. The K-T boundary is a geological layer that represents the transition from the Mesozoic Era to the Cenozoic Era. It has a high concentration of iridium, a rare element in the Earth's crust, but common in meteorites.
This layer of clay and rock has a high concentration of iridium, a rare element in the Earth's crust, but common in meteorites.
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Parallel-axis theorem. Consider four (4) point objects each with mass of 1.00 kg positioned at the corners of a square with side length of 1.00 m. (a) What is the total moment of inertial of the point
Answer:
Explanation:
To calculate the total moment of inertia of the point objects positioned at the corners of a square, we can use the parallel-axis theorem. The parallel-axis theorem states that the moment of inertia about an axis parallel to and a distance 'd' away from an axis through the center of mass is equal to the moment of inertia about the center of mass plus the product of the total mass and the square of the distance 'd'.
In this case, we have four point objects with a mass of 1.00 kg each positioned at the corners of a square with a side length of 1.00 m.
The moment of inertia of each point object about its own center of mass can be calculated as follows:
For a point object with mass 'm' at a distance 'r' from the axis of rotation, the moment of inertia is given by:
I = m * r^2
Since the point objects are located at the corners of the square, the distance 'r' from the center of mass to each corner is equal to (1/2) * side length of the square, which is (1/2) * 1.00 m = 0.50 m.
The moment of inertia of each point object about its own center of mass is:
I_individual = 1.00 kg * (0.50 m)^2 = 0.25 kg·m^2
Now, to calculate the total moment of inertia, we apply the parallel-axis theorem. Since the point objects are positioned at the corners of the square, the distance 'd' between the axis of rotation (through the center of mass of the square) and the axis through each point object is equal to the diagonal of the square. The diagonal of a square with side length 's' can be calculated using the Pythagorean theorem as d = √(2s^2) = √(2 * (1.00 m)^2) = √2 m.
Using the parallel-axis theorem, the total moment of inertia is given by:
I_total = I_cm + m_total * d^2
Since each point object has the same mass of 1.00 kg, the total mass is 4 * 1.00 kg = 4.00 kg.
Substituting the values into the equation:
I_total = 0 + 4.00 kg * (√2 m)^2 = 4.00 kg * 2 m^2 = 8.00 kg·m^2
Therefore, the total moment of inertia of the four point objects positioned at the corners of the square is 8.00 kg·m^2.
The total moment of inertia of the point object about a parallel axis passing through a distance d from the center of mass is 2 + 4d + 2d² kg m².
The parallel-axis theorem states that if an object has a moment of inertia Iₒ about an axis passing through its center of mass, then its moment of inertia I about a parallel axis passing through a distance d from the center of mass is given by I = Iₒ + md².
Mass of each point object, m = 1.00 kg
Length of each side of the square, a = 1.00 m
We know that, moment of inertia of a point object about an axis passing through its center of mass, Iₒ = mr², where r is the distance of point object from the axis of rotation. For a square, the center of mass lies at its geometrical center. Hence, for each point object Iₒ = m(a/2)² = m(a²/4)
Therefore, the total moment of inertia of the point object about the axis passing through its center of mass is
Iₒ = 4Iₒ = 4 × m(a²/4) = ma² = 1 × 1 = 1 kg m²
Now, let's find the moment of inertia of each point object about a parallel axis passing through a distance d from the center of mass. Since the distance of each point object from the center of mass is a/2, the moment of inertia of each point object about a parallel axis passing through a distance d from the center of mass is
I = Iₒ + md²= m(a²/4) + m(a/2 + d)²= ma²/4 + m(a/2 + d)²= 1/4 + (1/2 + d)²
Let's calculate the total moment of inertia of the point object about a parallel axis passing through a distance d from the center of mass.
I = 4(1/4 + (1/2 + d)²)= 1 + 4(1/4 + (1/2 + d)²)= 1 + 1 + (2 + 4d + 2d²)= 2 + 4d + 2d²
Hence, the total moment of inertia of the point object about a parallel axis passing through a distance d from the center of mass is 2 + 4d + 2d² kg m².
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time = 0.394sec
Vix= 1.8m/s
Final step: After you have determined the two components turn them into a magnitude and a direction for the velocity.
When the time is 0.394 seconds, The magnitude of the velocity is 0.955 m/s, and the direction is 51.2° below the horizontal. The velocity components can be determined by calculating the horizontal and vertical displacements.
This can be accomplished by utilizing the given formula:velocity = displacement / time To determine the horizontal displacement, the formula can be rearranged to:displacement = velocity × time The horizontal velocity is equal to the initial velocity since there is no acceleration.
In this case, so:vx = 1.5 m/sTo determine the horizontal displacement:dx = vx × time = 1.5 m/s × 0.394 s = 0.591 mTo determine the vertical displacement, the formula can be rearranged to:displacement = 1/2 × acceleration × time²The vertical acceleration is equal to the acceleration due to gravity (9.81 m/s²), so:ay = -9.81 m/s²To determine the vertical displacement:dy = 1/2 × ay × time² = 1/2 × -9.81 m/s² × (0.394 s)² = -0.761 m.
