A current of 6.67 A in a solenoid of length 13.6 cm creates a 0.373 T magnetic field at the center of the solenoid. How many turns does this solenoid contain?
Answer:
The value is [tex]N = 6051 \ turns[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 6.67 \ A[/tex]
The length is [tex]l = 13.6 \ cm = 0.136 \ m[/tex]
The magnetic field is [tex]B = 0.373 \ T[/tex]
Gnerally the number of turn per unit length is mathematically represented as
[tex]n = \frac{B}{\mu_o * I}[/tex]
Here [tex]\mu_o[/tex] is the permeability of free space with value [tex]\mu_o = 4*\pi*10^{-7}[/tex]
So
=> [tex]n = \frac{0.373}{ 4\pi * 10^{-7} * 6.67 }[/tex]
=> [tex]n = 44496 \ turns /m[/tex]
Generally the number of turn is mathematically represented as
[tex]N = n * l[/tex]
=> [tex]N = 44496 * 0.136[/tex]
=> [tex]N = 6051 \ turns[/tex]
The no of turns dos this solenoid contains will be 6051 turns.
What is a solenoid?A coil of wire that carries an electric current is a solenoid. A solenoid is an electromagnet formed by a helical coil of wire. Which generates a magnetic field when an electric current is passed through the coil .
The given data in the problem is;
B is the current =6.67 A
L is the length =13.6 cm=0.136 m
B is the magnetic field =0.373 T
The number of turns per unit length is found as;
[tex]\rm n= \frac{B}{\mu_0I} \\\\ \rm n= \frac{0.373}{4 \times 3.14 \times 10^{-7}\times 6.67} \\\\ \rm n = 44496 \ turn /m[/tex]
The number of turns is found as;
[tex]\rm N=n\times l \\\\ \rm N=44496\times 0.136 \\\\ \rm N=6051 \ turns[/tex]
Hence the no of turns dos this solenoid contains will be 6051 turns.
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A dentist causes the bit of a high-speed drill to accelerate from an angular speed of 1.34 x 104 rad/s to an angular speed of 4.37 x 104 rad/s. In the process, the bit turns through 3.10 x 104 rad. Assuming a constant angular acceleration, how long would it take the bit to reach its maximum speed of 7.43 x 104 rad/s, starting from rest?
Answer:
2.66 s
Explanation:
Given The initial angular speed ([tex]\omega_o[/tex]) = 1.34 x 10⁴ rad/s, the final angular speed (ω) = 4.37 x 10⁴ rad/s, the angular acceleration (α), distance (θ) = 3.10 x 10⁴ rad.
Using the formula:
[tex]\omega^2=\omega_o^2+2\alpha \theta\\\\Substituting:\\(4.37*10^4)^2=(1.34*10^4)^2+2(3.1*10^4)\alpha\\\\(4.37*10^4)^2-(1.34*10^4)^2=2(3.1*10^4)\alpha\\\\2(3.1*10^4)\alpha=1.73*10^9\\\\\alpha=2.79*10^4\ rad/s^2[/tex]
The constant acceleration = 2.79 * 10⁴ rad/s.
Since there is constant acceleration, to reach a final speed of 7.43 x 10⁴ rad/s, we use the formula:
[tex]\omega =\omega_o +\alpha t\\\\\omega_o=0,\omega=7.43*10^4\ rad/s,\alpha=2.79*10^4\ rad/s^2\\\\Substituting:\\\\7.43*10^4=0+2.79*10^4t\\\\7.43*10^4=2.79*10^4t\\\\t=2.66\ s[/tex]
What happens to the acceleration of a system when the mass of the system increases but the net force stays constant?
Explanation:
acceleration is inversely proportional to mass so acceleration will decrease if mass increases
If the mass of the system will increase, then the acceleration of the system will decrease because mass and acceleration are inversely proportional to each other.
What is Acceleration?In mechanics, acceleration refers to the rate at which the velocity of an object changes in relation to time. The magnitude of accelerations as a vector. The acceleration of an object depends on the direction of the net force acting on it.
It is a vector quantity because acceleration has both a magnitude and a direction. Additionally, velocity is a vector quantity. The change in the vector of velocity during a period of time divided by the period of time is the definition of acceleration.
