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vesna_86 [32]
3 years ago
9

A 70 kilogram mountaineer is standing on the summit of Mt. Everest. The distance between the mountaineer and the center of Earth

is 6.39 × 106 meters. What is the magnitude of the force of gravity acting on the mountaineer? (The value of G is 6.673 × 10-11 newton meter2/kilogram2. The mass of Earth is 5.98 × 1024 kilograms.)
90.50 newtons



111.5 newtons



684.1 newtons



4041.7 newtons
Physics
2 answers:
Ilia_Sergeevich [38]3 years ago
8 0
 <span>F = GMm/r² 

F = (6.673*10^-11)(5.98*10^24)(70)/(6.39*10^... 

F = 684 N (3sf)</span>
madreJ [45]3 years ago
5 0
The force due to gravity between two point masses is given by:
F = GMm/r²; where G is the gravitational constant, M is the larger mass, m is the smaller mass, and r is the distance between the two.
F =  (6.673 × 10⁻¹¹ × 5.98 × 10²⁴ × 70) / (6.39 × 10⁶)²
F = 684.1 N
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3 years ago
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Answer:

The answer is below

Explanation:

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\omega^2=\omega_o^2+2\alpha \theta\\\\19.67^2-50.13^2=2\alpha(272.9)\\\\19.67^2=50.13^2+2\alpha(272.9)\\\\2\alpha(272.9)=-2126.108\\\\\alpha=-3.89\ rad/s^2\\\\

b)

\omega=\omega_o+\alpha t\\\\19.67=50.13+(-3.89t)\\\\3.89t=50.13-19..67\\\\3.89t=30.46\\\\t=7.83\ s

c) θ = 95 rev * 2πr = 95 * 2π * 0.4572= 272.9 rad

a) When it stops, the final angular velocity is 0. Hence:

\omega^2=\omega_o^2+2\alpha \theta\\\\0=50.13^2+2(-3.89)\theta\\\\2(3.89)\theta=50.13^2\\\\2(3.89)\theta=2513\\\\\theta=323\ rad\\\\revolutions=\frac{\theta}{2\pi r}=\frac{323}{2\pi(0.4572)}  =112.4\ rev

θ = 323 rad

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