6. Differentiate Young's modulus from Bulk modulus?
1. 6. Differentiate Young's modulus from Bulk modulus?
Answer:
The basic difference between young and bulk modulus is shear modulus is that young modulus is ratio of tensile stress to tensile strain ,bulk modulus is ratio of volumetric stress to volumetric strain and shear modulus is ratio of shear stress to shear strain .
Explanation:
brainliest when?
2. unit of young modulus
Answer:
Young's modulus = stress/strain = (FL0)/A(Ln − L0). This is a specific form of Hooke's law of elasticity. The units of Young's modulus in the English system are pounds per square inch (psi), and in the metric system newtons per square metre (N/m2).
-honeyhoy
pahingi po ng ng iyong pag mamahal
3. Calculate for the following.Find the speed of sound in a bone whose density is 6,000kg/m^3 and whose Young’s modulus is 6.4x10^10 N/m^2.
Answer:
Ambot nimo Malay Kuba I'm not sure
4. A rod of length 1.5 m and of radius 0.4mm is stretched by 1.2mm on loading. If the Young's modulus of its material is 12.5x1010 N/m2. Find the stretching force.
Answer:
tignan mo nalang po sa pic
5. What is the speed of a sound such as a steel with a density of 9,000 kg/m3 and a young's modulus of 150 x 109 N/m2 Please answer ,I need it now. Be Honest.
Answer:
v = 4.08 × 10³ m/s
Explanation:
v = √(Y / ρ)
v = √(150 × 10⁹ N/m² / 9000 kg/m³)
v = 4.08 × 10³ m/s
#CarryOnLearning
6. 1. An iron bar is truck at one end by a rubber mallet, producing a sound wave. The Young’s modulus of iron is 21 x 1010 N/m2, and the density it’s is 7.8 x 103 kg/m3.
[tex]\green{\rule{1000pt}{900000pt}}[/tex]
Answer:
2 po
Explanation:
pa brainlest po ty
7. 1. Find the speed of sound in a steelrod assuming that the Young'smodulus is 2.2x1010 N/m2 andthe density is 1,000 kg/m3.
Explanation:
Sound
Sound waves are longitudinal. Our ears can hear the range of frequencies from 20Hz to 20,000Hz. This range is called the audible range. Frequencies above this range is called Ultrasonic that cats and dogs can hear. Ultrasound has medical application. By sending ultra sound wave to internal parts of body and analyzing the reflected sound waves, it is possible to create the image of a desired part by some imaging conversion process. Very high frequency (109Hz) ultrasound generated by electronic stimulation of quartz crystals is used in acoustic microscopy for generating sharp images.
The motion of sound waves in a gas, such as air, depends on the gas pressure and density. When a person speaks, the vibrations of his or her vocal cords cause air molecules to vibrate as well. Vibrations, at their best, propagate in air on spheres that expand and contract repeatedly transmitting their energy to bigger and bigger spheres thereby transmitting the sound energy via longitudinal waves. The shape of transmission in its ideal case is spherical and the sound energy propagates isotropically throughout the medium. Isotropic means same properties in all directions. We know that this is ideal and if you are standing in front of a speaker you hear it louder than standing at other places around it at the same distance. This means that the sound propagation from a speaker is directional and therefore, anisotropic.
The mechanism is that when a sphere of molecules expands, it creates a slight increase in pressure, ΔP, outward while leaving a slight relative vacuum, -ΔP, inward (or behind). This ripple in pressure causes a ripple in density that gets transmitted layer by layer, and the result is the transmission of sound waves.
Speed of Sound
The same way speed of waves in a string depends on the square roots of tensile force in the string and its mass per unit length, the speed of sound waves in a fluid depends on the square roots of the bulk modulus, B, of the fluid as well as its mass density, ρ. The bulk modulus of elasticity of a fluid is a measure of how compressible that fluid is when under a certain pressure. B is defined as the ratio of -ΔP over (ΔV / V). This simply calculates the change in pressure as a result of a relative change in volume. The Metric unit of B is N/m2. We may mathematically write:
Example 1:
The bulk modulus of elasticity of air is 1.41x105 N/m2 and its mass density is 1.29kg/m3. Calculate the speed of sound waves in air.
