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frozen [14]
3 years ago
5

A 50.0-kg bungee jumper steps off a bridge with a light bungee cord tied to her and to the bridge. The unstretched length of the

cord is 13.0 m. The jumper reaches reaches the bottom of her motion 40.0 m below the bridge before bouncing back. We wish to find the time interval between her leaving the bridge and her arriving at the bottom of her motion. Her overall motion can be separated into an 13.0-m free-fall and a 27.0-m section of simple harmonic oscillation.] (a) For the free-fall part, what is the appropriate analysis model to describe her motion.
particle under constant acceleration particle in simple harmonic motion particle under constant angular acceleration
Physics
1 answer:
lina2011 [118]3 years ago
8 0

Answer:

The the analysis for the free fall part should be done under the constant acceleration.

Explanation:

In the given problem, the jumper is falling under the free fall. Since, no external force is acting on the body therefore, the fall will be under the action gravity only. also, the acceleration due to gravity is always constant.

Therefore, the the analysis for the free fall part should be done under the constant acceleration.

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Answer:

Explanation:

Given

Pressure P_1=1.01\times 10^5 Pa

Volume of air V=60 m^3

Initial Temperature T_1=289 K

T_2=302 K

Initial moles is given by

PV=nRT

n_1=\frac{P_1V_1}{RT_1}

n_1=\frac{1.01\times 10^5\times 60}{R\cdot 289}

when some gas escape out

no of moles is equal to

n_2=\frac{P_2\times V_2}{R\cdot T_2}

n_2=\frac{1.01\times 10^5\times 60}{R\cdot 302}

remaining no of moles =n_1-n_2

=\frac{1.01\times 10^5\times 60}{R\cdot 289}-\frac{1.01\times 10^5\times 60}{R\cdot 302}

=\frac{1.01\times 10^5\times 60}{R}(\frac{1}{289}-\frac{1}{302})

Mass of air escape out

=(n_1-n_2)\times M

=(\frac{1.01\times 10^5\times 60}{R}(\frac{1}{289}-\frac{1}{302}))\times 28

=25.272\times 10^3\ g

m=25.272\ kg

7 0
4 years ago
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3 years ago
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3 years ago
A sphere of radius R has total charge Q. The volume charge density (C/m3) within the sphere is rho(r)=C/r2, where C is a constan
san4es73 [151]

Answer: C = Q/4πR

Explanation:

Volume(V) of a sphere = 4πr^3

Charge within a small volume 'dV' is given by:

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Volume(V) of a sphere = 4/3(πr^3)

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FOR TOTAL CHANGE 'Q', we integrate dq

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7 0
3 years ago
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Firlakuza [10]

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Explanation:

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3 years ago
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