You, a 70 kg person, leap from a 10 m tall building and land feet first on a trampoline. The center of the trampoline where you
1 answer:
Answer:
1373.4 N/m
Explanation:
Hooke's law states that the extension of a spring and force are related by the expression, F=kx where k is spring constant, x is extension of spring and F is the applied force. Making k the subject of the formula then
Also, F=gm hence the above formula is modified as
Taking g as 9.81 m/s2 , x as 0.5 m and m as 70 kg then
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Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C