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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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Plz i need help for the 5 problems. plz show the work!!!
Artemon [7]

Answer:

1.   3 m/s^{2}

2.   1.5 m/s^{2}

3.   3 seconds

4.   0 m/s^{2}

5.   2.2 seconds

Explanation:

(1)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=0 since it’s at rest, v=30m/s and t=10 seconds

a = \frac {30-0}{10}=3 m/s^{2}

(2)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=10m/s, v=22m/s and t=8 seconds

a = \frac {22-10}{8}=1.5 m/s^{2}

(3)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=0m/s since at rest, v=15m/s and a=5 \frac {m}{s^{2}}

= \frac {15-0}{5}=3s

(4)

When initial and final velocity are constant, there’s no acceleration as proven below

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making a the subject we have

a=\frac {v-u}{t}

Substituting u=20 since it’s at rest, v=20m/s and t=10 seconds

a = \frac {20-20}{10}=0 m/s^{2}

(5)

From v= u + at where v is final velocity, u is initial velocity, a is acceleration and t is time.

Making t the subject we have

t=\frac {v-u}{a}

Substituting u=9m/s since at rest, v=0m/s and a=-4.1 \frac {m}{s^{2}}

= \frac {0-9}{-4.1}=2.2s

8 0
3 years ago
A delivery truck leaves a warehouse and travels 2.60 km north. The truck makes a left turn and travels 1.25 km west before makin
yulyashka [42]

Answer:

4.19 km and 107.35 degrees north of east

Explanation:

So in the end, the truck is (2.6 + 1.4 = 4km) north and 1.25 km west from the warehouse. We can use the Pythagorean formula to calculate the magnitude and direction α of the truck displacement from the warehouse:

s = \sqrt{s_n^2 + s_w^2} = \sqrt{4^2 + 1.25^2} = \sqrt{16 + 1.5625} = \sqrt{17.5625} = 4.19 km

tan\alpha = \frac{s_n}{s_w} = \frac{4}{1.25} = 3.2

\alpha = tan^{-1}3.2 = 1.27 rad \approx 72.65 degrees north or west or (180 - 72.65) = 107.35 degrees north of east

3 0
3 years ago
If you want to make a flashlight and have two batteries and a light bulb how would you hook them up to make it bright and why?
ValentinkaMS [17]

Answer:

The positive end of battery 1 touching the negative end of battery 2 and wires connecting the negative of battery 1 to the light light bulb and the positive of battery 2 to the light bulb.

Explanation:

7 0
3 years ago
The components of vector A are:
Korvikt [17]

here as it is given that x component of the vector is positive while y component of the vector is negative so we can say the vector must inclined in Fourth quadrant.

So angle must be more than 270 degree and less than 360 degree

Now in order to find the value we can say that

tan\theta = \frac{opposite\: side}{adjacent\: side}

tan\theta = \frac{8.6}{6.1}

\theta = tan^{-1}1.41

\theta = 54.65^0

so it is inclined at above angle with X axis in fourth quadrant

Now if angle is to be measured counterclockwise then its magnitude will be

\theta = 360 - 54.65 = 305.3^0

so the correct answer will be 305 degree

3 0
3 years ago
The amount of light lost within a fiber-optic system is known as
densk [106]
Attenuation is the correct answer. 
6 0
3 years ago
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