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Mrrafil [7]
3 years ago
12

The intersection angle of a 3 degree curve is 45.2 degrees. What is the length of the curve? of Select one: O a. 455 m O b.573 m

C. 452 m O d. 25.9 km
Chemistry
1 answer:
ValentinkaMS [17]3 years ago
7 0

Explanation:

Relation between length of a curve and angle is as follows.

               l = R \times \Theta

where,   R = radius of curve

         \Theta = angle in radians

Also,     l = R \times \Theta \times \frac{\pi}{180}  .......... (1)

If curve has a degree of curvature D_{a} for standard length s, then

               R = \frac{s}{D_{a}} \times \frac{180}{\pi}   ........... (2)

Now, substitute the value of R from equation (2) into equation (1) as follows.

               l = \frac{s \times \Theta}{D_{a}}  

If s = 30 m, then calculate the value of l as follows.

                 l = \frac{s \times \Theta}{D_{a}}  

                   = 30 \times \frac{45.2}{3}              

                   = 452 m

thus, we can conclude that the length of the curve is 452 m.

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what is the volume of the air in a balloon that occupies 0.730 L at 28.0 c if the temperature is lowered to 0.00 C
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Answer:

The volume of the air is 0.662 L

Explanation:

Charles's Law is a gas law that relates the volume and temperature of a certain amount of gas at constant pressure. This law says that for a given sum of gas at a constant pressure, as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases because the temperature is directly related to the energy of the movement they have. the gas molecules. This is represented by the quotient that exists between volume and temperature will always have the same value:

\frac{V}{T}=k

If you have a certain volume of gas V1 that is at a temperature T1 at the beginning of the experiment and several the volume of gas to a new value V2, then the temperature will change to T2, and it will be true:

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In this case:

  • V1= 0.730 L
  • T1= 28 °C= 301 °K (0°C= 273°K)
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Solving:

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V2=0.662 L

<u><em>The volume of the air is 0.662 L</em></u>

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