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Inessa [10]
3 years ago
6

How fast must a cyclist climb a 6.0º hill to maintain a power output of 0.25 hp? Neglect work done by friction, and assume the m

ass of cyclist plus bicycle is 68 kg.
Physics
1 answer:
N76 [4]3 years ago
4 0

Answer:  2.7m/s

Explanation:

cyclist fight only against the component of weight along the road (no friction) , that is:

fx =Wsin6° = mgsin6°=68*9.8*sin6°=69.6N downward

so the force that moves the bike over the hill will be equal to fx but upward

Now we now that the power P = 0.25hp= 186.5W  is:

P=f*v  so v = P/fx  = 186.5/69.6 = 2.7m/s

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3 years ago
0.5-lbm of a saturated vapor is converted to asaturated liquid by being cooled in a weighted piston-cylinder device maintained a
klio [65]

Answer:

The boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

Explanation:

Given the data in the question;

Using the Clapeyron equation

(\frac{dP}{dT} )_{sat } = \frac{h_{fg}}{Tv_{fg}}

(\frac{dP}{dT} )_{sat } = \frac{\frac{H_{fg}}{m} }{T\frac{V_{fg}}{m} }

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we substitute

(\frac{dP}{dT} )_{sat } =( \frac{250Btu\frac{778Ibf-ft}{Btu} }{0.5}) / ( (15+460)\frac{1.5}{0.5})  

(\frac{dP}{dT} )_{sat } = 272.98 Ibf-ft²/R

Now,

(\frac{dP}{dT} )_{sat } = (\frac{P_2 - P_1}{T_2 - T_1})_{sat

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P₂ is the final pressure ( 60 psia )

T₁ is the initial temperature ( 15 + 460 )R

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we substitute;

T_2 = ( 15 + 460 ) + \frac{(60-50)psia(\frac{144in^2}{ft^2}) }{272.98}

T_2 = 475 + 5.2751\\

T_2 = 480.275 R

Therefore, boiling point temperature of this substance when its pressure is 60 psia is  480.275 R

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