0
9.8 m/s2
----------------------
Yes this is the average value of acceleration due to gravity near by the surface of the earth. As we go higher and higher level this g value decreases and becomes almost negligible. Same way as we go deeper and deeper the g value decreases and at the centre of the earth its value becomes zero.
Still curious? Ask our experts.
Chat with our AI personalities
Ezra
Coach
ViviAdd your answer:
Earn +20 pts
What is mass multiplied by gravity?
Force or weight Force= mass X acceleration gravity is an acceleration (9.8m/s2) Weight = mass X acceleration due to gravity
What is the difference between gravitational acceleration and acceleration?
Gravitational acceleration is simply acceleration due to gravity.
What are the effects of acceleration due to gravity on the time period of a pendulum?
The period of a pendulum (in seconds) is 2(pi)√(L/g), where L is the length and g is the acceleration due to gravity. As acceleration due to gravity increases, the period decreases, so the smaller the acceleration due to gravity, the longer the period of the pendulum.
How is it that acceleration due to gravity is EXACTLY thirty-two feet per second squared?
That it is exactly 32f/sec^2 not sure. That it is constant depends on the Earths' mass which is fairly constant.
When drawing the graph of distance vs time squared how do you calculate the the acceleration due to gravity?
Example: x axis = time, y axis = distance, plot values of s, when t = say 0 to 10, step 1 > If time is the variable, and distance the dependent, you should have been given a figure for acceleration (g), without which, you cant plot the graph. > Acceleration due to earths gravity (g) at earths surface radius is generally taken as = 9.82 metres per second / per second. > Use: s = (u*t) + (0.5 * g * t2) > where: s = distance u = initial velocity g = acceleration due to gravity (9.82 (m/s)/s) t = elapsed time