There are several types of acceleration :
Uniform AccelerationNon-Uniform AccelerationAverage AccelerationWhen the mass of the system will increase it will become heavy, and it will affect the acceleration of the body because due to this the acceleration of the body will decrease because both mass and acceleration are inversely proportional to each other.
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What is the wavelength emitted by compound BB that have spectral line detected at position 12.15 cmcm
Complete question is;
A chemist identifies compounds by identifying bright lines in their spectra. She does so by heating the compounds until they glow, sending the light through a diffraction grating, and measuring the positions of first-order spectral lines on a detector 15.0 cm behind the grating. Unfortunately, she has lost the card that gives the specifications of the grating. Fortunately, she has a known compound that she can use to calibrate the grating. She heats the known compound, which emits light at a wavelength of 461 nm, and observes a spectral line 9.95 cm from the center of the diffraction pattern.
What is the wavelength emitted by compound B that have spectral line detected at position and 12.15 cm?
Answer:
λ ≈ 525 nm
Explanation:
The diffraction grating equation of n-th order is given by;
nλ = dsinθₙ
For 1st order, n=1. Thus;L, we have;
λ = dsinθ₁
We are told that the first-order spectral lines on a detector 15.0 cm behind the grating. Also that, she observes a spectral line 9.95 cm from the center of the diffraction pattern.
Therefore, the angle of the known compound's 1st order is given by the relation:
tanθ₁ = 9.95/15
tanθ₁ = 0.6633
θ₁ = tan⁻¹(0.6633)
θ₁ = 33.56º
We are told that the unknown compound emits light at a wavelength of 461 nm.
Thus, from the first order equation above, we have;
d = λ/sinθ₁
d = 461nm/sin(33.56º)
d = 461nm/0.55281
d ≈ 834nm
λ = dsinθ₁= 870nm sin(29.68º) = 431nm
Now, spectral line is given as 12.15 cm from the center of the diffraction pattern. Thus;
θ₁ = tan⁻¹(12.15/15)
θ₁ = tan⁻¹(0.81)
θ₁ = 39°
Since λ = dsinθ₁
Thus;
λ = 834sin(39°)
λ = 834 × 0.6293
λ ≈ 525 nm
Describe and give an example of a normal force.
Answer:
Normal force is the force exerted on an object that is in contact with another non-moving object. Force is push and pull
Ex: When a book rests on a surface (like a table), the table is exerting a force on the book in order to support the weight of it.
Answer:
Normal Force is a contact force. When 2 objects are touching each other, they are in contact. Like when a cup is sitting on a table they are in contact cause the table has to hold the force of the cup up on the table.
A example of normal force is when a book is sitting on a table, and the table is holding the force of the book up.
Explanation:
Hope this helps!!
Have a blessed day/night! <33
A force of 1.46 N pushes against an object with a mass of 0.10 kg for 3.00 min. The object does not move. What is the work done on the object by the force
Answer:
0 J
Explanation:
Given data
force F=1.46N
mass m=0.1kg
time t= 3min
distance= 0 meters----"the object did not move"
We know that work-done is given as
work= Force * distance moved
Work= 1.46*0
work =0joules
We understand that when this is no distance move, then there is no work done, for work to be done, the object must move through a distance
What is the answer to this question?
Answer:
c
Explanation:
Answer:
b
Explanation:
Large igneous provinces are formed by ____________________, which can flow tens to hundreds of kilometers.
Answer:
Large igneous provinces are formed by flood basalt, which can flow tens to hundreds of kilometers.
Explanation:
which of the following statements about mass and weight is correct?
A. "Mass" is the scientific term for "weight''
B. Weight directly measures force while mass directly measures amount of matter.
C. Weight has unit of kilograms or grams. Mass has unit of newtons.
D. The mass of an object depends on the size of the object. The weight of an object dosent depend on the size of the object.
The correct answer is B. Weight directly measures force while mass directly measures the amount of matter.
Explanation:
Weight and mass are two related but not identical concepts. The first term "mass" can be found by measuring the amount of matter in a body or substance. For example, the mass in an apple is around 80 grams depending on its size. In this context, mass is measured using units such as grams or kilograms. On the other hand, weight considers not only the matter in an object but the force exerted in it, especially, the gravitational force. In the case of the apple, the weight is 1 N as newtons (N) are used to measure force and this is the result of multiplying the mass by gravity. According to this, option B is correct.
