The length of an elastic string is x m when the tension is 8n

Tension refers to the force that is transmitted through a string, rope, wire, or other similar object when it is pulled tight, trying to restore the object to its original, unstretched length. Learn how to solve for the strength of a tension force by using Newton’s Second Law of Motion. Created by David SantoPietro. No, not really. It is considered to be "convention". It is basically something which people universally agree. For example, if your name is Dan, is there any reason for your name to be Dan? No, its a convention and that's how everyone knows you. If you changed your name to Rick, would anything about your looks be different? Nope, its just convention; you can do whatever you want but it's better to work with universally accepted stuff and as a wise man has said - "perfectly balanced, as all things should be." As explained in the video, when the rope has mass, then one section of the rope will be pulling more mass (it will be pulling some rope and also the object) than the section farther from the object. So, close to the object, the rope pulls and exerts force on only the object and a small amount of rope. At the end of the rope (the furthest point from the object) the rope is exerting force on both the mass of the object and all of the rope between the object and the end of the rope. If there's a system wherein we have 2 masses connected by a thread (M and m) and there's one force pulling at one end (F1 at M) and another pulling at the other free end (F2 at m) ...

Hint:When force is applied to an elastic material in the elastic limit of that material then its length/area/volume changes. On removal of force, the material tends to return to its original size.Here we are concerned with one-dimensional string so its length will increase. Formula used: By the mathematical definition of Young’s modulus we can solve the above problem given as: \[Y = \dfrac = 5y - 4x \\ \] The correct answer is option B. Note: Most of the students get confused with the parameter $\Delta l$. It is the change in length i.e. the increased length – original length. It is commonly misinterpreted as the new length when force is applied because of the notation $l^`$ used in many books.

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https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F16%253A_Waves%2F16.04%253A_Wave_Speed_on_a_Stretched_String \( \newcommand\) • • • • • • • Learning Objectives • Determine the factors that affect the speed of a wave on a string • Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media The speed of a wave depends on the characteristics of the medium. For example, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness but may be made of similar material. They have different linear densities, where the linear density is defined as the mass per length, \[\mu = \frac\, kg. \nonumber\] The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed. Wave Speed on a String under Tension To see how the speed of a wave on a string depends on...

More • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • •

https://phys.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fphys.libretexts.org%2FBookshelves%2FUniversity_Physics%2FBook%253A_University_Physics_(OpenStax)%2FBook%253A_University_Physics_I_-_Mechanics_Sound_Oscillations_and_Waves_(OpenStax)%2F16%253A_Waves%2F16.04%253A_Wave_Speed_on_a_Stretched_String \( \newcommand\) • • • • • • • Learning Objectives • Determine the factors that affect the speed of a wave on a string • Write a mathematical expression for the speed of a wave on a string and generalize these concepts for other media The speed of a wave depends on the characteristics of the medium. For example, in the case of a guitar, the strings vibrate to produce the sound. The speed of the waves on the strings, and the wavelength, determine the frequency of the sound produced. The strings on a guitar have different thickness but may be made of similar material. They have different linear densities, where the linear density is defined as the mass per length, \[\mu = \frac\, kg. \nonumber\] The guitar also has a method to change the tension of the strings. The tension of the strings is adjusted by turning spindles, called the tuning pegs, around which the strings are wrapped. For the guitar, the linear density of the string and the tension in the string determine the speed of the waves in the string and the frequency of the sound produced is proportional to the wave speed. Wave Speed on a String under Tension To see how the speed of a wave on a string depends on...

Tension refers to the force that is transmitted through a string, rope, wire, or other similar object when it is pulled tight, trying to restore the object to its original, unstretched length. Learn how to solve for the strength of a tension force by using Newton’s Second Law of Motion. Created by David SantoPietro. No, not really. It is considered to be "convention". It is basically something which people universally agree. For example, if your name is Dan, is there any reason for your name to be Dan? No, its a convention and that's how everyone knows you. If you changed your name to Rick, would anything about your looks be different? Nope, its just convention; you can do whatever you want but it's better to work with universally accepted stuff and as a wise man has said - "perfectly balanced, as all things should be." As explained in the video, when the rope has mass, then one section of the rope will be pulling more mass (it will be pulling some rope and also the object) than the section farther from the object. So, close to the object, the rope pulls and exerts force on only the object and a small amount of rope. At the end of the rope (the furthest point from the object) the rope is exerting force on both the mass of the object and all of the rope between the object and the end of the rope. If there's a system wherein we have 2 masses connected by a thread (M and m) and there's one force pulling at one end (F1 at M) and another pulling at the other free end (F2 at m) ...

Hint:When force is applied to an elastic material in the elastic limit of that material then its length/area/volume changes. On removal of force, the material tends to return to its original size.Here we are concerned with one-dimensional string so its length will increase. Formula used: By the mathematical definition of Young’s modulus we can solve the above problem given as: \[Y = \dfrac = 5y - 4x \\ \] The correct answer is option B. Note: Most of the students get confused with the parameter $\Delta l$. It is the change in length i.e. the increased length – original length. It is commonly misinterpreted as the new length when force is applied because of the notation $l^`$ used in many books.

Views: 5,137 300 m , 150 m and 200 m. Qumetere, 30 cm , 20 cm, and 30 cm respectively. calculate discherge considering all bsses if, wuterlevel difference in 2 tanks is 15 meter. take f = 0.02 , 0.025 and 0.03 respectivut Given. watardifference calculate discharge consider all losses, Inlet + friction + contraction + friction + sudden texit. H = h i ​ + h f 1 ​ ​ + h c + h f 2 ​ + h exp + h f 3 ​ ​ + h exit. ​ . = 2 g 0.5 v 1 2 ​ ​ + h f 1 ​ ​ + 2 g 0.5 v 2 2 ​ ​ + h f 2 ​ ​ + 2 g ( V 2 ​ − V 3 ​ ) 2 ​ + h f 3 ​ ​ + 2 g . V 3 2 ​ ​ ​ Views: 5,101 50 mm diameter is cut out from a circuion from A. shown in Fig. 6.31. Find the centre of gravity of the section from A. 3. The ce distance (a) 2 h ​ Fig. 6.31. Fig. 6.32. 4. A circular hole 2. Find the centre of gravity of a semicircular section having outer and inner diameters of 200 mm and 160 mm respectively as shown in Fig. 6.32. [Ans. 57.5 mm from the base] 3. A circular sector of angle 4 5 ∘ is cut from the circle of radius 220 mm Determine the centre of gravity of the remainder from the centre of the sector. [Ans. 200 mm ] 4. A hemisphere of radius 80 mm is cut out from a right circular cylinder of diameter 80 mm and height 160 mm as shown in Fig. 6.33. Find the centre of gravity of the body from the base A B. [Ans. 77.2 mm ]

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