The woman weighs 100N now.
Explanation:
Given:
F is given as 400N
one full additional earth radius formed
To Find:
Weight of the woman=?
Solution:
We know that gravitational force is inversely proportional to the square of radius.
where
is the gravitational force
is the radius square
So when radius is doubled, the force becomes one fourth of the original.
Therefore
You can do animals that have 4 legs (dogs, horses), animals that fly (parrot, eagle), animals that live in the water (goldfish,bass)
Answer:
the super train is not completely within the tunnel.
Explanation:
From the given information:
The proper length = 1200 ft
The super train speed = 0.99c
By applying the concept of length contraction, the contracted length of the super-train can be determined by using the formula:
L = 169.3 ft
≅ 169 ft
Thus, the contracted length is 169 ft more than the proper length of the tunnel L' 50 ft.
As such, the super train is not completely within the tunnel.
Answer:
ΔF = 0.21 N
Explanation:
For this exercise that does not ask for the change in weight we must use the law of universal gravitation
F = G M m / r²
where r is the distance from the center of the Earth, in the lower part of the building is
r =
for the upper part of the building h = 1 mile = 1609.34 m
r =R_{e} + h
the weight or the force of attraction of gravity on the floor is F = 817 N, therefore the equation remains
817 = (G M m / R_{e}²)
let's find this force for the top of the building
F` = G M m / (R_{e} + h)²
let's take out R_{e} common factor
F ’= (G M m / R_{e}²) 1 / (1 + h / R_{e})²2
F ’= (G M m R_{e}²) (1 + h / R_{e})⁻²
as the quantity h /R_{e} = 1609 / 6.37 10⁶ << 1 we can make a series space
(1 + x)⁻² = 1 -2 x + ...
we substitute
F ’= (GMm /R_{e}²) (1 - 2 1609 / 6.37 10⁶)
F ’= (GMm /R_{e}²) (1 - 2.53 10⁻⁴) = (GMm / R_{e}²) 0.99974
F ’= 817 0.99974
F ’= 816.79 N
weight Change
ΔF = ΔW = 817 - 816.79
ΔF = 0.21 N
as we see this is a very small amount