The gravitational potential energy of a 88 kg person in a building that is 385 m tall is 332,024 J. The person then jogs up the stairs, an additional distance of 100 meters.
How would the potential energy of the person be affected?
A) There would be no change.
B) The potential energy would double.
C) The potential energy would increase almost 25%
D) The potential energy would decrease more than 50%.
Answer:
Part C
Explanation:
The potential energy would increase almost 25%
Photons with the lowest energy have the lowest
A. wavelength
B. self-esteem
C. frequency
D. speed
Answer:
c frequency
hope it helps:))
Explanation:
To solve this we must be knowing each and every concept related to photon. Therefore, the correct option is option C. Photons with the lowest energy have the lowest frequency.
What is photon?A photon, also known as a light quantum, is a small energy packet containing electromagnetic radiation. The idea came from Albert Einstein's explanation of something like the photoelectric phenomenon (1905), in which he hypothesized the possibility of energy levels packets during light transmission.
Photons all move just at speed of light. Photons are bosons that have no electric charge and rest mass one and unit of spin; they were field particle that are assumed to be the bearers of electromagnetic field. Photons with the lowest energy have the lowest frequency.
Therefore, the correct option is option C. Photons with the lowest energy have the lowest frequency.
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In which of the following situations shows no acceleration of the car?
O A car shortly after a stoplight turns green.
A car approaching a red light
O A car with the cruise control set at 80 km/h
O A car turning a curve at a constant speed
Other:
Answer:
I think A, a car shortly after a stoplight turns green OR if it's wrong it's B☺️
A car driver traveling on a linear segment of a level road at 60 miles per hour (mph) slows down to 40 miles per hour in a road construction zone. If the car traveled a distance of 360 ft from the original speed to the final speed then what is the car deceleration rate in ft per sec2
Answer:
a=5.98 ft/secExplanation:
given
initial velocity u=60mph to ft per sec= 88ft/s
final velocity v=40 mph to ft per sec= 58.6 ft/s
distance s=360ft
convert feet to mile 360ft is 0.06miles
we can apply the expression
v^2=u^2-2as
58.6^2=88^2-2*a*360
3433.96=7744-720a
3433.96-7744=-720a
-4310.04=-720a
a=4310.04/720
a=5.98 ft/sec
Which would increase the speed of a sound wave?
O A wave passes from a solid to a liquid while remaining the same temperature.
O The medium increases in temperature while remaining in the same phase.
O The medium decreases in temperature while remaining in the same phase.
O A wave passes from a liquid to a gas while remaining the same temperature.
(First correct answer with explanation gets Brainliest)
Answer:
B. The medium increases in temperature while remaining in the same phase.
Explanation: The most important factor influencing the speed of sound in air is temperature. The speed is proportional to the square root of the absolute temperature, giving an increase. Due to this information, B is the best answer.
What happens when the forces applied to an object at rest produce a net force of zero?
1.The object will not move at all.
2.The object will have positive acceleration.
3.The object will change its direction of motion.
Answer:
1. The object will not move at all.
Explanation:
The question above is related to "Newton's First Law of Motion." According to this law, an object that is at rest will remain in its position (at rest) unless acted upon by "unbalanced forces." If the forces applied to an object are balanced, the net force is zero. This means that the object will not move nor accelerate.
The law also mentioned that if the object is in motion, it will remain in same velocity (speed with direction) if there's no net force.
Two sailors, Popped-Eye and Mack Sparrow, are using identical pulleys to lift sunken treasure to their ship. Popped-Eye's treasure is three times heavier than Mack Sparrow's. Each sailor lifts his treasure an identical distance. Popped-Eye lifts his treasure the entire distance in 30s. Mack Sparrow requires twice as much time.
Which statement below is true regarding Popped-Eye's and Mack Sparrow's relative power?
A) Their power is equal.
B) Popped-Eye is 3 times more powerful than Mack Sparrow.
C) Popped-Eye is 6 times more powerful than Mack Sparrow.
D) Popped-Eye is 1.5 times more powerful than Mack Sparrow.
Answer:
Popped-Eye is 3 times more powerful than Mack Sparrow.
Explanation:
Popped-Eye is 6 times more powerful than Mack Sparrow.
What is power?In science and engineering, power is the rate at which work is completed or energy is delivered. It can be expressed as the product of the work completed (W) or the energy transferred (E) divided by the time interval (t), or W/t.
