Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s . The waves reflect from the wall, and the incoming and reflected waves overlap to make a lovely standing wave with an antinode at the wall. (Such waves are a common occurrence in certain places.) A kayaker is bobbing up and down with the water at the first antinode out from the wall.A) How far from the wall is she?B) What is the period of her up and down motion?
Answers
Answer:
a)15m
b)6.25s
Explanation:
A) She is ½ a wavelength away, or
d = λ/2 = 30/2 = 15 m
B)Speed of the wave:
V=fλ = λ/T
so,
T=λ/V= 30/4.8
T=6.25s
a) The distance from the wall is 15m
b) The period of her up and down motion is 6.25s
Calculation of the distance and period is:a.
Since Ocean waves of wavelength 30m are moving directly toward a concrete barrier wall at 4.8m/s .
Also,
She is ½ a wavelength away, or
d = λ/2
= 30/2
= 15 m
b)
Here the speed of wave should be used
T=λ/V
= 30/4.8
T=6.25s
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Related Questions
Bola bermassa 200 gram dilempar
ke bawah dari ketinggian 20 m
dengan kecepatan 2 m/s. Jika
percepatan gravitasi bumi 10
m/s2 energi kinetik pada
ketinggian 8 m adalah ......
Answers
Answer:
0.4
Explanation:
\( \frac{1}{2} mv ^{2} \)
kinetic energy formula , potential energy is not considered
0.5×0.2×2×2
Demario's biology class has a quiz every other Friday. This is the third time Demario has been so worried about other personal matters that he hasn’t done quite as well on the quizzes as he might have otherwise. What has Demario upset is the fact that the professor leaves the room while the students take the quiz, and over half the class is taking the opportunity to cheat. Demario knows personally several of the other students in the class, and some of the ones Demario hangout with are among those who are cheating. Demario knows that a failure to speak to the professor about the cheating will result in his own grade being lower, since the school grades on the curve system. But if Demario does say something to the professor he will be doing his friends and the others a great harm, since cheating is taken very seriously at the school and can lead to expulsion. If Demario doesn’t “turn in” the classmates, the only other alternative is he will get a worse grade than he deserves. He will also be labled as a snitch among is peers and a cheater at the school, either of which Demario has never done. Further, if the professor somehow discovers that Demario is hiding the cheating of others, Demario will be considered an accessory to the cheating, and may be reprimanded for not turning others in, since the school operates on an honor system.
Critical thinking About Ethics Right vs Wrong
What should Demario do?
Should Demario cheat?
Should Demario turn in the cheating classmates?
Should Demario say nothing and not cheat himself?
Be specific in your answer about exactly what Demario should do. (Remember to use one of the three moral theories to solve this dilemma.)
Answers
1. It is unethical for Demario to cheat
2. Demario should not engage in cheating himself
3. Demario should not turn in cheating classmates without careful consideration
4. Reporting classmates may harm relationships, create animosity, and potentially lead to severe disciplinary actions.
What is ethics?A virtue ethics perspective may also be useful in this circumstance. Demario should make an effort to respect moral principles like honesty, fairness, and integrity while taking the potential repercussions into account and exhibiting empathy for his fellow students.
By employing this strategy, Demario behaves responsibly, deals with the problem subtly, encourages fairness in the educational setting, and upholds his or her own moral standards.
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HELP!!! You have two substances, both of which have the same boiling point (or attraction between their molecules).
The first substance is made from molecules that are small (just a few atoms bonded together), and the second substance is made from mol that are larger (many atoms bonded together).
How can it be possible for two such different molecules to yield substances with the same boiling point? Describe the kinds of intermolecu attractions that must be involved and any other properties of the molecules that could cause this result.
Answers
The boiling point of a substance depends on the strength of the intermolecular forces between its molecules. In the case of the two substances you mentioned, even though they have different molecular sizes, they can still have the same boiling point if they have similar intermolecular attractions.
What is the boiling point about?For example, the smaller molecule may have a higher polarity due to the presence of polar bonds, which can result in dipole-dipole interactions between molecules. The larger molecule may have a similar dipole moment even though it has more atoms, or it may have a polar functional group that contributes to its intermolecular interactions.
In addition to dipole-dipole interactions, the molecules may also have London dispersion forces, which are present in all molecules due to the random fluctuations in electron density. These forces are proportional to the size of the molecule, so the larger molecule may have a stronger dispersion force even if it is less polar than the smaller molecule.
