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2 years ago

7. A package weighing 55 N is lifted to a height of 2.0 m by pulling it up a ramp

Physics

1 answer:

andrew-mc [135]2 years ago

6 0

1) Mechanical advantage: 8.5

2) Work output: 110 J

3) Work input: 28.6 J

4) Efficiency: 3.8

Explanation:

The (actual) mechanical advantage of a machine is the ratio between the output force (the load) and the input force (the effort):

$MA=\frac{F_{out}}{F_{in}}$

where

$F_{out}$ is the output force

$F_{in}$ is the input force

For the machine in this problem, we have

$F_{in}=6.5 N$ (force applied in input)

$F_{out}=55 N$ (output force, equal to the weight of the package which is lifted)

Therefore, the mechanical advantage is

$MA=\frac{55}{6.5}=8.5$

The work output in a machine that lifts an object is equal to the gravitational potential energy gained by the object:

$W_{out} = F_{out} h$

where:

$F_{out}$ is the output force, which is the weight of the object

h is the change in height of the object

In this problem,

$F_{out}=55 N$ (weight of the package)

h = 2.0 m (change in height of the package)

Therefore, the output work is:

$W_{out}=(55)(2.0)=110 J$

The work input is the amount of work done on the effort side of the machine. For a ramp, is given by:

$W_{in}=F_{in}L$

where

$F_{in}$ is the force in input (the effort)

L is the length of the ramp

For the ramp in this problem, we have:

$F_{in} = 6.5 N$ (input force)

L = 4.4 m (length of the ramp)

Therefore, the work input is:

$W_{in}=(6.5)(4.4)=28.6 J$

The efficiency of a machine is given by

$\eta=\frac{W_{out}}{W_{in}}$

where

$W_{out}$ is the output work

$W_{in}$ is the input work

The efficiency of a machine basically represents how much of the input work is transformed into useful output work.

For the ramp in this problem:

$W_{out}=110 J$

$W_{in}=28.6 J$

Therefore, the efficiency of this machine is

$\eta=\frac{110}{28.6}=3.8$

Note that the data given for this machine are unrealistic: in fact, for a real machine, the work output cannot be larger than the input work (and the efficiency cannot be larger than 1), so I assume there is a mistake in one of the numbers given in the problem.

Learn more about work:

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#LearnwithBrainly

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2 years ago

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2 years ago

This is really urgent

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.