Answer:
g_x = 3.0 m / s^2
Explanation:
Given:
- Change in length of spring [email protected] = 22.6 cm
- Time taken for 11 oscillations t = 19.0 s
Find:
- The value of gravitational free fall g_x at plant X:
Solution:
- We will assume a simple harmonic motion of the mass for which Time is:
T = 2*pi*sqrt(k / m ) ...... 1
- Sum of forces in vertical direction @equilibrium is zero:
F_net = k*x - m*g_x = 0
(k / m) = (g_x / x) .... 2
- substitute Eq 2 into Eq 1:
2*pi / T = sqrt ( g_x / x )
g_x = (2*pi / T )^2 * x
- Evaluate g_x:
g_x = (2*pi / (19 / 11) )^2 * 0.226
g_x = 3.0 m / s^2
Answer:
-2.5 is the answer to your question
I’m gonna have to say “Ocean waves” as the answer
To solve this problem it is necessary to apply the concepts related to the Centrifugal Force and the Gravitational Force. Since there is balance on the body these two Forces will be equal, mathematically they can be expressed as
Where,
m = Mass
G =Gravitational Universal Constant
M = Mass of the Planet
r = Distance/Radius
Re-arrange to find the velocity we have,
At the same time we know that the period is equivalent in terms of the linear velocity to,
If our values are that the radius of mars is 3400 km and the distance above the planet is 100km more, i.e, 3500km we have,
Replacing we have,
Therefore the correct answer is C.