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34kurt
2 years ago
14

How many significant digits are in the measurement 4,032,010 m/s?

Physics
2 answers:
choli [55]2 years ago
5 0
Since there is no decimal point in the number given above, the counting for the number of the significant figures will start from the left. Then, the first zero from the left is insignificant. Therefore, in this number there are 6 significant figures. 
3241004551 [841]2 years ago
4 0

Answer:

6

Explanation:

The rules which we have to use to find the number of significant figures here are given below:

1. If a number does not contain a decimal point, then the zeros between two non zero digits are significant.

2. the trailing zeros are not significant.

The number is

4,032, 010 m/s

so, the zero between 4 and 3 is significant

the zero between 2 and 1 is significant

the zero at the last is not significant

So, the number of significant figures are 6.

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Answer:

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5 0
3 years ago
series RC circuit is built with a 15 kΩ resistor and a parallel-plate capacitor with 18-cm-diameter electrodes. A 18 V, 36 kHz s
andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

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\displaystyle X_c=\frac{1}{wC}

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\displaystyle C=\frac{1}{wX_c}

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\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

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\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

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\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

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\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

8 0
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Answer:

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