Therefore,, it is possible for two different molecules to have the same boiling point if they have similar intermolecular attractions, such as dipole-dipole interactions and London dispersion forces. The size of the molecule may also play a role in determining the strength of these forces, but it is not the only factor.
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i
e. network executives make hasty
When winding an old clock, it is important not to overwind it. Over-
winding occurs when the mainspring is almost fully wound, but the
operator continues to turn the winding key. This causes the main
spring to coil too tight, and might even break it.
110. This paragraph best supports the statement that
a. clocks have changed over the years.
b. old-fashioned clocks become fragile with age.
c. old-fashioned clocks were operated by an internal spring.
d. overwinding clocks used to be a common mistake.
e. time flies when you're having fun.
Answers
The paragraph primarily discusses the concept of overwinding old clocks and its consequences, indicating that overwinding clocks used to be a common mistake. Here option D is the correct answer.
The paragraph explains that overwinding occurs when the mainspring is almost fully wound, but the operator continues to turn the winding key, resulting in the spring coiling too tightly or even breaking.
This suggests that overwinding was a mistake commonly made in the past when operating old-fashioned clocks. The other options, such as clocks changing over the years or clocks becoming fragile with age, are not directly addressed in the paragraph and are therefore less supported.
The option e. "time flies when you're having fun" is unrelated to the paragraph and can be disregarded as an irrelevant answer choice. Hence option D is the correct answer.
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Which statement best describes the circular flow model?
Answers
As part of astronaut training, a prospective astronaut is spun around in a human centrifuge such that the candidate experiences a centripetal acceleration that is 2.5 times the acceleration due to gravity on the surface of the earth. If the candidate is 7.27 m from the center, determine the candidate's speed in meters per second.
Answers
Due to the relationship between linear and angular speed, the candidate's speed is 13.3 m/s
Circular MotionA satellite orbiting the earth is one of the example of a circular motion. The satellite will experience gravitational attraction towards the center of the earth. This attraction force provides the centripetal force
Given that a prospective astronaut is spun around in a human centrifuge such that the candidate experiences a centripetal acceleration that is 2.5 times the acceleration due to gravity on the surface of the earth. If the candidate is 7.27 m from the center, the parameters to work on are
Centripetal acceleration a = 2.5g = 2.5 × 9.8 = 24.5 m/s²Orbital radius r = 7.27 mSpeed v = ?To determine the candidate's speed in meters per second, let us first find its angular speed w in rad/s by using the formula
a = w²r
24.5 = 7.27w²
w² = 24.5 / 7.27
w² = 3.37
w = √3.37
w = 1.84 rad/s
The relationship between linear and angular speed is
v = wr
v = 1.84 × 7.27
v = 13.3 m/s
Therefore, the candidate's speed in meters per second is 13.3 m/s
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two forces f1=(8i+3j)N and f2=(4i+6j) are acting on 5kg object then what is the magnitude and the direction of the resultant force
what is its acceleration of x and y component
what is the magnitude of acceleration of the object
Answers
Two forces f1=(8i+3j)N and f2=(4i+6j) are acting on 5kg object then the magnitude of the resultant force is 15 N and the direction of the resultant force is approximately 36.87 degrees from the positive x-axis.
The acceleration of the object in the x-component (\(a_x\)) is 2.4 \(m/s^{2}\), and the acceleration in the y-component (\(a_y\)) is 1.8 \(m/s^{2}\).
The magnitude of the acceleration of the object is 3 \(m/s^{2}\).
To find the magnitude and direction of the resultant force, we need to add the two given forces together.
Given:
f1 = (8i + 3j) N
f2 = (4i + 6j) N
To find the resultant force (\(F_res\)), we simply add the corresponding components:
\(F_res\) = f1 + f2
= (8i + 3j) + (4i + 6j)
= (8 + 4)i + (3 + 6)j
= 12i + 9j
The magnitude of the resultant force (\(|F_res|\)) can be found using the Pythagorean theorem:
\(|F_res|\)= \(\sqrt{(12^2) + (9^2)}\)
= \(\sqrt{144 + 81}\)
= \(\sqrt{225}\)
= 15 N
So, the magnitude of the resultant force is 15 N.
