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2 years ago

8

The blue curve is the plot of the data. The straight orange line is tangent to the blue curve at t = 40 s. A plot has the concen

tration of A in molar on the y axis and time in seconds on the x axis. A curve contains the points (0, 0.52), (20, 0.43), (40, 0.35), (60, 0.29), (80, 0.24), and (100, 0.20). A line touches the curve at (40, 0.35) and has a y intercept of (0, 0.48). Approximate the instantaneous rate of this reaction at time t = 40 s.

2 answers:

victus00 [196]2 years ago

7 0

Answer:

.00325

Explanation:

YOu have to look at the red graph

Sever21 [200]2 years ago

5 0

Answer:

0.00325 moles/liter/second

Explanation:

The tangent line has a slope of (y2 -y1)/(x2 -x1) = (0.35-0.48)/(40-0) = -0.00325.

The rate of the reaction is about 0.00325 moles/liter/second.

_____

This is the rate of decrease of the concentration of A.

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Hydraulic press is called an instrument for multiplication of force. Why?

Lisa [10]

Answer:

Hydraulic press is called an instrument for multiplication of force. Why? Because it uses Pascal's idea and principle: F=p*S. If we apply small force to small piston you will generate a pressure. According to Pascal's law pressure is the same everywhere in closed system so the same pressure will act on large piston on the other side too.

Explanation:

4 0

2 years ago

a bicycle pump contains 20cm3 of air at a pressure of 100kpa the air is then pumped in a single stroke through a valve into a ty

riadik2000 [5.3K]

If we assume also that the temperature of the air does not change, we can use Boyle's Law:
p₁V₁ = p₂V₂

Now, we know:
p₁ = 100kPa
V₂ = 100cm³ (the volume of the tyre)
V₁ = 120cm³ (becuse the air is contained inside the tyre AND the pump)

We can solve for p₂:
p₂ = (p₁V₁)/V₂
= (100×120)/100
= 120kPa

Therefore your answer is: 120kPa

8 0

2 years ago

answers Collision derivation problem. If the car has a mass of 0.2 kg, the ratio of height to width of the ramp is 12/75, the in

Natasha2012 [34]

Answer:

4.8967m

Explanation:

Given the following data;

M = 0.2kg

∆p = 0.58kgm/s

S(i) = 2.25m

Ratio h/w = 12/75

Firstly, we use conservation of momentum to find the velocity

Therefore, ∆p = MV

0.58kgm/s = 0.2V

V = 0.58/2

V = 2.9m/s

Then, we can use the conservation of energy to solve for maximum height the car can go

E(i) = E(f)

1/2mV² = mgh

Mass cancels out

1/2V² = gh

h = 1/2V²/g = V²/2g

h = (2.9)²/2(9.8)

h = 8.41/19.6 = 0.429m

Since we have gotten the heigh, the next thing is to solve for actual slant of the ramp and initial displacement using similar triangles.

h/w = 0.429/x

X = 0.429×75/12

X = 2.6815

Therefore, by Pythagoreans rule

S(ramp) = √2.68125²+0.429²

S(ramp) = 2.64671

Finally, S(t) = S(ramp) + S(i)

= 2.64671+2.25

= 4.8967m

3 0

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.

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

The gravitational force produce between any two object kept 2.5×10 to the power 4 km apart is 580N.At what distance should they

timofeeve [1]

Answer:

d = 3.54 x 10⁴ Km

Explanation:

Given,

The distance between the two objects, r = 2.5 x 10⁴ Km

The gravitational force between them, F = 580 N

The gravitational force between the two objects is given by the formula

F = GMm/r² newton

When the gravitational force becomes half, then the distance between them becomes

Let us multiply the above equation by 1/2 on both sides

( 1/2) F = (1/2) GMm/r²

= GMm/2r²

= GMm/(√2r)²

Therefore, the distance becomes √2d, when the gravitational force between them becomes half

d = √2r = √2 x 2.5 x 10⁴ Km

= 3.54 x 10⁴ Km

Hence, the two objects should be kept at a distance, d = 3.54 x 10⁴ Km so that the gravitational force becomes half.

3 0

2 years ago