k = spring constant of the spring = 100 N/m
m = mass hanging from the spring = 0.71 kg
T = Time period of the spring's motion = ?
Time period of the oscillations of the mass hanging is given as
T = (2π) √(m/k)
inserting the values in the above equation
T = (2 x 3.14) √(0.71 kg/100 N/m)
T = (6.28) √(0.0071 sec²)
T = (6.28) (0.084) sec
T = 0.53 sec
hence the correct choice is D) 0.53
Answer:
In crash one, because the pink soccer ball had more mass, so it would have a stronger force.
Answer:
Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
Explanation:
The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.
When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.
The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.
The velocity of the orbit is given by the relation,

Where
V - velocity of the orbit at a height h from the surface
R - Radius of the second object
G - Gravitational constant
h - height from the surface
The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.
We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"
- it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
From the question we are told
the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?
Generally the equation for the Force is mathematically given as
F=\frac{F}{dx}
Therefore
F=-kdx
k=600Nm^{-1}
now
K.E=0.5x ds^2
K.E=600*(-0.1^2)
K.E=3J
Therefore
the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line
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