Given parameters:
Popped-Eye's treasure is three times heavier than Mack Sparrow's.
Mack Sparrow requires twice as much time.
So, Popped-Eye can lift times heavier treasure in half of time compere to Mack Sparrow .
Hence, Popped-Eye's power is 3÷(1/2) = 3×2 = 6 times more than Mack Sparrow's.
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A space vehicle is traveling at 4140 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 99 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation
Answer:
V_ce = 1128 m/s
Explanation:
We are given;
Speed of space vehicle relative to the earth; V_i = 4140 km/h = 1150 m/s
relative speed between the motor and the command module; V_mc = 99 km/h = 27.5 m/s
mass of the motor is four times the mass of the module.
If the mass of the module is m
Mass of motor = 4m
From conservation of momentum, since the system is isolated, then;
Initial momentum = final momentum.
Thus;
MV_i = 4m(V_me) + m(V_ce) - - - (eq 1)
Where;
M is mass of space vehicle
V_me is speed of motor relative to the earth
V_ce is speed of command module relative to the earth
Now, relative speed between the motor and the command module is given by the formula;
V_mc = V_me - V_ce
Thus, V_me = V_mc + V_ce
Putting (V_mc + V_ce) for V_me in eq 1, we have;
MV_i = 4m(V_mc + V_ce) + m(V_ce)
Now, the mass of the space vehicle will be the sum of the mass of the module and mass of the motor.
Thus;
M = m + 4m
M = 5m
Thus;
5mV_i = 4m(V_mc + V_ce) + m(V_ce)
Divide through by m to get;
5V_i = 4(V_mc + V_ce) + (V_ce)
5V_i = 4V_mc + 4V_ce + V_ce
5V_i = 4V_mc + 5V_ce
We want to find the speed of the command module relative to Earth just after the separation. Thus, let's make V_ce the subject;
V_ce = (5V_i - 4V_mc)/5
V_ce = ((5 × 1150) - (4 × 27.5))/5
V_ce = 1128 m/s
An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m when the object is accelerating at 1.4 m/s2, what is the mass of the object?
Answer:
Approximately [tex]28.3\; \rm kg[/tex], assuming that the spring was pulled horizontally.
Explanation:
Let [tex]k[/tex] denote the spring constant of this spring. Calculate the size of tension in this spring that corresponds to the [tex]\Delta x = 0.88\; \rm m[/tex] stretch:
[tex]F(\text{tension}) = k\cdot \Delta x = 45\; \rm N \cdot m^{-1} \times 0.88\; \rm m = 39.6\; \rm N[/tex].
(The negative sign is ignored.)
If the spring is pulled horizontally, then the tension that the spring exerts would be the only unbalanced force on this object. The net force n this object would be equal to the tension that the spring had exerted: [tex]39.6\; \rm N[/tex].
Newton's Second Law of motion suggests the following relation between the net force [tex]\sum F[/tex] on the object, the mass [tex]m[/tex] of the object, and the acceleration [tex]a[/tex] of the object:
[tex]\sum F = m \cdot a[/tex].
For this object, [tex]a = 1.4\; \rm m \cdot s^{-2}[/tex]. It has been found that the net force on this object is [tex]\sum F = 39.6\; \rm N[/tex]. Rearrange the original equation [tex]\sum F = m \cdot a[/tex] and solve for [tex]m[/tex], the mass of this object:
[tex]\displaystyle m = \frac{\sum F}{a} = \frac{39.6\; \rm N}{1.4\; \rm m \cdot s^{-2}} \approx 28.3\; \rm kg[/tex].
In other words (based on all these assumptions,) the mass of this object should be approximately [tex]28.3\; \rm kg[/tex].
What is the answer to this question?
Answer:
C I just did it. so yeah....
A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?
Answer:
20 m/s
Explanation:
velocity after 2 sec.
v=40+(−10)t
=40−10∗2=20m/s
We have that for the Question "A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?" it can be said that the balls velocity after 2 seconds
v=67.44m/s
From the question we are told
A ball is thrown up in the air with an initial velocity of 40m/s what is the balls velocity after 2 seconds?