Solution: Using the formula v = (B / ρ )0.5, we get: v = [1.41x105 / 1.29] m/s = 331 m/s. (Speed of sound in air at STP)
Hope You Find This Example Useful!
:D
8. Find the speed of sound in a bone whose density is 6,000kg/m^3 and whose Young’s modulus is 6.4x10^10 N/m^2.
Answer:
huh? sorry i didnt understand im very sorry
9. What is shear modulus
Explanation:
just press the picture above
10. Find the speed of sound in a rod that has a density of 500 kg/m3 and a young's modulus of 1.2x10 5th power N/m2.
Answer:
wait lang po solving muna
11. Determine the density of an aluminum rod if the speed of sound through it is 5 100 m/s. Young's modulus 7 x 10 N/m.
Answer:
The Formula is y ÷ v²
(young modulus ÷ velocity²)
(7x10^10 N/m^2 ÷ 5100² m/s)
Explanation:
12. an iron bar is struck at one end by a rubber mallet producing a sound wave. the young's modulus of iron is 21x10
Answer:
210
Explanation:
basta di ko Ren nagets eh
13. A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0 kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young’s modulus for nylon?
Explanation:
To determine Young's modulus for nylon, you will need to use the formula for Young's modulus, which is:
Young's modulus = (stress) / (strain)
In this problem, the stress is the force acting on the rope due to the weight of the climber, and the strain is the elongation of the rope.
To find the stress, you need to first determine the weight of the climber in Newtons. To do this, you can use the formula for weight, which is:
weight = mass * acceleration due to gravity
The acceleration due to gravity on Earth is approximately 9.8 m/s^2. Therefore, the weight of the climber in Newtons is:
weight = 65.0 kg * 9.8 m/s^2 = 637 N
The rope has a cross-sectional area of approximately 0.000386 m^2, which you can calculate using the formula for the area of a circle:
area = pi * radius^2
where the radius is the radius of the rope, which you can find by dividing the diameter by 2.
Next, you can determine the stress by dividing the weight of the climber by the cross-sectional area of the rope:
stress = (weight) / (cross-sectional area)
= (637 N) / (0.000386 m^2)
= approximately 1.65 x 10^7 Pa
To find the strain, divide the elongation of the rope by its original length:
strain = (elongation) / (original length)
= (1.10 m) / (45.0 m)
= approximately 0.024
Now that you have both the stress and the strain, you can use the formula for Young's modulus to find the Young's modulus for nylon:
Young's modulus = (stress) / (strain)
= (1.65 x 10^7 Pa) / (0.024)
= approximately 6.9 x 10^8 Pa
This is the Young's modulus for nylon under the given conditions.
#CarryOnLearning14. Quiz #1 A steel bar is clamped to two vertical supports at 18 degrees Celsius. The coefficient of linear expansion for Aluminum is 11x10-6/C and the Young's Modulus of Steel is 200 x 10⁹ N/m². (a) Calculate the thermal stress on the steel bar at -30 degrees Celsius. (b) Calculate the thermal strain. (c) Find the Modulus of Elasticity of Steel.
Answer:
(a) To calculate the thermal stress on the steel bar at -30 degrees Celsius, we can use the formula:
σ = EαΔT
where σ is the thermal stress, E is the Young's Modulus of Steel, α is the coefficient of linear expansion for Aluminum, and ΔT is the change in temperature.
Since the steel bar is clamped to the vertical supports, it is prevented from expanding or contracting freely. Therefore, any change in temperature will result in a thermal stress being developed in the bar.
Using the given values, we have:
ΔT = (-30°C) - (18°C) = -48°C
σ = (200 x 10^9 N/m²) x (11 x 10^-6 /°C) x (-48°C)
= -105.6 N/m²
Therefore, the thermal stress on the steel bar at -30 degrees Celsius is -105.6 N/m².
(b) To calculate the thermal strain, we can use the formula:
ε = αΔT
where ε is the thermal strain, α is the coefficient of linear expansion for Aluminum, and ΔT is the change in temperature.
Using the given values, we have:
ε = (11 x 10^-6 /°C) x (-48°C)
= -0.000528
Therefore, the thermal strain is -0.000528.
