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ValentinkaMS [17]

2 years ago

7

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight. Use this co

ncept, along with the variable mass value, the ruler, and the moveable line, to estimate the spring constant k of the spring. Set the damping to "Lots" so that the mass will come to rest quickly after being changed, and make sure the gravity is set to "Earth." Finally, set the spring constant to "Small." Estimate the spring constant several times (using different values of mass) and average together for the most accurate calculation.

2 answers:

Reika [66]2 years ago

8 0

Answer:

3.0 N*m

Explanation:

F = −k*x = g

Using PHeT spring constant simulation.

natima [27]2 years ago

5 0

Recall that in the equilibrium position, the upward force of the spring balances the force of gravity on the weight is given below.

Explanation:

Measure unstretched length of spring, L. E.g. L = 0.60m.

Set mass to a convenient value (e.g. m = 0.5kg).

Hang mass.

Measure new spring length, L'. E.g. L' = 0.70m.

Calculate extension: e = L' - L = 0.70 – 0.60 = 0.10m

Use mg = ke (in equilibrium weight = tension)

k = mg/e

Don't know what value you are using for example. Suppose it is 10N/kg (same thing as 10m/s²).

k = 0.5*10/0.10 = 50 N/m

Repeat for a few different masses. (L always stays the same.)

Take the average of your k values.

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hodyreva [135]

20) When light passes from air to glass and then to air

21) When a light ray enters a medium with higher optical density, it bends towards the normal

22) Index of refraction describes the optical density

23) Light travels faster in the material with index 1.1

24) Glass refracts light more than water

25) Index of refraction is $n=\frac{c}{v}$

26) Critical angle: $[tex]sin \theta_c = \frac{n_2}{n_1}$ [/tex]

27) Critical angle is larger for the glass-water interface

Explanation:

20)

It is possible to slow down light and then speed it up again by making light passing from a medium with low optical density (for example, air) into a medium with higher optical density (for example, glass), and then make the light passing again from glass to air.

This phenomenon is known as refraction: when a light wave crosses the interface between two different mediums, it changes speed (and also direction). The speed decreases if the light passes from a medium at lower optical density to a medium with higher optical density, and viceversa.

21)

The change in direction of light when it passes through the boundary between two mediums is given by Snell's law:

$n_1 sin \theta_1 = n_2 sin \theta_2$

with

$n_1, n_2$ are the refractive index of 1st and 2nd medium

$\theta_1, \theta_2$ are the angle of incidence and refraction (the angle between the incident ray (or refracted ray) and the normal to the boundary)

The larger the optical density of the medium, the larger the value of n, the smaller the angle: so, when a light ray enters a medium with higher optical density, it bends towards the normal.

22)

The index of refraction describes the optical density of a medium. More in detail:

A high index of refraction means that the material has a high optical density, which means that light travels more slowly into that medium
A low index of refraction means that the material has a low optical density, which means that light travels faster into that medium

Be careful that optical density is a completely different property from density.

23)

As we said in part 22), the index of refraction describes the optical density of a medium.

In this case, we have:

A material with refractive index of 1.1
A material with refractive index of 2.2

As we said previously, light travels faster in materials with a lower refractive index: therefore in this case, light travels more quickly in material 1, which has a refractive index of only 1.1, than material 2, whose index of refraction is much higher (2.2).

24)

Rewriting Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$ (1)

For light moving from air to water:

$n_1 \sim 1.00$ is the index of refraction of air

$n_2 = 1.33$ is the index of refraction ofwater

In this case, $\frac{n_1}{n_2}=\frac{1.00}{1.33}=0.75$

For light moving from air to glass,

$n_2 = 1.51$ is the index of refraction of glass

And so

$\frac{n_1}{n_2}=\frac{1.00}{1.51}=0.66$

From eq.(1), we see that the angle of refraction $\theta_2$ is smaller in the 2nd case: so glass refracts light more than water, because of its higher index of refraction.

