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fenix001 [56]
3 years ago
5

Hooke’s law describes the linear relationship between stress and strain through Young’s modulus. Given two materials under the s

ame elastic stress, the material with the higher modulus will stretch _________ the material with a lower modulus.
Physics
2 answers:
allsm [11]3 years ago
6 0

Answer:

The material with higher modulus will stretch less than the material with lower modulus

Explanation:

This is because Young's modulus, E = stress/strain = σ/ε. Now, ε = σ/Y. Since σ = constant for both materials, ε ∝ 1/Y. So the strain is inversely proportional to the Young's modulus. Thus, the material with the higher modulus will stretch less than the material with lower modulus.

stiks02 [169]3 years ago
3 0

Answer:

The material with higher modulus will stretch less than

The material with lower modulus

Explanation:

A material with a higher modulus is stiffer and has better resistance to deformation. The modulus is defined as the force per unit area required to produce a deformation or in other words the ratio of stress to strain.

E= stress/stain

Hooks law states that provided the elastic limit is not exceeded the extension e of a spring is directly proportional to the load or force attached

F=ke

Where k is the constant which gives the measure of the spring under tension

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Points a and b lie in a region where the y-component of the electric field is Ey=α+β/y2. The constants in this expression have t
Drupady [299]

Answer:

V_{a} - V_{b} = 89.3

Explanation:

The electric potential is defined by

         V_{b} - V_{a} = - ∫ E .ds

In this case the electric field is in the direction and the points (ds) are also in the direction and therefore the angle is zero and the scalar product is reduced to the algebraic product.

         V_{b} - V_{a} = - ∫ E ds

We substitute

         V_{b} - V_{a} = - ∫ (α + β/ y²) dy

We integrate

          V_{b} - V_{a} = - α y + β / y

We evaluate between the lower limit A  2 cm = 0.02 m and the upper limit B 3 cm = 0.03 m

           V_{b} - V_{a} = - α (0.03 - 0.02) + β (1 / 0.03 - 1 / 0.02)

            V_{b} - V_{a} = - 600 0.01 + 5 (-16.67) = -6 - 83.33

            V_{b} - V_{a} = - 89.3 V

As they ask us the reverse case

             V_{b} - V_{a} = - V_{b} - V_{a}

             V_{a} - V_{b} = 89.3

3 0
3 years ago
According to universal gravitation, both mass and air resistance affect the gravitational attraction between objects
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A dog walks 12 meters to the east and then 16 meters back to the west for this motion what is the distance moved What is the mag
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The distance is 28 meters, and the displacement is -4.

For the distance it would be 12 + 16 = 28.

For the displacement it would be 12 - 16 = -4.

would really appreciate a brainliest! Hope this helped!

6 0
3 years ago
A truck has shock absorbers with a spring constant of 24200 N/m. When it hits a bump, it oscillates at 0.429 Hz. What is the mas
siniylev [52]

Answer:

3331.5 kg

Explanation:

Given:

Spring constant of the spring (k) = 24200 N/m

Frequency of oscillation (f) = 0.429 Hz

Let the mass be 'm' kg.

Now, we know that, a spring-mass system undergoes Simple Harmonic Motion (SHM). The frequency of oscillation of SHM is given as:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}

Rewrite the above equation in terms of 'm'. This gives,

2\pi f=\sqrt{\frac{k}{m}}\\\\Squaring\ both\ sides,\ we\ get:\\\\(2\pi f)^2=\frac{k}{m}\\\\m=\frac{k}{4\pi^2 f^2}

Now, plug in the given values and solve for 'm'. This gives,

m=\frac{24200\ N/m}{4\pi^2\times (0.429\ Hz)^2 }\\\\m=\frac{24200\ N/m}{4\pi^2\times 0.184\ Hz^2}\\\\m\approx3331.5\ kg

Therefore, the mass of the truck is 3331.5 kg.

3 0
3 years ago
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