To solve this problem it is necessary to apply the concepts related to the Force from Hooke's law, the force since Newton's second law and the potential elastic energy.
Since the forces are balanced the Spring force is equal to the force of the weight that is
Where,
k = Spring constant
x = Displacement
m = Mass
g = Gravitational Acceleration
Re-arrange to find the spring constant
Just before launch the compression is 40cm, then from Potential Elastic Energy definition
Therefore the energy stored in the spring is 63.72J
The equator would be warmer than the poles. :)
I am thinking that maybe the problem is not with the calibration. It might be that the buffered solution is already expired since at this point the solution is already not stable and will give a different pH reading than what is expected.
This means that there is same current flow in both the circuit, or the circuit one has twice the power of circuit two.
According to ohm's law, the resistance is given as
I=V/R
Since the circuit one has twice the voltage, and resistance
I1=I2
The original width was 94.71 cm
<span>The area decreased 33.1% </span>
<span>The equation for the final size is </span>
<span>2X^2 = 1.2 m^2 </span>
<span>X^2 - 0.6 m^2 </span>
<span>X^2 = 10000 * .6 cm </span>
<span>X = 77.46 cm (this is the width) </span>
<span>The length is 2 * 77.46 = 154.92 cm </span>
<span>The original length was 154.92 + 34.5 = 189.42 cm </span>
<span>The original width was 189.42 / 2 = 94.71 cm </span>
<span>The original area was 94.71 * 189.92 = 17939.9 cm^2 </span>
<span>The new area is 79.46 * 154.92 = 12000.1 cm^2 </span>
<span>The difference between the original and current area is 17939.9 - 12000.1 = 5939.86 cm^2 </span>
<span>The percentage the area decreased is 5939.86 ' 17939.9 = 33.1%</span>