To find the direction of the resultant force, we can use trigonometry. The direction can be represented by the angle θ between the positive x-axis and the resultant force vector. We can calculate θ using the inverse tangent function:
θ = arctan(9/12)
= arctan(3/4)
≈ 36.87 degrees
Therefore, the direction of the resultant force is approximately 36.87 degrees from the positive x-axis.
Now let's calculate the acceleration of the object in the x and y components. We know that force (F) is related to acceleration (a) through Newton's second law:
F = ma
For the x-component:
\(F_x\)= 12 N
m = 5 kg
Using \(F_x\)= \(ma_x\), we can solve for \(a_x\):
12 N = 5 kg * \(a_x\)
\(a_x\)= 12 N / 5 kg
\(a_x\) = 2.4 \(m/s^{2}\)
For the y-component:
\(F_y\) = 9 N
m = 5 kg
Using \(F_y\) = \(ma_y\), we can solve for \(a_y\):
9 N = 5 kg * \(a_y\)
\(a_y\) = 9 N / 5 kg
\(a_y\)= 1.8 \(m/s^{2}\)
So, the acceleration of the object in the x-component (\(a_x\)) is 2.4 \(m/s^{2}\), and the acceleration in the y-component (\(a_y\)) is 1.8 \(m/s^{2}\).
To find the magnitude of the acceleration (|a|), we can use the Pythagorean theorem:
|a| = \(\sqrt{(a_x^2) + (a_y^2)}\)
= \(\sqrt{(2.4^2) + (1.8^2}\)
= \(\sqrt{5.76 + 3.24}\)
= \(\sqrt{9}\)
= 3 \(m/s^{2}\)
Therefore, the magnitude of the acceleration of the object is 3 \(m/s^{2}\)
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laser light (λ = 568.7 nm) is incident on a single slit. what is the maximum width of the slit (in nm) for which no diffraction minima are observed? round to the closest integer.
Answers
The maximum width of the slit (in nm) for which no diffraction minima are observed is 569 nm.
What is the maximum width of the slit?The maximum width of the slit (in nm) for which no diffraction minima are observed is calculated by applying the following equation.
sinθ = mλ/a
where;
m is = 1 order integerthe maximum angle for diffraction to occur = 90⁰λ is the wavelength = 568.7 nma is the maximum width of the slitSubstitute the given parameters and solve for the maximum width of the slit (in nm) for which no diffraction minima are observed.
sin(90) = (1 x 568.7 nm) / a
1 = (568.7 nm) / a
a = (568.7 nm) /1
a = 568.7 nm
to the closest integer = 569 nm
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What is the magnitude and direction of the electric field intensity when the distance of a point charge from the center of a dipole is very large as compared to the length of the dipole?
Answers
To calculate electric field created by a dipole on the axial line, magnitude and direction of the electric field intensity when the distance of a point charge from the center .
All the measurement of distances are to be taken from the centre(O).
Let the distance between O to +q and O to –q be ‘l’. So, total length between +q and –q will be ‘2l’.
Take a point ‘p’ on the axial line at the distance ‘r’ from the center as shown in figure.
Electric field on the axial line of dipole is not 0
magnitude =E = 14πε0 2Pr2
An electric field is the physical field that surrounds electrically charged particles and exerts force on all other charged particles in the field, either attracting or repelling them.
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One train is at the 20 Km position, another is at the 200Km position, both are approaching, the first with a speed of 30Kmh-1 and the second with a speed of 48 Kmh-1. If they leave at 7:12, what time do they cross? Answer: t = 2.30 H Time: 9:30 am.
Answers
Explanation:
The distance between them is 200 km − 20 km = 180 km.
The relative velocity is 30 km/h − (-48 km/h) = 78 km/h.
The time it takes is 180 km / (78 km/h) = 2.31 hours, or 2 hrs 18 min.
Therefore, the trains meet at 9:30 AM.
A rocket moving around the earth at height "H", If the gravitational acceleration "g1" at height
His of gravitational acceleration 'g at earth surface. If Earth radius is "R", find "H"
using R
Answers
Answer:
At the earth's surface g = G M / R^2 where G is the gravitational constant
at H g1 = G M / (R + H)^2 using Gauss' theorem for enclosed mass
g1 = G M / (R^2 + 2 R H) ignoring H^2 as it is small compared to R^2
g / g1 = (R^2 + 2 R H) / R^2 = 1 + 2 R H
g = g1 + 2 R H g1
g1 - g = - 2 R H or H = (g1 - g) / 2 R
Can someone help me solve this problems please? It's a physics problem.
