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astraxan [27]
4 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]4 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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A 1.4 kg particle moves along an x axis, being propelled by a variable force directed along that axis. Its position is given by
mihalych1998 [28]

Answer:

The value of c is 27.4 m/s²

Explanation:

Hi there!

Let´s write the position function:

x = 3.0 m + (4.0 m/s) · t + c · t² - (1.6 m/s³) · t³

The velocity of the particle is given by the derivative of the position function with respect to time:

dx/dt = v = 4.0 m/s + 2 · c · t - 4.8 m/s³ · t²

The acceleration of the particle is the derivative of the velocity function with respect to t:

dv/dt = a = 2 · c - 9.6 m/s³ · t

The applied force at t = 3.0 s is calculated as follows:

F = m · a

Where:

F = applied force.

m = mass of the particle.

a = acceleration.

Then:

F = m · a

36 N = 1.4 kg · a

36 N / 1.4 kg = a

a = 26 m/s²

We have derived the equation of the acceleration above:

a = 2 · c - 9.6 m/s³ · t

Then, using a = 26 m/s² and t = 3.0 s, we can solve the equation for c:

26 m/s² = 2 · c - 9.6 m/s³ · 3.0 s

26 m/s² + 9.6 m/s³ · 3.0 s  = 2 · c

54.8 m/s² = 2 · c

54.8 m/s² / 2 = c

c = 27.4 m/s²

The value of c is 27.4 m/s²

4 0
4 years ago
A student pushes a 2.85 kg cart causing it to accelerate at a rate of 4.9 m/s squared .What amount of force must the student hav
uysha [10]
In order to find the force (F), you would have to use the formula for it:
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7 0
4 years ago
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