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:
Lifetime = 4.928 x 10^-32 s
Explanation:
(1 / v2 – 1 / c2) x2 = T2
T2 = (1/ 297900000 – 1 / 90000000000000000) 0.0000013225
T2 = (3.357 x 10^-9 x 1.11 x 10^-17) 1.3225 x 10^-6
T2 = (3.726 x 10^-26) 1.3225 x 10^-6 = 4.928 x 10^-32 s
Evaporation (or another word to use is water vapor.)
Is the variable you change, independent, I, something I change.
Answer: 
Explanation:
This problem can be solved by the following equation:
Where:
is the change in kinetic energy
is the electric potential difference
is the electric charge
Finding :
Finally:
