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svlad2 [7]
3 years ago
11

1,632 rounded by the nearest ten is ____

Physics
2 answers:
maksim [4K]3 years ago
8 0

Answer:

1,632 rounded by the nearest ten is 1630

783 rounded by the nearest hundred is 800

43,761 rounded by the nearest thousand is 44,00

Explanation:

Anna71 [15]3 years ago
4 0
1,632 rounded by the nearest ten is 1,630 because 2 is not greater or equal to 5.
783 rounded by the nearest hundred is 800 because 83 is greater than or equal to 50.
43,761 rounded by the nearest thousand is 44,00 because 761 is greater than or equal to 500.

Step-by-Step
If it is round by the nearest 10, there is one 0 so you only count the last number. For example, 437. The last number is 7, and we are only using 7 in this case because there is only 1 zero in 10. Next step is see if 7 is greater than or equal to 5. We can see that it is, so we will round up in this case making 437 change into 440. Another example, 11,439 rounded to the nearest 1,000 th. Well there is 3, zeros so we will count the last three numbers only. Is 439 greater than or equal to 500? No. So we will round down in this case and the answer will be 11,000. Hope this helps you to understand :)
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A magnetic field of 37.2 t has been achieved at the mit francis bitter national magnetic laboratory. Find the current needed to
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Answer:

Here is the complete question:

https://www.chegg.com/homework-help/questions-and-answers/magnetic-field-372-t-achieved-mit-francis-bitter-national-magnetic-laboratory-find-current-q900632

a) Current for long straight wire  =3.7\ MA

b) Current at the center of the circular coil =2.48\times 10^{5}\ A

c) Current near the center of a solenoid 236.8\ A

Explanation:

⇒ Magnetic Field due to long straight wire is given by (B),where B=\frac{\mu\times I}{2\pi(r) },so\ I=\frac{B\ 2\pi(r)}{\mu}

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Plugging the values,

Conversion 1\times 10^6 A = 1\ MA,and 2cm=\frac{2}{100}=0.02\ m

I=\frac{37.2\times \ 2\pi(0.02)}{4\ \pi \times (10^{-7})}=3.7\ MA

⇒Magnetic Field at the center due to circular coil (at center) is given by,B=\frac{\mu\times I (N)}{2(a)}

So I= \frac{2B(a)}{\mu\ N} = \frac{2\times 37.2\times 0.42}{4\pi\times 10^{-7}\times 100}=2.48\time 10{^5}\ A

⇒Magnetic field due to the long solenoid,B=\mu\ nI=\mu (\frac{N}{l})I

Then I=\frac{B}{\mu(\frac{N}{L})} \approx 236.8\A  

So the value of current are  3.7 MA,2.48\times 10^{5} A and 236.8\ A respectively.

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