<span> gravitational force varies based on 1/r^2
when you're double the distance =10,000 to 20,000, the force is 4 times smaller so on and so forth.
</span><span>As force is proportional to 1 / {distance squared}, the force will be 1 / 2^2 (i.e. 1/4) of the force at the reference distance (i.e. 1/4 * 600 = 150 lb)
</span>hope this helps
The spring is initially stretched, and the mass released from rest (v=0). The next time the speed becomes zero again is when the spring is fully compressed, and the mass is on the opposite side of the spring with respect to its equilibrium position, after a time t=0.100 s. This corresponds to half oscillation of the system. Therefore, the period of a full oscillation of the system is

Which means that the frequency is

and the angular frequency is

In a spring-mass system, the maximum velocity of the object is given by

where A is the amplitude of the oscillation. In our problem, the amplitude of the motion corresponds to the initial displacement of the object (A=0.500 m), therefore the maximum velocity is
Answer:
Yes. A 200 kg bucket of cement = About 440.925 pounds of cements.
Explanation:
Answer:
The acceleration of the both masses is 0.0244 m/s².
Explanation:
Given that,
Mass of one block = 602.0 g
Mass of other block = 717.0 g
Radius = 1.70 cm
Height = 60.6 cm
Time = 7.00 s
Suppose we find the magnitude of the acceleration of the 602.0-g block
We need to calculate the acceleration
Using equation of motion

Where, s = distance
t = time
a = acceleration
Put the value into the formula



Hence, The acceleration of the both masses is 0.0244 m/s².