Generally the Newtons equation for the Motion is mathematically given as
[tex]v=u+at\\\\Therefore\\\\v=40+9.8*2.8\\\\v=67.44m/s\\\\[/tex]
Therefore
the balls velocity after 2 seconds
v=67.44m/s
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two forces act on a 6 kg object. one of the forces is 10 N. if the object accelerates at 2 m/s^2, what is the greatest possible magnitude of the other force
Answer:
22NExplanation:
Let the two forces be acting in the horizontal direction on the object.
Given
Mass m = 6kg
One of the force = 10N
acceleration = 2m/s²
Taking the sum of forces along the x component, we will have:
[tex]\sum F_x = ma_x\\Ff - 10 = 6(2)\\Ff - 10 = 12\\Ff = 12+10\\Ff = 22N[/tex]
Note that Ff is the greatest possible magnitude of the other force.
Hence the greatest possible magnitude of the other force is 22N
You are originally 1.0 m beneath the surface of a pool. If you dive to 2.0 m beneath the surface, what happens to the absolute pressure on you
Answer:
The absolute pressure on me increases to 130,900 Pa.
Explanation:
Since I dive to a depth of 2.0 m, there is a depth change of 2.0 m which will cause a corresponding pressure change. The pressure change is thus ΔP = ρgΔh where ρ = density of water = 1000 kg/m³, g = acceleration due to gravity = 9.8m/s² and Δh = depth change = 2.0 m
ΔP = ρgΔh = 1000 kg/m³ × 9.8m/s² × 2.0 m = 19,600 Pa.
Since the absolute pressure P at 1.0 m is the pressure due to atmosphere plus the pressure of 1.0 m of water, we have
P = P₀ + ρgΔh' where P₀ = atmospheric pressure = 101,500 Pa and Δh' = 1.0 m
P = 101,500 Pa + 1000 kg/m³ × 9.8m/s² × 1.0 m = 101,500 Pa + 9800 Pa = 111,300 Pa
So, the absolute pressure at 2.0 m is P' = P + ΔP = 111,300 Pa + 19,600 Pa = 130,900 Pa.
So the absolute pressure on me increases to 130,900 Pa.
The position vector r of a particle relative to a certain point has a magnitude of 3m , and the force F on the particle has a magnitude of 4N . What is the angle between the directions of r and F if the magnitude of the associated torque equals
Answer:
θ = 0
Explanation:
As per the question Given that
Magnitude = 2m
Force F contains MAgnitude = 4N
Based on the above information
T = rFsinθ
Now if
T = 0,
so r and F would not be zero so that sinθ should be equivalent to zero for holding an equation
Therefore
θ = 0
The same is to be considered as it is relevant as per the question
What is the difference between fast-twitch and slow-twitch muscle fibers?
A. They are almost identical.
B. Fast-twitch fibers can deliver a quick burst of power. Slow-twitch
fibers can maintain a contraction for a longer time.
C. Fast-twitch fibers are what keep us going on a long bike ride or
hike. Slow-twitch fibers are what we use while dashing for the bus
or jumping up for a rebound.
D. Fast-twitch fibers take a long time to warm up, but are stronger
than slow-twitch fibers once they get going.
Answer:
B. fast-twitch fibers can deliver a quick burst of power. Slow-twitch fibers can maintain a contraction for a longer time.
Answer:
the person above is right, just came for answers lol :')
Explanation:
What is the normal force acting on a mountain goat that weighs 550 N and is standing on such a slope
Answer:
Explanation:
We have that the normal force acting on a mountain goat that weighs 550 N and is standing on such a slope is
N=275N
From the question we are told
Mountain goats can easily scale slopes angled at 60∘ from horizontal. What is the normal force acting on a mountain goat that weighs 550 N and is standing on such a slope?Generally the equation for the normal force acting on a mountain goat is mathematically given as
N=mgcos
N=550*cos60
N=275N
Therefore
The normal force acting on a mountain goat that weighs 550 N and is standing on such a slope is
N=275N
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A person in a grocery store is standing carrying a big sack of rice (that has a mass of 10 kg) on the top of his head. Where is his CG located relative to his original CG (i.e when he is standing without the sack of rice on his head)
a) Higher than his original CG
b) Lower than his original CG
c) At the same point as his original CG
d) Insufficient information
Answer:
The CG is located higher than his original CG.
Explanation:
Centre of gravity (CG), in physics, an imaginary point in a body of matter where, for convenience in certain calculations, the total weight of the body may be thought to be concentrated. In a uniform gravitational field the centre of gravity is identical to the centre of mass, a term preferred by physicists.