(c) To find the Modulus of Elasticity of Steel, we can rearrange the formula for thermal stress:
E = σ/αΔT
Using the values we calculated earlier, we have:
E = (-105.6 N/m²)/(11 x 10^-6 /°C x -48°C)
= 180 x 10^9 N/m²
Therefore, the Modulus of Elasticity of Steel is 180 x 10^9 N/m².
15. Fractional change in Volume of oil is 1 percent. when a pressure of 2x10^7 Nm^-2 is applied. Calculate the bulk modulus and its compressibility:
Answer:
#Work Hard
Brainliest me pls
thank you ^w^
16. A sodium silicate glass has no surface defects as etching has removed them. Calculate the Fracture Toughness using Griffith formula for that glass. Having Young’s modulus = 70 GNm-2, surface energy Ɣs= 0.56 Jm-2and Fracture Strength: 900 MNm-2.
Answer:
Dikonmaintindahn
Explanation:
wala me alam dyan
kayabmo nayan ah
God bless you
makapasa ka sana hah
gdd
17. Calculate the speed of sound through a platinum rod. The Young's modulus of platinum is 168 x 10° N/m' and its density is 21.4 x 103 kg/m°. Pa answer po ASAP
Answer:
44022
103
× 214
_________
412 *may Keri sa 0 *1*
301
206
____________________
44022
18. Estimate the speed of sound in solid such as steel (p= 8000kg/m3) which has a Young's modulus of 200X10 9 N/m2
Answer:
Correct option is
C
2×10
−4
s
velocity of sound v=
3
y
=
8000
2×10
11
=
2
10
4
Time taken to traverse 1 m, t=
v
d
=
2
10
4
1
=2×10
−4
s
19. modulus and elasticity
Answer:
An elastic modulus is a quantity that measures an object or substance's resistance to being deformed elastically when a stress is applied to it. The elastic modulus of an object is defined as the slope of its stress–strain curve in the elastic deformation region: A stiffer material will have a higher elastic modulus.
20. bolk modulus of water
Answer:
The bulk modulus is a factor in the speed of seismic waves from earthquakes. A common statement is that water is an incompressible fluid.
Explanation:
21. A nylon rope used by mountaineers elongates 1.10 m under the weight of a 65.0 kg climber. If the rope is 45.0 m in length and 7.0 mm in diameter, what is Young’s modulus for nylon?
Explanation:
To determine Young's modulus for nylon, you will need to use the formula for Young's modulus, which is:
Young's modulus = (stress) / (strain)
In this problem, the stress is the force acting on the rope due to the weight of the climber, and the strain is the elongation of the rope.
To find the stress, you need to first determine the weight of the climber in Newtons. To do this, you can use the formula for weight, which is:
weight = mass * acceleration due to gravity
The acceleration due to gravity on Earth is approximately 9.8 m/s^2. Therefore, the weight of the climber in Newtons is:
weight = 65.0 kg * 9.8 m/s^2 = 637 N
The rope has a cross-sectional area of approximately 0.000386 m^2, which you can calculate using the formula for the area of a circle:
area = pi * radius^2
where the radius is the radius of the rope, which you can find by dividing the diameter by 2.
Next, you can determine the stress by dividing the weight of the climber by the cross-sectional area of the rope:
stress = (weight) / (cross-sectional area)
= (637 N) / (0.000386 m^2)
= approximately 1.65 x 10^7 Pa
To find the strain, divide the elongation of the rope by its original length:
strain = (elongation) / (original length)
= (1.10 m) / (45.0 m)
= approximately 0.024
Now that you have both the stress and the strain, you can use the formula for Young's modulus to find the Young's modulus for nylon:
Young's modulus = (stress) / (strain)
= (1.65 x 10^7 Pa) / (0.024)
= approximately 6.9 x 10^8 Pa
This is the Young's modulus for nylon under the given conditions.
#Hopethishelps22. A sodium silicate glass has no surface defects as etching has removed them. Calculate the Fracture Toughness using Griffith formula for that glass. Having Young’s modulus = 70 GNm-2, surface energy Ɣs= 0.56 Jm-2and Fracture Strength: 900 MNm-2.
