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Lelu [443]
3 years ago
6

Suppose a 96.10 kg hiker has ascended to a height of 1841 m above sea level in the process of climbing Mt. Washington. By what p

ercent has the hiker's weight changed from its value at sea level as a result of climbing to this elevation? Use g=9.807 m/s2 and R=6.371×106 m . Keep four significant figures in your weight calculation to find the percent difference, and enter your answer as a positive value.
Physics
1 answer:
lbvjy [14]3 years ago
6 0

Answer:

0.029%

Explanation:

Weight on the surface is given by,

F_W = mg = (96.1Kg)(9.8m/s^2)=941.78N

The acceleration due to gravity at given height is,

g_h=\frac{gR^2}{R+h^2}

The weight in another height is equal to

F_h=mg_h

F_h= m*\frac{gR^2}{R+h^2}

F_h = \frac{F_WR^2}{(R+h)^2}

F_h=\frac{(941.78)(6.371*10^6)}{(6.371*10^6+1841)}

F_h = 941.50N

The change of the Weight is,

\frac{F-F_h}{F}*100=\frac{941.78-941.50}{941.78}*100=0.029\%

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