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
<span>Because the moon has no atmosphere, it is not possible for geologic events to occur on the moon.</span>
Answer:
<u>Given</u><em> </em><em>-</em><em> </em><u>M</u><u> </u><u>=</u><u> </u>20 kg
k = 0.4
F = 200 N
<u>To </u><u>find </u><u>-</u><u> </u> acceleration
<u>Solution </u><u>-</u><u> </u>
F= kMA
200 = 0.4 * 20 * acceleration
200 = 8 * a
a = 8/200
a = 0.04 m s²
<h3>a = 0.04 m s²</h3>
The interaction of electric currents or fields and magnetic fields.