Answers
Answer:
i cant see
Explanation:
but im smart
A man is standing away from the School
Building at a distance of
300m . He claps his hands and hears an echo calculate the time interval of him hearing his echo
Answers
The time interval between the man clapping and hearing his echo is approximately 1.75 seconds.
What do you mean by echo?An echo is a repetition or reflection of a sound or signal. It can be caused by sound waves bouncing off a surface, signal interference, or the repetition of a message in communication.
The speed of sound in air at room temperature is approximately 343 meters per second. When a person claps, the sound waves propagate outward in all directions and reach the school building, where they bounce off and return to the person as an echo. The time it takes for the sound to travel the distance to the building and back to the person is the time interval between the clap and the echo.
To calculate the time interval, we can use the following formula:
time = distance / speed
where distance is the total distance traveled by the sound (twice the distance from the person to the school building), and speed is the speed of sound in air.
distance = 2 x 300m = 600m
speed = 343 m/s
time = 600m / 343 m/s = 1.75 seconds (rounded to two decimal places)
Therefore, the time interval between the man clapping and hearing his echo is approximately 1.75 seconds.
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Based on the diagram, why does the lightbulb light when the loop rotates, and what is the energy change involved?
When the wire moves in an electric field, electrons in the wire move and become mechanical energy. The mechanical energy causes the light to glow. Electrical energy used to rotate the loop is converted to light energy.
When the wire moves in an electric field, electrons in the wire move and become mechanical energy. The mechanical energy causes the light to glow. Electrical energy used to rotate the loop is converted to light energy.
When the wire moves in a magnetic field, electrons in the wire move and become an electric current. The current causes the light to glow. Mechanical energy used to rotate the loop is converted to electrical energy.
When the wire moves in a magnetic field, electrons in the wire move and become an electric current. The current causes the light to glow. Mechanical energy used to rotate the loop is converted to electrical energy.
When the wire moves in an electric field, electrons in the wire move and become mechanical energy. The mechanical energy causes the light to glow. Mechanical energy used to rotate the loop is converted to electrical energy.
When the wire moves in an electric field, electrons in the wire move and become mechanical energy. The mechanical energy causes the light to glow. Mechanical energy used to rotate the loop is converted to electrical energy.
When the wire moves in a magnetic field, electrons in the wire move and become an electric current. The current causes the light to glow. Mechanical energy used to rotate the loop is converted to light energy.
Answers
Answer:
Based on the information provided, the lightbulb lights when the loop rotates because the movement of the wire in an electric or magnetic field causes electrons in the wire to move and become either mechanical energy or an electric current. This energy causes the light to glow. The energy change involved is the conversion of electrical or mechanical energy used to rotate the loop into either light or electrical energy
Explanation:
if a bus drives 2km due east,5km 45 degrees north of east, 4km at 30 degrees north of west, then 2km due south. what is the bus's resultant displacement?
Answers
The bus resultant displacement is approximately 4.14 km at an angle of 59.5 degrees north of east.
The first movement is 2 km due east, which means it has an x-component of 2 km and a y-component of 0 km.
The second movement is 5 km at 45 degrees north of east. This can be broken down into x and y components using trigonometry: x = 5 km * cos(45) = 3.54 km y = 5 km * sin(45) = 3.54 km
The third movement is 4 km at 30 degrees north of west. This can also be broken down into x and y components using trigonometry: x = -4 km * cos(30) = -3.46 km (negative because it’s towards the west) y = 4 km * sin(30) = 2 km
The fourth movement is 2 km due south, which means it has an x-component of 0 km and a y-component of -2 km.
Adding up all the x and y components, we get: x_total = 2 + 3.54 + (-3.46) + 0 = 2.08 km y_total = 0 + 3.54 + 2 + (-2) = 3.54 km
The magnitude of the resultant displacement can be calculated using the Pythagorean theorem: resultant_displacement = sqrt(x_total^2 + y_total^2) = sqrt(2.08^2 + 3.54^2) ≈ 4.14 km
The direction of the resultant displacement can be calculated using the arctan function: direction = arctan(y_total / x_total) ≈ 59.5 degrees north of east
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Which statement is true about the theory of plate tectonics and the theory of continental drift?
