If a person weighs 882 N on the surface of the earth, at what altitude above the earth’s surface must they be for their weight t
1 answer:
To calculate weight we use formula:
When we have two bodies we use general gravity formula:
Where:
G = gravity constant =
Mp = mass of person = 882 / 9.81= 89.9kg
ME = mass of Earth =
r = distance between Earth and person
From these two equations we find that left sides are equal so the right sides must be equal too.
We solve this for r:
r=6371116m = 6371.116km
This is distance from center of Earth. Radius of Earth is 6370km and height above surface is 6371.116 - 6370 = 1.116km or 1116m.
When we have two bodies we use general gravity formula:
Where:
G = gravity constant =
Mp = mass of person = 882 / 9.81= 89.9kg
ME = mass of Earth =
r = distance between Earth and person
From these two equations we find that left sides are equal so the right sides must be equal too.
We solve this for r:
r=6371116m = 6371.116km
This is distance from center of Earth. Radius of Earth is 6370km and height above surface is 6371.116 - 6370 = 1.116km or 1116m.
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Answer:
x = v₀ cos θ t , y = y₀ + v₀ sin θ t - ½ g t2
Explanation:
This is a projectile launch exercise, in this case we will write the equations for the x and y axes
Let's use trigonometry to find the components of the initial velocity
sin θ = / v₀
cos θ = v₀ₓ / v₀
v_{y} = v_{oy} sin θ
v₀ₓ = vo cos θ
now let's write the equations of motion
X axis
x = v₀ₓ t
x = v₀ cos θ t
vₓ = v₀ cos θ
Y axis
y = y₀ + t - ½ g t2
y = y₀ + v₀ sin θ t - ½ g t2
v_{y} = v₀ - g t
v_{y} = v₀ sin θ - gt
= v_{oy}^2 sin² θ - 2 g y
As we can see the fundamental change is that between the horizontal launch and the inclined launch, the velocity has components