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Elena-2011 [213]
2 years ago
13

The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m3. Convert this volume to liters

and express the result in standard exponential notation?
Physics
1 answer:
maksim [4K]2 years ago
3 0
<h2>Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3.67 x 10⁹ L</h2>

Explanation:

The Vehicle Assembly Building at the Kennedy Space Center in Florida has a volume of 3,666,500 m³.

Volume = 3,666,500 m³

1 m³ = 1000 L

So volume = 3,666,500 x 1000 = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3666500000 L

Volume of vehicle Assembly Building at the Kennedy Space Center in Florida = 3.67 x 10⁹ L

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Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²

v = 5.4 m/s



Case B :

B .60 kg 35 N/m .9 m

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k = spring constant of the spring = 35 N/m

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v = 4.6 m/s




Case C :

C .55 kg 40 N/m 1.1 m

m = mass of car = 0.55 kg

k = spring constant of the spring = 40 N/m

h = height of the hill = 1.1 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²

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Case D :

D .84 kg 32 N/m .95 m

m = mass of car = 0.84 kg

k = spring constant of the spring = 32 N/m

h = height of the hill = 0.95 m

x = compression of spring = 0.25 m

Using conservation of energy between Top of hill and Bottom of hill

Total energy at Top of hill = Total energy at Bottom of hill

spring energy + potential energy = kinetic energy

(0.5) k x² + mgh = (0.5) m v²

(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²

v = 4.6 m/s


hence closest is in case C at 5.1 m/s




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\Delta h=L(1-\cos \theta )

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mgL(1-\cos \theta )=\frac{mv^2}{2}

v=\sqrt{2gL(1-\cos \theta )}

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Answer:

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thickness = 12

moves the entire = 12.0 cm

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a = \frac{-150}{9.8}

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