Answer to A spring<span> is </span>stretched<span> to a </span>displacement<span> of </span>3.4 m<span> from </span>equilibrium<span>. </span>Then<span> the </span>spring<span> is</span>released<span> and ... </span>Then<span> the </span>spring<span> is </span>released<span> and </span>allowed<span> to </span>recoil<span> to a </span>displacement<span> of </span>1.9 m<span> from</span>equilibrium<span>. The </span>spring constant<span> is </span>11 N/m<span>. What </span>best describes<span> the </span>work involved<span> as the </span>spring recoils<span>? A)87 J of </span>work<span> is performed ...</span>
Your answer will be Radio Waves .
That seems to be the only to make sense. Hope that helps u
Answer:
option (D) is correct.
Explanation:
According to the work energy theorem, the work done by all forces is equal to the change in kinetic energy of the body.
the kinetic energy of a body is directly proportional to the square of the speed of the body.
As the kinetic energy change, the speed of the body also change.
Option (D) is correct.
The density of seawater at a depth where the pressure is 500 atm is 
Explanation:
The relationship between bulk modulus and pressure is the following:

where
B is the bulk modulus
is the density at surface
is the variation of pressure
is the variation of density
In this problem, we have:
is the bulk modulus

is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)
Therefore, we can find the density of the water where the pressure is 500 atm as follows:

Learn more about pressure in a fluid:
brainly.com/question/9805263
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