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astraxan [27]
3 years ago
15

The temperature of a chemical reaction ranges between 40 degrees Celsius and 180 degrees Celsius. The temperature is at its lowe

st point when t = 0, and the reaction completes 1 cycle during a 12-hour period. What is a cosine function that models this reaction?
f(t) = 70 cos 12t + 110
f(t) = 110 cos 12t + 70
f(t) = −70 cos pi over 6t + 110
f(t) = −110 cos pi over 6t + 70
Physics
1 answer:
Vikki [24]3 years ago
3 0
The best and most correct answer among the choices provided by the question is <span>f(t) = −70 cos pi over 6t + 110</span><span>.
</span>


Hope my answer would be a great help for you.    
If you have more questions feel free to ask here at Brainl
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The planet Mars has a mass of 6.1 × 1023 kg and radius of 3.4 × 106 m. What is the acceleration of an object in free fall near t
Oksanka [162]

The acceleration due to gravity of Mars is 3.5\ m / s^{2}

<u>Explanation:</u>

As per universal law of gravity, the gravitational force is directly proportional to the product of masses and inversely proportional to the square of the distance between them. But in the present case, the gravity need to be determined between Mars and the object on Mars. Since the mass of Mars is greater than the mass of any object. Thus,

      \text {Gravitational force of planet}=\frac{G M m}{R^{2}}

Here, G is the gravitational constant, R is the radius of Mars and M, m is the mass of Mars and the object respectively..

Also, according to Newton’s second law of motion, the acceleration of any object will be equal to the ratio of force exerted on it to the mass of the object.

So in order to determine the acceleration due to gravity of Mars, divide the gravitational force of Mars by mass of object on the surface of Mars.

   Acceleration\ due\ to\ gravity\ of\ mars =\frac{\text {Gravitational force of Mars}}{m \text { of object }}

  Acceleration\ due\ to\ gravity\ of\ mars =\frac{G M m}{R^{2} \times m}=\frac{G M}{R^{2}}

  \text { Acceleration due to gravity of mars }=\frac{6.67259 \times 10^{-11} \times 6.1 \times 10^{23}}{3.4 \times 3.4 \times 10^{12}}=\frac{40.703 \times 10^{12}}{11.56 \times 10^{12}}

 \text { Acceleration }=3.5\ \mathrm{m} / \mathrm{s}^{2}

3 0
3 years ago
You throw a basketball and a tennis ball across the classroom so that each ball has the exact same momentum. Explain how this is
Helga [31]

Answer:

It is possible by increasing the speed of the tennis ball by a factor of (Mass of the tennis ball)/(Mass of the basketball)

Explanation:

The momentum of a body = The bod's mass × The body's velocity

Therefore, the momentum of a given mass of an object, such as a tennis ball can be increased by increasing the velocity or speed of the object. Whereby the speed of the ball, v₁, is increased such that the momentum of the basketball and the tennis ball will be the same, is given by the following equation

Mass of the basketball × v₂ = Mass of the tennis ball × v₁

Therefore, v₁/v₂ = (Mass of the tennis ball)/(Mass of the basketball)

7 0
3 years ago
Light propagate faster through medium “a” than medium “b”
dangina [55]

1) Medium "b" has more optical density

2) Light must hit the interface between the two mediums perpendicularly

Explanation:

1)

Refraction occurs when light propagates from a medium into a second medium.

The optical density of a medium is given by its index of refraction, which is defined as:

n=\frac{c}{v}

where

c is the speed of light in a vacuum

v is the speed of light in a medium

Higher index of refraction means higher optical density, and light propagater slower into a medium with higher optical density.

In this problem, light propagates faster through medium "a" than medium "b": this means that medium "a" has lower refractive index of medium "b", and so "b" has more optical density.

2)

We can answer this part by referring to Snell's law, which gives the relationship between the direction of the incident ray and of the refracted ray when light passes through the interface between two media:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the two mediums

\theta_1, \theta_2 are the angle of incidence and of refraction (the angle that light makes with the normal to the surface in medium 1 and medium 2)

Here we want the direction of propagation of the light ray not to change: this means that it must be

sin \theta_1 = sin \theta_2 (1)

However, here we have two mediums "a" and "b" with different index of refraction, so

n_1\neq n_2

Therefore the only angle that can satisfy eq.(1) is

\theta_1 = \theta_2 = 0

So, the light must hit the surface perpendicular to the interface between the two mediums.

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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