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2 years ago

6

A person using an oxygen mask is breathing air that is 33% oxygen. what is the partial pressure of the o2, when the air pressure

in the mask is 110 kpa

1 answer:

yawa3891 [41]2 years ago

5 0

The partial pressure of the O2 is 36.3 kiloPascal when the air pressure in the mask is 110 kiloPascal based on the isotherm relation. This problem can be solved by using the isotherm relation equation which stated as Vx/Vtot = px/ptot, where V represents volume, p represents the pressure, x represents the partial gas, and tot represents the total gas<span>. Calculation: 33/100 = px/110 --> px = 36.3</span>

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At what time t is the turtle second time a distance of 10.0 cm from its starting point?

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Answer:

10 cm.

Explanation:

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2 years ago

Joel uses a claw hammer to remove a nail from a wall. He applies a force of 40 newtons on the hammer. The hammer applies a force

jarptica [38.1K]

Hi!

Mechanical advantage is defined as the<em> ratio of force produced by an object to the force that is applied to it.</em>

In our case, this would be the ratio of the force applied by the claw hammer on the nail to the force Joel applies to the claw hammer, which is

160:40 or 4:1

So the mechanical advantage of the hammer is four.

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3 0

2 years ago

Read 2 more answers

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.

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

Brian and Jack decided to investigate which water fountain at school has the coldest water. The two boys take measurements using

qaws [65]

Answer:

b

Explanation:

he keep the thermoter in the water longer

3 0

2 years ago

A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv

Juli2301 [7.4K]

Answer:

<em>The final charge on the 6.0 mF capacitor would be 12 mC</em>

Explanation:

The initial charge on 4 mF capacitor = 4 mf x 50 V = 200 mC

The initial Charge on 6 mF capacitor = 6 mf x 30 V =180 mC

Since the negative ends are joined together the total charge on both capacity would be;

q = $q_{1} -q_{2}$

q = 200 - 180

q = 20 mC

In order to find the final charge on the 6.0 mF capacitor we have to find the combined voltage

q = (4 x V) + (6 x V)

20 = 10 V

V = 2 V

For the final charge on 6.0 mF;

q = CV

q = 6.0 mF x 2 V

q = 12 mC

Therefore the final charge on the 6.0 mF capacitor would be 12 mC

5 0

2 years ago

Read 2 more answers