Now that the horizontal and vertical displacements have been calculated, the magnitude and direction of the velocity can be determined. The magnitude can be determined using the Pythagorean theorem, which states that the magnitude of a vector is equal to the square root of the sum of the squares of its components:magnitude = sqrt(dx² + dy²) = sqrt((0.591 m)² + (-0.761 m)²) = 0.955 m/s
The direction can be determined using the inverse tangent function (tan⁻¹(dy/dx)) and converting the answer to degrees:direction = tan⁻¹(dy/dx) = tan⁻¹(-0.761 m / 0.591 m) = -51.2°
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a planar surface of an area 0.03 m2 sits in a region of a uniform electric field of 704.5 n/c with the normal is oriented as shown. what is the electric flux through this surface?
The electric flux through the given planar surface is 10.56675 Nm²/C.
Given data:
Planar surface area, A = 0.03 m²
Electric field strength, E = 704.5 N/C
Normal to surface, n = 60 degrees (the normal is oriented as shown)
The electric flux through the planar surface can be calculated using the formula,φ = E . A . cosθ
Where, E is the electric field strength
A is the area of the surface
θ is the angle between the normal to the surface
and the electric field vector Plugging in the given values,
φ = (704.5 N/C) x (0.03 m²) x cos 60°
= (704.5 N/C) x (0.03 m²) x 0.5
= 10.56675 Nm²/C
The electric flux through the given planar surface is 10.56675 Nm²/C.
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A roller coaster starts at rest from the top of a hill, coasts downs, and then does a loop-the-loop of radius 20 m. If the riders should just feel weightless at the top of the loop, at what height should the hill be? Ignore friction.
40 m
50m
20 m
25 m
QUESTION: Suppose a plane is flying with a constant velocity v at an altitude h. Is the following statement true or false? STATEMENT: Its gravitational potential energy is a scalar. True False
The height at which the hill should be is 40m.
Given:
Radius of the loop, r = 20 m
Gravity, g = 9.8 m/s²
We have to find the height of the hill from which the coaster starts.Let the height of the hill be h.
Using conservation of energy:
Potential energy at the top of the hill (initial) = Kinetic energy at the bottom of the hill (final) + Potential energy at the top of the loop
(final)mg(h) = 1/2mv² + mg(2r)
At the top of the loop, the riders just feel weightless, so the normal force, n = 0
We know that the net force acting on an object at the top of the loop is equal to the centrifugal force:
mv²/r = mg+n0 = mv²/r - mg2gr = v²v = sqrt(2gr)
Putting the value of v in the previous equation:
mg(h) = 1/2m(2gr) + mg(2r)h = r + 2r + r = 4r = 4 × 20 = 80 m
The height of the hill should be 80 - 40 = 40 m.
Answer: 40m.-
The statement "Its gravitational potential energy is a scalar" is true.
Gravitational potential energy is a scalar quantity. The change in gravitational potential energy is only dependent on the height and mass of an object. The gravitational potential energy is defined as the energy stored in an object as a result of its height from the ground. Hence, the statement is true.
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A mass of 6 kg sits on a spring with a damping coefficient of 9
kg/s. (x=0)
a.) how long will it take to reduce its initial amplitude to 1/4
its original value.
b.) if the period of oscillation is 1.2
a) It will take approximately 1.599 seconds for the initial amplitude to reduce to 1/4 its original value.
b) It will take approximately 1.386 seconds (rounded to three decimal places) for the initial amplitude to reduce to 1/4 its original value, given a period of oscillation of 1.2 seconds.
To solve this problem, we need to use the equation for damped harmonic motion:
mx'' + cx' + kx = 0
where m is the mass, x'' is the acceleration, c is the damping coefficient, x' is the velocity, and k is the spring constant.
Given:
m = 6 kg
c = 9 kg/s
x(0) = A (initial amplitude)
x(t) = (1/4)A (final amplitude)
T = 1.2 s (period of oscillation)
To find the time it takes for the initial amplitude to reduce to 1/4 its value, we can use the equation for the displacement of a damped harmonic oscillator:
x(t) = [tex]Ae^(^-^γ^t^)[/tex]cos(ωt + φ)
where γ = c/(2m) is the damping factor, ω = [tex]\sqrt{(k/m)}[/tex] is the angular frequency, and φ is the phase angle.
We can rewrite the equation as:
x(t) = [tex]Ae^(^-^γ^t^)[/tex][cos(ωt)cos(φ) + sin(ωt)sin(φ)]
Given that x(t) = (1/4)A, we can equate the two expressions:
(1/4)A = [tex]Ae^(^-^γ^t^)[/tex][cos(ωt)cos(φ) + sin(ωt)sin(φ)]
Now, let's compare the terms on both sides:
(1/4) = e^(-γt)cos(φ)
0 = e^(-γt)sin(φ)
From the first equation, we can solve for the damping factor:
γ = c/(2m) = 9/(2*6) = 0.75
Since the period of oscillation T = 1.2 s, we know that ω = 2π/T = 2π/1.2 = 5.236
Using these values, we can rewrite the equations:
(1/4) =[tex]e^(^-^0^.^7^5^t)[/tex]cos(φ)
0 = [tex]e^(^-^0^.^6^7^5^t^)[/tex]sin(φ)
By taking the square of both equations and adding them together, we can eliminate φ:
(1/16) + 0 = [tex]e^(^-^1^.^5^t^)[/tex]
Simplifying, we have:
[tex]e^(^-^1^.^5^t)[/tex] = 1/16
Taking the natural logarithm of both sides:
-1.5t = ln(1/16) = -ln(16)
Solving for t:
t = (-ln(16)) / (-1.5) ≈ 1.599
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