25)

The index of refraction of a material is

$n=\frac{c}{v}$

c is the speed of light in a vacuum

v is the speed of light in the material

So, the index of refraction is inversely proportional to the speed of light in the material:

The higher the index of refraction, the slower the light
The lower the index of refraction, the faster the light

26)

From Snell's law,

$sin \theta_2 = \frac{n_1}{n_2}sin \theta_1$

We notice that when light moves from a medium with higher refractive index to a medium with lower refractive index, $n_1 > n_2$ , so $\frac{n_1}{n_2}>1$ , and since $sin \theta_2$ cannot be larger than 1, there exists a maximum value of the angle of incidence $\theta_c$ (called critical angle) above which refraction no longer occurs: in this case, the incident light ray is completely reflected into the original medium 1, and this phenomenon is called total internal reflection.

The value of the critical angle is given by

$sin \theta_c = \frac{n_2}{n_1}$

For angles of incidence above this value, total internal reflection occurs.

27)

Using:

$sin \theta_c = \frac{n_2}{n_1}$

For the interface glass-air,

$n_1 \sim 1.51\\n_2 = 1.00$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.00}{1.51})=41.5^{\circ}$

For the interface glass-water,

$n_1 \sim 1.51\\n_2 = 1.33$

The critical angle is

$\theta_c = sin^{-1}(\frac{n_2}{n_1})=sin^{-1}(\frac{1.33}{1.51})=61.7^{\circ}$

So, the critical angle is larger for the glass-water interface.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

7 0

2 years ago

Complete the passage to summarize factors affecting the speed of a wave. The material or substance that a wave moves through is

pantera1 [17]

the first one is medium, the second one is type, and the third one is temperature . if i gave the correct answer, please give best answer x

6 0

2 years ago

Read 2 more answers

An airplane pilot wishes to fly directly westward. According to the weather bureau, a wind of 75.0 km/hour is blowing southward.

Alex17521 [72]

Answer:

The speed of the plane relative to the ground is 300.79 km/h.

Explanation:

Given that,

Speed of wind = 75.0 km/hr

Speed of plane relative to the air = 310 km/hr

Suppose, determine the speed of the plane relative to the ground

We need to calculate the angle

Using formula of angle

$\sin\theta=\dfrac{v'}{v}$

Where, v'=speed of wind

v= speed of plane

Put the value into the formula

$\sin\theta=\dfrac{75}{310}$

$\theta=\sin^{-1}(\dfrac{75}{310})$

$\theta=14.0^{\circ}$

We need to calculate the resultant speed

Using formula of resultant speed

$\cos\theta=\dfrac{v''}{v}$

Put the value into the formula

$\cos14=\dfrac{v''}{310}$

$v''=\cos14\times310$

$v''=300.79\ km/h$

Hence, The speed of the plane relative to the ground is 300.79 km/h.

6 0

2 years ago

A box slides down a frictionless plane inclined at an angle θ ¸ above the horizontal. The gravitational force on the box is dire

DedPeter [7]

<h2>Answer: at an angle $\theta$ below the inclined plane. </h2>

If we draw the <u>Free Body Diagram</u> for this situation (figure attached), taking into account only the gravity force in this case, we will see the weight $W$ of the block, which is directly proportional to the gravity acceleration $g$ :

$W=m.g$

This force is directed vertically at an angle $\theta$ below the inclined plane, this means it has an X-component and a Y-component:

$W=W_{X}+W_{Y}$

$W_{X}=m.g.cos(\theta)$

$W_{Y}=m.g.sin(\theta)$

Therefore the correct option is c

6 0

2 years ago

Read 2 more answers

A 4.0-m-diameter playground merry-go-round, with a moment of inertia of 350 kg⋅m2 is freely rotating with an angular velocity of

Flauer [41]

Answer:

$v = 4.375\,\frac{m}{s}$

Explanation:

The situation of the system Ryan - merry-go-round is modelled after the Principle of the Angular Momentum Conservation:

$(350\,kg\cdot m^{2})\cdot (1.5\,\frac{rad}{s} ) - (2\,m)\cdot (60\,kg)\cdot v = 0\,kg\cdot \frac{m^{2}}{s}$

The initial speed of Ryan is:

$v = 4.375\,\frac{m}{s}$

5 0

2 years ago

Read 2 more answers