Answer:
the difference is due to resistance tolerance
Explanation:
In mathematical calculations, either done by hand or in a computer program, the heat taken from the resistors is the nominal value, which is the writing in its color code, so all calculations give a result, but the Resistors have a tolerance, indicated by the last band that is generally 5%, 10%, 20% and in the expensive precision resistance can reach 1%.
This tolerance or fluctuation in the resistance value is what gives rise to the difference between the computation values and the values measured with the instruments, multimeters.
Another source of error also occurs due to temperature changes in the circuit that affect the nominal resistance value, there is a very high resistance group that indicates the variation with the temperature, they are only used in critical circuits, due to their high cost
In summary, the difference is due to resistance tolerance.
4 m/s because it is THR fastest voice
Answer:
The distance of the second dark band away from the central bright spot be located is 
Explanation:
Given that,
Wave length = 500 nm
Radius 
Distance from the hair sample D= 6 m
We need to calculate the distance of the second dark band away from the central bright spot be located


Using formula for dark fringe

Put the value into the formula




Hence, The distance of the second dark band away from the central bright spot be located is 
Answer:
a. a = 
b. T = -0.81 N
Explanation:
Given,
- weight of the lighter block =

- weight of the heavier block =

- inclination angle =

- coefficient of kinetic friction between the lighter block and the surface =

- coefficient of kinetic friction between the heavier block and the surface =

- friction force on the lighter block =

- friction force on the heavier block =

Let 'a' be the acceleration of the blocks and 'T' be the tension in the string.
From the f.b.d. of the lighter block,

From the f.b.d. of the heavier block,

From eqn (1) and (2), we get,

![\Rightarrow a\ =\ \dfrac{g(w_1sin\theta\ -\ \mu_1w_1cos\theta\ +\ w_2sin\theta\ +\ \mu_2w_2cos\theta)}{w_1\ +\ w_2}\\\Rightarrow a\ =\ \dfrac{g[sin\theta(w_1\ +\ w_2)\ +\ cos\theta(\mu_2w_2\ -\ \mu_1w_1)]}{w_1\ +\ w_2}\\](https://tex.z-dn.net/?f=%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bg%28w_1sin%5Ctheta%5C%20-%5C%20%5Cmu_1w_1cos%5Ctheta%5C%20%2B%5C%20w_2sin%5Ctheta%5C%20%2B%5C%20%5Cmu_2w_2cos%5Ctheta%29%7D%7Bw_1%5C%20%2B%5C%20w_2%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7Bg%5Bsin%5Ctheta%28w_1%5C%20%2B%5C%20w_2%29%5C%20%2B%5C%20cos%5Ctheta%28%5Cmu_2w_2%5C%20-%5C%20%5Cmu_1w_1%29%5D%7D%7Bw_1%5C%20%2B%5C%20w_2%7D%5C%5C)
![\Rightarrow a\ =\ \dfrac{9.81\times [sin30^o\times (3.0\ +\ 7.0)\ +\ cos30^o\times (0.31\times 7.0\ -\ 0.13\times 3.0)]}{3.0\ +\ 7.0}\\\Rightarrow a\ =\ 6.4\ m/s.](https://tex.z-dn.net/?f=%5CRightarrow%20a%5C%20%3D%5C%20%5Cdfrac%7B9.81%5Ctimes%20%5Bsin30%5Eo%5Ctimes%20%283.0%5C%20%2B%5C%207.0%29%5C%20%2B%5C%20cos30%5Eo%5Ctimes%20%280.31%5Ctimes%207.0%5C%20-%5C%200.13%5Ctimes%203.0%29%5D%7D%7B3.0%5C%20%2B%5C%207.0%7D%5C%5C%5CRightarrow%20a%5C%20%3D%5C%206.4%5C%20m%2Fs.)
part (b)
From the eqn (2), we get,
Answer:
d. ovaries, uterus, and Fallopian tubes
Explanation: