Work is done when spring is extended or compressed. Elastic potential energy is stored in the spring. Provided inelastic deformation has not happened, the work done is equal to the elastic potential energy stored.
Answer:
The intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Explanation:
Given;
intensity of the sound level, dB = 60 dB
The intensity of the sound in W/m² is calculated as;
![dB = 10 Log[\frac{I}{I_o} ]\\\\](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C)
where;
I₀ is threshold of hearing = 1 x 10⁻¹² W/m²
I is intensity of the sound in W/m²
Substitute the given values and for I;
![dB = 10 Log[\frac{I}{I_o} ]\\\\60 = 10 Log[\frac{I}{I_o} ]\\\\6 = Log[\frac{I}{I_o} ]\\\\10^6 = \frac{I}{I_o} \\\\I = 10^6 \ \times \ I_o\\\\I = 10^6 \ \times \ 1^{-12} \ W/m^2 \\\\I = 1\ \times \ 10^{-6} \ W/m^2](https://tex.z-dn.net/?f=dB%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C60%20%3D%2010%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C6%20%3D%20%20Log%5B%5Cfrac%7BI%7D%7BI_o%7D%20%5D%5C%5C%5C%5C10%5E6%20%3D%20%5Cfrac%7BI%7D%7BI_o%7D%20%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%20I_o%5C%5C%5C%5CI%20%3D%2010%5E6%20%5C%20%5Ctimes%20%5C%201%5E%7B-12%7D%20%5C%20W%2Fm%5E2%20%5C%5C%5C%5CI%20%3D%201%5C%20%5Ctimes%20%5C%2010%5E%7B-6%7D%20%5C%20W%2Fm%5E2)
Therefore, the intensity of the sound in W/m² is 1 x 10⁻⁶ W/m².
Answer:
Explanation:
Given that,
A point charge is placed between two charges
Q1 = 4 μC
Q2 = -1 μC
Distance between the two charges is 1m
We want to find the point when the electric field will be zero.
Electric field can be calculated using
E = kQ/r²
Let the point charge be at a distance x from the first charge Q1, then, it will be at 1 -x from the second charge.
Then, the magnitude of the electric at point x is zero.
E = kQ1 / r² + kQ2 / r²
0 = kQ1 / x² - kQ2 / (1-x)²
kQ1 / x² = kQ2 / (1-x)²
Divide through by k
Q1 / x² = Q2 / (1-x)²
4μ / x² = 1μ / (1 - x)²
Divide through by μ
4 / x² = 1 / (1-x)²
Cross multiply
4(1-x)² = x²
4(1-2x+x²) = x²
4 - 8x + 4x² = x²
4x² - 8x + 4 - x² = 0
3x² - 8x + 4 = 0
Check attachment for solution of quadratic equation
We found that,
x = 2m or x = ⅔m
So, the electric field will be zero if placed ⅔m from point charge A, OR ⅓m from point charge B.