His CG is located higher than his original CG when he is standing without the sack of rice on his head.
A system is said to be in stable equilibrium if, when displaced from equilibrium, it experiences a net force or torque in a direction opposite to the direction of the displacement. A system is in unstable equilibrium if, when displaced, it experiences a net force or torque in the same direction as the displacement from equilibrium. A system is in neutral equilibrium if its equilibrium is independent of displacements from its original position. A marble on a flat horizontal surface is an example.
You throw a ball downward from a window at a speed of 1.7 m/s. How fast will it be moving when it hits the sidewalk 2.5 m below?
Answer:
7.43m/s
Explanation:
Given parameters:
Initial velocity = 1.7m/s
Height = 2.5m
Unknown:
Final velocity = ?
Solution:
To solve this problem, we apply one of the motion equation. The most fitting one given the parameters is;
V² = U² + 2gh
where V is the final velocity
U is the initial velocity
g is the acceleration due to gravity = 9.8m/s
h is the height
Input the parameters and solve;
V² = 2.5² + 2(9.8)(2.5)
V = 7.43m/s
gas is escaping from a sperical ballom 4 cm3 per minute .how fast is the surface area shrinking when radius is 24cm
Answer:
dS/dt = 0.33 cm²/min
Explanation:
Volume of sphere is given by;
V = (4/3)πr³
We are told that gas is escaping from a spherical balloon 4 cm³ per minute. Thus; dV/dt = 4 cm³/min
Now, let's find the rate at which the radius is changing.
dr/dt = (dV/dt) ÷ (dV/dr)
Now, dV/dr = 4πr²
Thus;
dr/dt = 4/(4πr²)
dr/dt = 1/(πr²)
Now, surface area of sphere is given by;
S = 4πr²
Thus, dS/dr = 8πr
Now, let's find the rate at t which area is shrinking.
dS/dt = (dS/dr) × (dr/dt)
Plugging relevant terms, we have;
dS/dt = 8πr × 1/(πr²)
dS/dt = 8/r
At r = 24 cm, we have;
dS/dt = 8/24
dS/dt = 1/3
dS/dt = 0.33 cm²/min
Consider the situation a book a sliding to a stop while moving across the classroom floor. Of the forces listed identify which act upon the book
The forces which act upon the book are Normal, Gravity, Air resistance, and Friction. Therefore, the (A), (B), (D), and (E) options are correct.
What are Normal force, Gravity, Air resistance, and Friction?The normal force is a contact force that surfaces exert to prevent objects from passing through each other. If two surfaces are not in direct contact, they are able to exert a normal force on each other. The book and the floor are exerting normal on each other.
Gravity is the force by which the earth draws objects toward its center. The force of gravity keeps everything on the surface of the earth. Therefore, the sliding book on the floor experiences the force of gravity.
Air resistance is a force that arises due to air. The force acts in the opposite direction to the motion of an object through the air. As the book slides across the floor, it experiences resistance due to air.
Friction is the resistance caused by the surfaces that are in contact when they slide over each other. Friction arises due to the irregularities on the two surfaces. The book and floor both have irregularities on their surfaces which is caused friction between them.
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Your question is incomplete, but most probably the complete question was,
Consider the situation a book a sliding to a stop while moving across the classroom floor. Of the forces listed identify which act upon the book:
A)Normal,
B) Gravity
C)Applied
D) Air resistance,
E) Friction
F) Tension
elastic limit is equal to:(a) stress (b)strain (c) young's modulus (d)modulus of rigidity
Answer:
A. stress
Explanation:
Hope this helped. Have an amazing day/night!!
We have that for the Question "elastic limit is equal to" it can be said that
Young's modulusThis speaks to a body retaining it shape after changes occur
Option C
From the question we are told
elastic limit is equal to
(a) stress
(b)strain
(c) young's modulus
(d)modulus of rigidity
Where
Where
Stress is the force acting on a material
and
strain
The effect of force acting on a material
Young's modulus
This speaks to a body retaining it shape after changes occur
Modulus of rigidity
m=sheer stress/shear strain
Now having that
Elastic limit is a point where an elastic body does not return to its original shape
The option chosen is'
Young's modulus
This speaks to a body retaining it shape after changes occur
Option C
Generally the equation for the is mathematically given as
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