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fgiga [73]
3 years ago
14

How many significant digits are measurement 0.00210 mg?

Physics
2 answers:
jek_recluse [69]3 years ago
8 0
There are 3 significant figures, if that answers the question.
madam [21]3 years ago
5 0
3 it two one zero even though the rest of the zeros don't really count lol
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When you strike two tuning forks simultaneously, you hear three beats per second. The frequency of the first tuning fork is 440
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Answer:4

Explanation:

Given

Frequency of beats is 3 beats per second

Frequency of first tuning fork is f_1=440\ Hz

Beat frequency is difference in the frequency of tuning forks i.e.

either f_1-f_2 or f_2-f_1 =3

so f_2=3+440=443\ Hz

or

f_2=440-3=437\ Hz

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Two stationary positive point charges, charge 1 of magnitude 3.90 nC and charge 2 of magnitude 1.80 nC, are separated by a dista
soldi70 [24.7K]

Answer:

v = 7793150 m/s

Explanation:

First, we are going to calculate the electrical potential in the point middle between the two charges

Remember that the electrical potential can be calculated as:

v = \frac{kQ}{r}

                 Where     k = 8.9874 x 10^{9} \frac{Nm^{2} }{C^{2} }

and it is satisfy the superposition principle, thus

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.23} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.23}

v = 222.73v

The electrical potential at 10 cm from charge 1 is:

v = \frac{8.9874x10^{9}(3.90x10^{-9} ) }{0.1} +  \frac{8.9874x10^{9}(1.80x10^{-9} ) }{0.36}

v = 395.44 v

Since the work - energy theorem, we have:

q\Delta v = \frac{mv^{2} }{2}

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Therefore:

v = \sqrt{\frac{2q\Delta v}{m} }

v = 7793150 m/s

6 0
2 years ago
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