In order to determine the acceleration of the block, use the following formula:

Moreover, remind that for an object attached to a spring the magnitude of the force acting over a mass is given by:

Then, you have:

by solving for a, you obtain:

In this case, you have:
k: spring constant = 100N/m
m: mass of the block = 200g = 0.2kg
x: distance related to the equilibrium position = 14cm - 12cm = 2cm = 0.02m
Replace the previous values of the parameters into the expression for a:

Hence, the acceleration of the block is 10 m/s^2
Answer:
ΔU = - 310.6 J (negative sign indicates decrease in internal energy)
W = 810.6 J
Explanation:
a.
Using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Absorbed = 500 J
ΔU = Change in Internal Energy of Gas = ?
W = Work Done = PΔV =
P = Pressure = 2 atm = 202650 Pa
ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³
Therefore,
Q = ΔU + PΔV
500 J = ΔU + (202650 Pa)(0.004 m³)
ΔU = 500 J - 810.6 J
<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>
<u></u>
b.
The work done can be simply calculated as:
W = PΔV
W = (202650 Pa)(0.004 m³)
<u>W = 810.6 J</u>
Answer:
Anything below 7.0 is acidic, so the range would be 0 to 7.
Neutral is simply 7, in the middle of the scale.
Lastly, anything above 7.0 is basic or alkaline, so that would be 7 to 14.
Good luck, I hope this helps
As per the given Figure attached here we know that both charges q1 and q2 will apply same force on charge q3 and hence the resultant force due to both charges will be along Y axis vertically upwards
So here we know that

now from the above equation


so both of the charges will apply 0.288 N force on q3 charge along the line joining them
now the net force due to vector sum is given by

here we know that angle is

now we have


so net force on q3 is 0.46 N vertically upwards along +Y axis