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3 years ago

How do Swati and Banks adjust their body position during a skydiving jump so they can fall at the same rate?

1 answer:

3 0

Sky diving involves free fall under gravity along with the drag due to air molecules pushing against the body slowing the rate of fall of a body. This is actually a significant amount of force. The drag force depends on the contact surface area and weight of the body. More the surface area in contact, more would be the drag. The sitting position of the skydiver would experience less drag than the chest down position. This is because of the less contact surface area of the body with the air molecules while in the former case. Since no two persons have identical body shape and weight, the rate of fall can be made nearly equal but not exactly equal. This is would be possible when they are having same body position.

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Give two explains food that are good sources of protein

Ad libitum [116K]

Answer:

Chicken and peanut butter lol

Explanation:

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3 years ago

An ideal spring hangs from the ceiling. A 1.25-kg mass is hung from the spring. After all vibrations have died away, the spring

ch4aika [34]

The kinetic energy of the mass at the instant it passes back through its equilibrium position is about 1.20 J

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<h3>Further explanation</h3>

Let's recall Elastic Potential Energy formula as follows:

$\boxed{E_p = \frac{1}{2}k x^2}$

where:

<em>Ep = elastic potential energy ( J )</em>

<em>k = spring constant ( N/m )</em>

<em>x = spring extension ( compression ) ( m )</em>

Let us now tackle the problem!

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<u>Given:</u>

mass of object = m = 1.25 kg

initial extension = x = 0.0275 m

final extension = x' = 0.0735 - 0.0275 = 0.0460 m

<u>Asked:</u>

kinetic energy = Ek = ?

<u>Solution:</u>

<em>Firstly , we will calculate the spring constant by using </em><em>Hooke's Law</em><em> as follows:</em>

$F = k x$

$mg = k x$

$k = mg \div x$

$k = 1.25(9.8) \div 0.0275$

$k = 445 \frac{5}{11} \texttt{ N/m}$

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<em>Next , we will use </em><em>Conservation of Energy</em><em> formula to solve this problem:</em>

$Ep_1 + Ek_1 = Ep_2 + Ek_2$

$\frac{1}{2}k (x')^2 + mgh + 0 = \frac{1}{2}k x^2 + Ek$

$Ek = \frac{1}{2}k (x')^2 + mgh - \frac{1}{2}k x^2$

$Ek = \frac{1}{2}k ( (x')^2 - x^2 ) + mgh$

$Ek = \frac{1}{2}(445 \frac{5}{11}) ( 0.0460^2 - 0.0275^2 ) + 1.25(9.8)(0.0735)$

$\boxed {Ek \approx 1.20 \texttt{ J}}$

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<h3>Learn more</h3>

Kinetic Energy : brainly.com/question/692781

Acceleration : brainly.com/question/2283922

The Speed of Car : brainly.com/question/568302
Young Modulus : brainly.com/question/9202964
Simple Harmonic Motion : brainly.com/question/12069840

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<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Elasticity

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