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qaws [65]
3 years ago
8

What is the sprinters output at 2.0 s, 4.0 s and 6.0 s?

Physics
1 answer:
kari74 [83]3 years ago
5 0

A 59 kg sprinter, starting from rest, runs 47 m in 7.0 s at constant acceleration.?  

What is the sprinter's power output at 2.0 s, 4.0 s, and 6.0 s?  

Instantaneous Power is the force times velocity  

P = Fv  

Because the acceleration is constant, the force will be constant as well  

F = ma  

P = mav  

for constant acceleration, the velocity at each time is found using  

v = at  

P = ma(at) = ma²t  

find the acceleration using kinematic equation  

s = ½at²  

a = 2s/t²  

a = 2(47) / 7.0²  

a = 1.918 m/s²  

P(2.0) = 59(1.918²)2.0 = 434.25 W = 0.43 kW  

P(4.0) = 59(1.918²)4.0 = 868.51 W = 0.87 kW  

P(6.0) = 59(1.918²)6.0 = 1302.76 W = 1.3 kW  

I hope this helped.  


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38.4 mol of krypton is in a rigid box of volume 64 cm^3 and is initially at temperature 512.88°C. The gas then undergoes isobari
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Answer:

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Final Pressure second process P_{3} = 1,317 * 10^{10} Pa

Explanation:

Using the Ideal Gases Law yoy have for pressure:

P_{1} = \frac{n_{1} R T_{1} }{V_{1} }

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P is the pressure, in Pa

n is the nuber of moles of gas

R is the universal gas constant: 8,314 J/mol K

T is the temperature in Kelvin

V is the volumen in cubic meters

Given that the amount of material is constant in the process:

n_{1} = n_{2} = n

In an isobaric process the pressure is constant so:

P_{1} = P_{2}

\frac{n R T_{1} }{V_{1} } = \frac{n R T_{2} }{V_{2} }

\frac{T_{1} }{V_{1} } = \frac{T_{2} }{V_{2} }

V_{2} = \frac{T_{2} V_{1} }{T_{1} }

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Replacing on the ideal gases formula the pressure at this piont is:

P_{2} = 3,92 * 10^{9} Pa

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T = \frac{P V }{n R }

For the second process you have that T_{2} = T_{3}  So:

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P_{2} V_{2}  = P_{3} V_{3}

P_{3} = \frac{P_{2} V_{2}}{V_{3}}

P_{3} = 1,317 * 10^{10} Pa

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