Answer:
i dunno
Explanation:
23. An object weighs 4.5 x 10 5 N is supported by steel cable, with cross-sectional area of 6.0 cm 2, the Young’s modulus of cable is 2.0 x 1011 N/m 2. Calculate the elongation of the cable if the weight hangs 30 m from support.
Answer:
sorry I didn't get a chance to meet you at
24. 1. An iron bar is truck at one end by a rubber mallet, producing a sound wave. The Young’s modulus of iron is 21 x 1010 N/m2, and the density it’s is 7.8 x 103 kg/m3.
Answer:
Nalilito ako sa question na ito
Explanation:
Sorry po hindi ko talaga alam ang answer
25. Complete the following statement: young's modulus cannot be applied to
claim is the most important part of the text or it is also called as the
central argument or thesis statement of the text. It is a sentence that summarizes
the most important thing that the writer wants to say as a result of his/her
thinking, reading, or writing. Claims in texts are significant proposi
26. A 2.00m long and 3.00 cm radius copper rod experiences a tensional load of 45.6 N and increase its length to 2.20m. What is its young’s modulus of elasticity?
Answer:
Fish Cake
Step-by-step explanation:
Wash and peel oranges. Remove the white membrane and cut each orange to 2. Remove the seeds with the help of a fork. Add them to a blender jar or juicer chute. If using ginger, peel the skin and rinse well. If using herbs, spray some vinegar and salt. Freeze your oranges for an hour if you prefer cold juice.
27. what is the bulk modulus of water where the speed of sound is 1450 m/s
Answer:
Sound
Sound waves are longitudinal. Our ears can hear the range of frequencies from 20Hz to 20,000Hz. This range is called the audible range. Frequencies above this range is called Ultrasonic that cats and dogs can hear. Ultrasound has medical application. By sending ultra sound wave to internal parts of body and analyzing the reflected sound waves, it is possible to create the image of a desired part by some imaging conversion process. Very high frequency (109Hz) ultrasound generated by electronic stimulation of quartz crystals is used in acoustic microscopy for generating sharp images.
The speed of sound waves in a gas, such as air, depends on the gas pressure and density. When a person speaks, the vibrations of his or her vocal cords cause air molecules to vibrate as well. Vibrations, at their best, propagate in air on spheres that expand and contract repeatedly transmitting their energy to bigger and bigger spheres thereby transmitting the sound energy via longitudinal waves. The shape of transmission in its ideal case is spherical and the sound energy propagates isotropically throughout the medium. Isotropic means same properties in all directions. We know that this is ideal and if you are standing in front of a speaker you hear it louder than standing at other places around it at the same distance. This means that the sound propagation from a speaker is directional and therefore, anisotropic.
The mechanism is that when a sphere of molecules expands, it creates a slight increase in pressure, ΔP, outward while leaving a slight relative vacuum, -ΔP, inward (or behind). This ripple in pressure causes a ripple in density that gets transmitted layer by layer, and the result is the transmission of sound waves.
Speed of Sound
The same way speed of waves in a string depends on the square root of tensile force in the string as well as its mass per unit length, the speed of sound waves in a fluid depends on the square root of the bulk modulus B of the fluid as well as its mass density ρ. The bulk modulus of elasticity of a fluid is a measure of how compressible that fluid is when under a certain pressure. B is defined as the ratio of -ΔP over ΔV/V. This simply calculates the change in pressure as a result of a relative change in volume. The SI unit of B is N/m2. We may mathematically write:
Explanation:
28. Find the speed of sound in a wood whose density is 55 kg/m and whose Young's modulus is55,000 N/m
Answer:
1000 m/s²
Step-by-step explanation:
55,000÷55
Sana correct po
Delete if wrong
Pa brainliest po kung correct ako
29. A 15-cm-long tendon was found to stretch “[C]”mm by a force of “[H]”N. The tendon was approximately round with a average diameter of 8.5mm. Calculate Young’s modulus of this tendon.
Answer:
sesamoid bones
Tiny,round,speck of bone found in a tendon are called sesamoid bones.
30. A 90-kg mountain climber hangs from a nylon rope and stretches it by 25.0 cm. If the rope was originally 30.0 m long and its diameter is 1.0 cm, what is Young’s modulus for the nylon?
Answer:
ehh sorry po di k po alam p