A) The theory of plate tectonics proves the theory of continental drift completely wrong.
B) The theory of plate tectonics tells exactly where the continents were before Pangaea divided.
C) The theory of plate tectonics gives the method by which continents can move as part of plates.
D) The theory of plate tectonics does not explain how continental movements could occur.
Answers
The statement that is true about the theory of plate tectonics and the theory of continental drift C. The theory of plate tectonics gives the method by which continents can move as part of plates .
What is theory of plate tectonics and the theory of continental drift ?According to the scientific hypothesis of plate tectonics, the underground movements of the Earth create the primary landforms. By explaining a wide range of phenomena, including as mountain-building events, volcanoes, and earthquakes, the theory, which became firmly established in the 1960s, revolutionized the earth sciences.
The scientist Alfred Wegener is most closely connected with the concept of continental drift. Wegener wrote a paper outlining his notion that the continents were "drifting" across the Earth, occasionally crashing through oceans and into one another, in the early 20th century.
According to tectonic theory, the Earth's surface is dynamic and can move up to 1-2 inches every year. The numerous tectonic plates constantly move and interact. The outer layer of the Earth is altered by this motion. The result is earthquakes, volcanoes, and mountains.
Therefore, option C is correct.
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Put the following in order from greatest to least.
Centimeter (cm)
Meter (m)
Kilometer (km)
Astronomical Unit (AU)
Answers
Answer:
Astronomical Unit (AU)
Kilometer (km)
Meter(m)
Centimeter(cm)
A 200 kg bumper car travelling at 1 m/s hits a 300 kg bumper car travelling at "-2" m/s. If the velocity of 200 kg bumper car after the collision is "-3" m/s and the force exerted on the 200 kg bumper car was 500 N, what was the magnitude of the force exerted on the 300 kg bumper car?
A.) 200 N
B.) 500 N
C.) 750 N
D.) 333N
Answers
The magnitude of the force exerted on the 300 kg bumper car is 500 N.
option B.
What is Newton's third law of motion?
Newton's third law of motion states that for every action force, there is an equal and opposite reaction force. That is action force and reaction force are equal in magnitude but opposite in direction.
Mathematically, the formula for Newton's third law of motion is given as;
Fa = - Fb
where;
Fa is the force exerted on 200 kg bumperFb is the reaction force of 300 kg bumperFb = - 500 N
| Fb | = 500 N
Thus, the force exerted by 200 kg bumper is to the force exerted by 300 kg bumper.
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. Two astronauts are 1.90 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electromagnetic waves. The time it takes for sound waves to travel at 346 m/s through the air between the astronauts equals the time it takes for the electromagnetic waves to travel to the earth. How far away from the earth is the spaceship
Answers
Answer:
d= 1650 km.
Explanation:
Assuming that sound waves travel along a straight line at a constant speed of 346 m/s, we can find the time needed for the sound to travel the distance of 1.9 m between the astronauts, just applying the definition of average velocity, as follows:\(t = \frac{\Delta x}{v} =\frac{1.9m}{346m/s} = 5.5 msec (1)\)
The electromagnetic waves travel in free space at the same speed of light in vacuum ( since light is a electromagnetic wave indeed), i.e., 3*10⁸ m/s.Applying the same formula than in (1) we can solve for the distance d, as follows:\(d = v*t = 3e8 m/s*5.5e-3s = 1650 km (2)\)
A force of 42 N changes the length of Elastic X by 1.2 m while a force of 24 N changes the length of Elastic Y by 2.1
m. Calculate the work done on both Elastic X and Elastic Y. Solve this problem by showing your work in an
organized step-by-step fashion and include proper significant digits. (3 marks)
Answers
The work done on both Elastic X and Elastic Y is 25.2 J.
Work done on elastic XThe work done on the Elastic X is calculated as follows;
W = ¹/₂fx
where;
f is the applied force to material Xx is the extension of material XW = ¹/₂(42)(1.2)
W = 25.2 J
Work done on elastic YThe work done on the Elastic Y is calculated as follows;
W = ¹/₂fx
where;
f is the applied force to material Yx is the extension of material YW = ¹/₂(24)(2.1)
W = 25.2 J
Thus, the work done on both Elastic X and Elastic Y is 25.2 J.
