Answer:
16,18,22
Or
1,3,7
Explanation:
The detailed explanation is contained in the image attached. The lengths are found using Pythagoras theorem and the two lengths reflects the two values of x yielded by the quadratic equation
an astronaut weighs 8.00x10^2 on the surface of earth. what is the weight of the astronaut 6.37x10^6 meters above the surface of
1 answer:
To solve this problem, we use the Law of Universal Gravitation:
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
F = Gm1m2/d^2
where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:
F = k/d^2
k = Fd^2
The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1
(8E+2)(d1)^2 = F2(2d1)^2
(8E+2)(d1)^2 = 4F2(d1)^2
(8E+2)=4F2
F2 = 200 Newtons
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