Answer:
1069.38 gallons
Explanation:
Let V₀ = 1.07 × 10³ be the initial volume of the gasoline at temperature θ₁ = 52 °F. Let V₁ be the volume at θ₂ = 97 °F.
V₁ = V₀(1 + βΔθ) β = coefficient of volume expansion for gasoline = 9.6 × 10⁻⁴ °C⁻¹
Δθ = (5/9)(97°F -52°F) °C = 25 °C.
Let V₂ be its final volume when it cools to 52°F in the tank is
V₂ = V₁(1 - βΔθ) = V₀(1 + βΔθ)(1 - βΔθ) = V₀(1 - [βΔθ]²)
= 1.07 × 10³(1 - [9.6 × 10⁻⁴ °C⁻¹ × 25 °C]²)
= 1.07 × 10³(1 - [0.024]²)
= 1.07 × 10³(1 - 0.000576)
= 1.07 × 10³(0.999424)
= 1069.38 gallons
The minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
<h3>
What is average speed?</h3>
The average speed of an object is the ratio of total distance traveled by the object to the total time of motion of the object.
<h3>Total time taken by the car during the entire race</h3>
time = distance/average speed
time = (1.41 km) / (278 km/h)
time = 0.0051 hr
The car travels the first half of the race, d (¹/₂ x 1410 m) at 210 km/h;
d = 705 m = 0.705 km
t1 = 0.705/210
t1 = 0.0034 hr
<h3>time for the second half</h3>
t2 = 0.0051 - 0.0034 hr
t2 = 0.0017 hr
<h3>minimum average speed of the second half</h3>
v = d/t
v = 0.705 km / 0.0017 hr
v = 414.7 km/hr
Thus, the minimum average speed it must have in the second half of the event in order to qualify is 414.7 km/h.
Learn more about average speed here: brainly.com/question/4931057
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Answer: 2.72727272......g/cm3^
Explanation:
D = g/cm^3
soo d = 15/5.5
