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saw5 [17]
3 years ago
14

An astronaut measures her mass by means of a device consisting of a chair attached to a large spring.Her mass can be determined

from the period of the oscillations she undergoes when set into motion.If her mass together with the chair is 170 kg, and the spring constant is 1250 N/m, what time is required for her to undergo 10 full oscillations?
Physics
1 answer:
Olenka [21]3 years ago
6 0

Answer:

23 seconds

Explanation:

Step one:

given data

mass m= 170kg

The spring constant is 1250 N/m

Required

The period required for 10 oscillations

Step two:

The expression relating period, mass, and spring constant is

T=2 \pi \sqrt{\frac{m}{k} } \\

substituting our data we have

T=2 *3.142 \sqrt{\frac{170}{1250} } \\\\T=6.284* \sqrt{0.136}\\\\T=6.284*0.3687\\T=2.3s

A Period is defined as the time required to complete one full oscillation

hence, having found the period to be 2.3 seconds, the time required for 10 oscillations will be 2.3*10= 23seconds

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Answer:

Explanation:

All the rest of the information is extraneous. The only 2 things you have to know are

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A 52 kg and a 95 kg skydiver jump from an airplane at an altitude of 4750 m, both falling in the pike position. Assume all value
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Answer: 52 kg skydiver: 9.09 m/s and 522.55 s

              95 kg skydiver: 12.3 m/s and 386.2 s

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For skydivers, when falling through the air, the forces acting on it are gravitational and drag forces. At a certain point, drag force equals gravitational force, which is constant on any part of the planet, producing a net force that is zero. Since there is no net force, there is no acceleration and, consequently, velocity is constant. When that happens, the person reached the <u>Terminal</u> <u>Velocity</u>.

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F_{G}=F_{D}

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m is mass in kg

g is acceleration due to gravitational force in m/s²

ρ is density of the fluid in kg/m³

C is drag coefficient

A is area of the object in the fluid in m²

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The 52kg skydiver has terminal velocity of:

v_{T}=\sqrt{\frac{2(52)(9.8)}{(1.21)(0.7)(0.14)} }

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The 95kg skydiver's terminal velocity is

v_{T}=\sqrt{\frac{2(95)(9.8)}{(1.21)(0.7)(0.14)} }

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The 52 kg and 95kg skydivers' terminal velocity are 9.09m/s and 12.3m/s, respectively.

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52 kg at 9.09 m/s:

t=\frac{4750}{9.09}

t = 522.5

95 kg at 12.3 m/s:

t=\frac{4750}{12.3}

t = 386.2

The 52 kg and 95kg skydivers' time to reach the floor are 522.5 s and 386.2 s, respectively.

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