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:
Explanation:
The relation between time period of moon in the orbit around a planet can be given by the following relation .
T² = 4 π² R³ / GM
G is gravitational constant , M is mass of the planet , R is radius of the orbit and T is time period of the moon .
Substituting the values in the equation
(.3189 x 24 x 60 x 60 s)² = 4 x 3.14² x ( 9380 x 10³)³ / (6.67 x 10⁻¹¹ x M)
759.167 x 10⁶ = 8.25 x 10²⁰ x 39.43 / (6.67 x 10⁻¹¹ x M )
M = .06424 x 10²⁵
= 6.4 x 10²³ kg .
Answer:
the acceleration is 130.3m/s²
Explanation:
Given data
Force F= 18.9N
Mass of ball m= 0.145kg
Acceleration a=?
Applying the Newton's second law of motion
"The rate of change of momentum of a body is proportional to the external force".
F=ma
a= F/m
a= 18.9/0.142
a= 130.3m/s²
