The spring constant (k) is equal to the slope multiplied by the acceleration due to gravity. Show why this is the case. Note: (K=slope*g) & slope of particular problem is 0.2621.
The spring constant (k) is equal to the slope multiplied by the acceleration due to gravity....
Are the parameters 'acceleration due to gravity g' and 'spring constant k' different on Earth and Moon or are they not? Elaborate and explain.
In this experiment, the acceleration due to gravity ("g") was equal to the slope of the line obtained by plotting 4π2L on the vertical axis (y) and which of the following on the horizontal (x) axis? Question 2 options: T T squared T cubed L L squared
At what altitude above the Earth's surface is the acceleration due to gravity equal to g/132?
The value of the acceleration due to gravity is not constant, decreasing with height above the surface of the Earth.What is the value of the acceleration due to gravity, in m s-2, in the ionosphere at a height 3.9x105 m above the Earth's surface? Give your answer to 3d.p.
Pre-Lab Activity - Predictions of velocity, acceleration and force for a moving spring Question 1: (1pt) For our experiment, we will change madded, the mass added to the spring, and measure Ay, the displacement after the spring achieves equilibrium, that will change in response. Because the spring force will balance the force of gravity of the mass hanging on the spring, Newton's 2nd law results in equation 3 from the Introduction: kAy madded9 (3) Where g is the acceleration due...
At the surface of the moon, the acceleration due to the gravity of the moon is a. What is the acceleration due to the gravity of the moon at a diatance fron the center of the moon equal to 4 times the radius of the moon?
The acceleration due to gravity, g, is constant at sea level on the Earth's surface. However, the acceleration decreases as an object moves away from the Earth's surface due to the increase in distance from the center of the Earth. Derive an expression for the acceleration due to gravity at a distance h above the surface of the Earth, 9h. Express the equation in terms of the radius R of the Earth, g, and h. 9A Suppose a 74.35 kg...
a particular planet has a radius of 1000km and the acceleration due to gravity at its surface is 4m/s^2. what is the acceleration due to gravity 1000km above its surface?
o) The acceleration due to gravity on the moon is 1/6 of what it is on Earth. We are going to do clock experiments on the moon. We have one clock that runs because of the oscillation of a ass (m) on an ideal spring (spring constant k), which oscillates with an amplitude A, We have a second clock that runs because of a simple pendulum made of a mass (m) on a wire of length L, which oscillates with...
(a) Explain the relationship between the universal gravitational constant G and the acceleration due to gravity at the earth's surface g. Therefore, calculate g from G using the relationships given below. Justify the choice of units for G (Nm kg?). F= mg The mass of the earth is 5.98 x 1024 kg, it's radius is Tg = 6.38 x 10 m, and G = 6.67 x 10-11 Nm?kg (10 marks) (b) Explain, including performing the integral, how the work done...