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valkas [14]
3 years ago
6

A roller coaster is released from the top of a track that is 57 m high. Find the roller coaster speed when it reaches ground lev

el.
Physics
1 answer:
Art [367]3 years ago
7 0

Hi there!

Assuming the track is frictionless:

E_i = E_f\\\\PE = KE\\\\mgh = \frac{1}{2}mv^2

Cancel out the masses and rearrange to solve for velocity:

gh = \frac{1}{2}v^{2}\\\\v = \sqrt{2gh}\\\\

Plug in the given height and let g = 9.8 m/s²:

v = \sqrt{2(9.8)(57)} = \boxed{33.42 m/s}

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The answer is 125 Joules

The first thing to take note of is the work equation: W=F×D

Since we already have our force and our distance that will help make this problem easier.

So, W=25*5

W=125

Therefore, our answer is 125 Joules since work is measured in joules

Hope this helped!! :)


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3 years ago
What type of motion does this graph represent?
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The aorta is the main artery from the heart. a typical aorta has an inside diameter of 1.8 cm and carries blood at speeds of up
defon

Answer:

Explanation:

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a₁v₁  = a₂ v₂

π r₁² x v₁ = π r₂² x v₂

(0.9 x 10⁻²)² x .35 = ( .45 x 10⁻² )² x v₂

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3 0
3 years ago
A 11.0 kg satellite has a circular orbit with a period of 1.80 h and a radius of 7.50 × 106 m around a planet of unknown mass. I
Anuta_ua [19.1K]

Answer:

Explanation:

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GM = 4 x 3.14² x 7.5³ x 10¹⁸ / 6480²

= 3.96 X 10¹⁴

Expression for acceleration due to gravity

g = GM / R² where R is radius of satellite

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R² = 3.96 X 10¹⁴ / 20

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8 0
3 years ago
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Rufina [12.5K]

Answer:

Glucose and Oxygen

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Cellular respiration can be simply expressed as shown below:

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The reactants are glucose and oxygen.

The products are CO₂, water and ATP

7 0
3 years ago
Read 2 more answers
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