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Dmitriy789 [7]
3 years ago
5

You are getting bored as you are stuck at home, and want to find out the effective 'spring constant' of random objects. You find

a spare sock seems to stretch nice and easily. So you place a roll of quarters in your sock (approximate mass 0.23 kg) and notice that it stretches about an inch --- 2.54cm . What is the spring constant of your sock (in N/m)
Physics
2 answers:
Luda [366]3 years ago
4 0

Given Information:  

Mass of sock = 0.23 kg

Stretched length of sock = x = 2.54 cm = 0.0254 m

Required Information:  

Spring constant = k = ?

Answer:  

Spring constant = k = 88.82 N/m

Explanation:  

We know from the Hook's law that

F = kx

Where k is spring constant, F is the applied force and x is length of sock being stretched.

k = F/x

Where F is given by

F = mg

F = 0.23*9.81

F = 2.256 N

So the spring constant is

k = 2.256/0.0254

k = 88.82 N/m

Therefore, the spring constant of the sock is 88.82 N/m

sdas [7]3 years ago
4 0

Answer:

the spring constant of the sock = 88.74 N

Explanation:

According to Hooke's law  

F=kx

Where:

force F = mg  

k = spring constant

x =  stretching

We can say mg = kx

Making the spring constant k the subject of the formula; we have :

k = \frac{mg}{x}

Given that:

mass m = 0.23 kg

g = acceleration due to gravity = 9.81

x = 2.54 cm = 0.0254 m

Then plugging our values into the above derived equation; we have;

k= \frac{0.23*9.8}{0.0254 }\\ \\ \  k = 88.74 N/m

Thus, the spring constant of the sock = 88.74 N

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Select the situation for which the torque is the smallest.
alina1380 [7]

Answer:

e. The torque is the same for all cases.

Explanation:

The formula for torque is:

τ = Fr

where,

τ = Torque

F = Force = Weight (in this case) = mg

r = perpendicular distance between force an axis of rotation

Therefore,

τ = mgr

a)

Here,

m = 200 kg

r = 2.5 m

Therefore,

τ = (200 kg)(9.8 m/s²)(2.5 m)

<u>τ = 4900 N.m</u>

<u></u>

b)

Here,

m = 20 kg

r = 25 m

Therefore,

τ = (20 kg)(9.8 m/s²)(25 m)

<u>τ = 4900 N.m</u>

<u></u>

c)

Here,

m = 8 kg

r = 62.5 m

Therefore,

τ = (8 kg)(9.8 m/s²)(62.5 m)

<u>τ = 4900 N.m</u>

<u></u>

Hence, the correct answer will be:

<u>e. The torque is the same for all cases.</u>

6 0
3 years ago
The normal eye, myopic eye and old age
yanalaym [24]

Answer:

1)    f’₀ / f = 1.10, the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) the two diameters have the same order of magnitude and are very close to each other

Explanation:

You have some problems in the writing of your exercise, we will try to answer.

1) The equation to be used in geometric optics is the constructor equation

          \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

where p and q are the distance to the object and the image, respectively, f is the focal length

* For the normal eye and with presbyopia

the object is at infinity (p = inf) and the image is on the retina (q = 15 mm = 1.5 cm)

        \frac{1}{f'_o} = 1/ inf + \frac{1}{1.5}

        f'₀ = 1.5 cm

this is the focal length for this type of eye

* Eye with myopia

the distance to the object is p = 15 cm the distance to the image that is on the retina is q = 1.5 cm

           1 / f = 1/15 + 1 / 1.5

           1 / f = 0.733

            f = 1.36 cm

this is the focal length for the myopic eye.

In general, the two focal lengths are related

         f’₀ / f = 1.5 / 1.36

         f’₀ / f = 1.10

The question of the relationship between the focal length (f'₀) and the distance to the retina (image) is given by the constructor's equation

2) For this second part we have a diffraction problem, the point diameter corresponds to the first zero of the diffraction pattern that is given by the expression for a linear slit

          a sin θ= m λ

the first zero occurs for m = 1, as the angles are very small

          tan θ = y / f = sin θ / cos θ

for some very small the cosine is 1

          sin θ = y / f

where f is the distance of the lens (eye)

           y / f = lam / a

in the case of the eye we have a circular slit, therefore the system must be solved in polar coordinates, giving a numerical factor

           y / f = 1.22 λ / D

           y = 1.22 λ f / D

where D is the diameter of the eye

          D = 2R₀

          D = 2 0.1

          D = 0.2 cm

           

the eye has its highest sensitivity for lam = 550 10⁻⁹ m (green light), let's use this wavelength for the calculation

         

* normal eye

the focal length of the normal eye can be accommodated to give a focus on the immobile retian, so let's use the constructor equation

      \frac{1}{f} = \frac{1}{p} + \frac{1}{q}

sustitute

       \frac{1}{f} = \frac{1}{25} + \frac{1}{1.5}

       \frac{1}{f}= 0.7066

        f = 1.415 cm

therefore the diffraction is

        y = 1.22  550 10⁻⁹  1.415  / 0.2

        y = 4.75 10⁻⁶ m

this is the radius, the diffraction diameter is

       d = 2y

       d_normal = 9.49 10⁻⁶ m

* myopic eye

In the statement they indicate that the distance to the object is p = 15 cm, the retina is at the same distance, it does not move, q = 1.5 cm

       \frac{1}{f} = \frac{1}{15} + \frac{1}{ 1.5}

        \frac{1}{f}= 0.733

         f = 1.36 cm

diffraction is

        y = 1.22 550 10-9 1.36 10-2 / 0.2 10--2

        y = 4.56 10-6 m

the diffraction diameter is

        d_myope = 2y

         d_myope = 9.16 10-6 m

         \frac{d_{normal}}{d_{myope}} = 9.49 /9.16

        \frac{d_{normal}}{d_{myope}} =  1.04

we can see that the two diameters have the same order of magnitude and are very close to each other

8 0
3 years ago
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Rufina [12.5K]

Luego de utilizar la fórmula del volumen de un líquido encontramos que si el acohol tiene 450 gramos entonces su volumen es de

La fórmula del volumen de la densidad de un líquido es la siguiente:

Densidad= Masa / Volumen

Volumen =Masa / Densidad

Es conocido que la densidad del alcohol es 789 kg/m³, y como sabemos por dato que tiene 450gr el siguiente paso es la sustitución:

Volumen=0.45/ 789

Volumen = 0,00057 m³    

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3 years ago
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dlinn [17]

<u>Answer</u>

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<u>Answer</u>

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<em>By applying the powder, you are trying to make the hand dry hence increasing the friction for a better grip. </em>

If the gymnast doesn't do this the parallel bars may slip from the hands and injure himself or herself.

4 0
3 years ago
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cricket20 [7]

Answer:

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