Rubber. The other three are metals, and therefore good conductors.
Answer:
I think the answer is C.
Explanation:
A primary source is a first hand account of an event while a secondary source is a retelling or second hand account meaning as many details will be prevalent.
Answer:
a) 
b) 
c) 
d)
or 18.3 cm
Explanation:
For this case we have the following system with the forces on the figure attached.
We know that the spring compresses a total distance of x=0.10 m
Part a
The gravitational force is defined as mg so on this case the work donde by the gravity is:

Part b
For this case first we can convert the spring constant to N/m like this:

And the work donde by the spring on this case is given by:

Part c
We can assume that the initial velocity for the block is Vi and is at rest from the end of the movement. If we use balance of energy we got:

And if we solve for the initial velocity we got:

Part d
Let d1 represent the new maximum distance, in order to find it we know that :

And replacing we got:

And we can put the terms like this:

If we multiply all the equation by 2 we got:

Now we can replace the values and we got:


And solving the quadratic equation we got that the solution for
or 18.3 cm because the negative solution not make sense.
Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion.Displacement<span> is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.
</span>To calculate displacement<span>, simply draw a vector from your starting point to your final position and solve for the length of this line. If your starting and ending position are the same, like your circular 5K route, then your </span>displacement<span> is 0. In physics, </span>displacement<span> is represented by Δs.
For me to solve this I would need to know the time, but I can give you a handy displacement calculator I used that helped me.
https://www.easycalculation.com/physics/classical-physics/constant-acc-displacement.php
Hope I helped.
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