Answer:
A, B, and E
Explanation:
The springs are identical, and are compressed the same amount, so they have the same initial elastic potential energy. (E is true)
Energy is conserved, so the darts have the same amount of kinetic energy. (A is true, C is false)
The lighter dart has the same energy as the heavier dart. Since it has less mass, it must have a greater speed. (B is true, D is false)
Answer:
Explanation:
Given
mas of block 
speed of block 
spring constant 
As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring




Well we can't see the picture that the teacher has, and we have to figure out what's in the picture from some clues in the answer choices. The picture seems to show an atom with 3 protons and 3 neutrons in the nucleus, and 3 electrons whizzing around the nucleus.
If that's what's in the picture, then Mike is correct (first choice), because the atomic number shows the number of protons in all atoms. There are 3 protons and the atomic number of lithium is 3.
Second choice . . . False, because electrons are not involved in the atomic mass.
Third and fourth choices . . . both false; sadly, Joan is woefully unclear on the concepts.
Hello!
A stretched spring has 5184 J of elastic potential energy and a spring constant of 16,200 N/m. What is the displacement of the spring?
0.57 m
0.64 m
0.80 m
1.25 m
Data:



For a spring (or an elastic), the elastic potential energy is calculated by the following expression:

Where k represents the elastic constant of the spring (or elastic) and x the deformation or displacement suffered by the spring.
Solving:









Answer:
The displacement of the spring = 0.8 m (or 0.80 m)
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I Hope this helps, greetings ... Dexteright02! =)
Answer:
The magnitude of the electric field at a point equidistant from the lines is 
Explanation:
Given that,
Positive charge = 24.00 μC/m
Distance = 4.10 m
We need to calculate the angle
Using formula of angle



We need to calculate the magnitude of the electric field at a point equidistant from the lines
Using formula of electric field

Put the value into the formula



Hence, The magnitude of the electric field at a point equidistant from the lines is 