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How dose the Sum Of Forces help you
Answers
Answer:
The purpose of adding force vectors is to determine the net force acting upon an object. In the above case, the net force (vector sum of all the forces) is 0 Newton. ... We would say that the object is at equilibrium. Any object upon which all the forces are balanced (Fnet = 0 N) is said to be at equilibrium.
Explanation:
hope that helps you
i didnt want my question public i made a mistake i want it taken down
Answers
When a star starts with nuclear fussion, what is the composition of its core
Answers
A 175,000 kg space probe is landing on an alien planet with a gravitational acceleration of 8.25. If its fuel is ejected from the rocket motor at 35,000 m/s what must the mass rate of change of the space ship (delta m)/(delta t) be to achieve at upward acceleration of 2.00 m/s^2? Remember to use the generalized form of Newton's Second Law.
answer with correct units
Answers
The mass rate of change of the space probe is approximately 28.49 kg/s .
What is the mass rate of the space probe?To solve this problem, we can use the generalized form of Newton's Second Law, which states that the force acting on an object is equal to its mass times its acceleration:
F = ma
In this case, the force acting on the space probe is the thrust force generated by the rocket motor, which is equal to the rate of change of momentum of the ejected fuel:
F = (Δ m /Δt) * v
where;
Δ m /Δt t is the mass rate of change of the space ship, and v is the velocity of the ejected fuel, which is given as 35,000 m/s.Since the space probe is landing on the planet, the net force acting on it should be equal to the force of gravity pulling it down minus the upward thrust force generated by the rocket motor. So we can write:
F_net = m * g - (Δ m /Δt) * v
Plugging in the values and solving for delta m / delta t, we get:
2.00 m/s² = (175,000 kg * 8.25 m/s²) - (Δ m / Δt) * 35,000 m/s
Δ m / Δt = (175,000 kg * 8.25 m/s² - 2.00 m/s² * 35,000 m/s) / 35,000 m/s
Δm / Δt ≈ 28.49 kg/s
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the efficiency of a machine cannot be 100 percent why?
Answers
Answer:
A machine cannot be 100 percent efficient because output of a machine is always less than input. A certain amount of work done on a machine is lost to overcome friction and to lift some moving parts of the machine
friction and heat and wearing down and fuel sources problems are why machine cannot be 100% efficient.
Hope this helps...
While tying his shoes, a person holds the
shoestrings at the angles indicated in the
illustration below and pushes downward
with a force of 20 lbs. Use the
component method to determine the
tensions T1 and T1 in each shoestring.
Verify your answer using the force table.
Answers
To find the tensions T1 and T2 in the shoestrings, we use the component method and set up equations for the x- and y-components of the forces. By assuming that T1 and T2 have the same magnitude, we can solve for T and find that T1 and T2 are both approximately 8.66 lbs. To verify this, with weights representing the forces and confirm that the table balances.
Let T1 and T2 be the tensions in the shoestrings. Since the person pushes downward at an angle of 60 degrees to the right with a force of 20 lbs, we can represent this force as a vector F with components
Fx = 20 cos(60) = 10 lbs
Fy = -20 sin(60) = -17.32 lbs (negative because it points downwards)
The tensions in the shoestrings can be represented by vectors T1 and T2. Since the shoestrings are tied to the shoes, the tensions must be perpendicular to the shoes. Therefore, we can represent T1 and T2 as follows
T1x = T1 cos(90) = 0
T1y = T1 sin(90) = T1
T2x = T2 cos(90) = 0
T2y = T2 sin(90) = T2
Using the component method, we can add the x-components and y-components of the forces to get
ΣFx = T1x + T2x + Fx = 0 + 0 + 10 = 10 lbs
ΣFy = T1y + T2y + Fy = T1 + T2 - 17.32 = 0
Since the shoestrings are tied together at the shoes, we know that T1 and T2 have the same magnitude
T1 = T2 = T
Substituting this into the second equation, we get
T + T - 17.32 = 0
Solving for T, we get
T = 8.66 lbs
Therefore, the tensions in the shoestrings are both approximately 8.66 lbs.
To verify this, we can set up the forces on the table to match the forces acting on the shoestrings. We can use a weight of 20 units to represent the force applied by the person, and two weights of 8.66 units each to represent the tensions in the shoestrings.
We can arrange the weights so that one tension weight is at an angle of 60 degrees to the right and the other is at an angle of 30 degrees to the left. If the forces are in equilibrium, the force table should balance. By adjusting the weights, we can confirm that the force table does indeed balance, verifying our calculated tensions.
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--The given question is incomplete, the complete question is given
" While tying his shoes, a person holds the
shoestrings at the angles at 60 degree to right and pushes downward
with a force of 20 lbs. Use the
component method to determine the
tensions T1 and T1 in each shoestring.
Verify your answer "--
A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.7 rad/s in 2.9 s
Answers
Answer:
The torque exerted on the merry-go-round is 766.95 Nm
Explanation:
Given;
mass of the merry-go-round, m = 416 kg
radius of the disk, r = 1.7 m
angular speed of the merry-go-round, ω = 3.7 rad/s
time of motion, t = 2.9 s
The torque exerted on the merry-go-round is calculated as;
\(\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm\)
Therefore, the torque exerted on the merry-go-round is 766.95 Nm
What fraction of the total kinetic energy is rotational for the following objects rolling without slipping on a horizontal sur- face? (a) a uniform solid cylinder; (b) a uniform sphere; (c) a thin- walled, hollow sphere; (d) a hollow cylinder with outer radius R and inner radius R>2.
Answers
The ratio of their rotational kinetic energies is 4/5 or 0.8.
Let's denote the mass and radius of the cylinder and sphere as "m" and "r", respectively. At the top of the incline, both objects have only potential energy, which is then converted to kinetic energy. At the bottom of the incline, both objects have both translational and rotational kinetic energy.
For a uniform solid cylinder, the rotational inertia is\(1/2 * m * r^2\). For a uniform sphere, the rotational inertia is\(2/5 * m * r^2\). Therefore, the ratio of their rotational kinetic energies is:
(rotational kinetic energy of sphere) / (rotational kinetic energy of cylinder)
\(= (2/5 * m * r^2 * (v/r)^2) / (1/2 * m * r^2 * (v/r)^2)\)
= (4/5)
Therefore, rotational kinetic energy of sphere is 80% of rotational kinetic energy of the cylinder at bottom of the incline.
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--The complete Question is, A uniform solid cylinder and a uniform sphere with the same mass and radius are released from rest at the top of an incline. They both roll without slipping down the incline and reach the bottom with the same translational speed. What is the ratio of their rotational kinetic energies at the bottom of the incline?--
which of the following statements is true? During heat flow, energy is converted to matter.
Answers
Answer:
During heat flow, much of the energy is dissipated and cannot be used for useful work.
Explanation:
Which of the following statements is true?
During heat flow, much of the energy is lost.
During heat flow, energy is converted to matter.
During heat flow, much of the energy is dissipated and cannot be used for useful work.
a 1.8-cmcm-tall object is 22 cmcm to the left of a lens with a focal length of 11 cmcm . a second lens with a focal length of -5 cmcm is 37 cmcm to the right of the first lens.
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A lens with an 11 cm focal length is 22 cm to the left of a 1.8 cm tall object. The image's height has decreased by 0.25 pixels when a second lens with a focal length of -5cm is placed 37cm to the right of the first lens.
Let's determine the distance at which the image is formed from the first lens since the image of the first lens serves as a virtual object for the second lens. Use the lens manufacturer's formula:
1/o + 1/i = 1/f, where I am the image distance and o is the object distance,
1/i = 1/f - 1/o = 1/(11cm) - 1/(22cm) = 0.045 1/cm, so
i = 22cm
the virtual object for the second lens is left at a distance of 37 cm - 22 cm, or 15 cm, from the second lens. Utilize the lens manufacturer's formula once more:
1/i = 1/f - 1/o (This time, the object distance is the separation between the second lens and the virtual object.)
1/i = 1/(-5cm) - 1/(15cm) = -0.26 1/cm, so
i = -15/4cm away from the second lens.
The sum of the two magnifications from the two lenses is the total magnification: m = m1m2, where m = -i/o, so
m = (-22 cm/22 cm)( 15/4cm/15cm) = -0.25, Consequently, the image's height is 0.25 pixels shorter than it was before.
It is the inverse of the system's optical power that determines how strongly an optical system diverges or converges light: the focal length. The convergence or divergence of light is indicated by a positive focal length, whilst the opposite is true for a negative focal length.
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