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bija089 [108]
4 years ago
8

What portion (division) of a meter stick is a centimeter?

Physics
2 answers:
Anastasy [175]4 years ago
8 0
A meter is 100 meters. So a hundredth of a meter stick is a centimeter.<span />
Nataliya [291]4 years ago
7 0

Answer

\frac{1}{100}

Explanation:

A meter stick is a tool used to measure length and distances. We know that one meter is equal to 100cm.

So,

100 cm = 1 meter

1 cm = \frac{1}{100} meter

Thus we can see that one hundredth of a meter stick is one centimeter.

The centimeter scale can be used to measure smaller lengths.

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Keith_Richards [23]
I think the answer would be B
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4 years ago
Two long, parallel wires are separated by a distance of 2.60 cm. The force per unit length that each wire exerts on the other is
Alex777 [14]

Answer:

<em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires</em>

Explanation:

Force per unit length on the wires = 4.30×10^−5 N/m

distance between wires = 2.6 cm = 0.026 m

current through one wire = 0.52 A

current on the other wire = ?

Recall that the force per unit length of two wires conducting and lying parallel and close to each other is given as

F/l = \frac{u_{0}I_{1} I_{2}  }{2\pi r }

where F/l is the force per unit length on the wires

u_{0} = permeability of vacuum = 4π × 10^−7 T-m/A

I_{1} = current on the first wire = 0.520 A

I_{2} = current on the other wire = ?

r = the distance between the two wire = 0.026 m

substituting the value into the equation, we have

4.30×10^−5 = \frac{4\pi *10^{-7}*0.520*I_{2}  }{2\pi *0.026} =  \frac{ 2*10^{-7}*0.520*I_{2}  }{0.026}

4.30×10^−5 = 4 x 10^-6 I_{2}

I_{2} = (4.30×10^-5)/(4 x 10^-6) = <em>10.75 A</em>

<em>The current is in opposite direction since it causes a repulsion force between the wires.</em>

6 0
4 years ago
ured length.  Measurement errors are another type of uncertainty involved in single measurements. Suppose we measure the diamet
ryzh [129]

Answer:

  T = (99.78 ±0.05) ºC

Explanation:

This problem refers to the procedure to perform the measurements and their correct reporting.

Suppose that the thermometer used has an appreciation of 0.1ºC and the values ​​of 5 measurements resulted in

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 1      99.8

 2     99.7

 3     99.7

 4     99.8

 5     99.9

the processing calculate mean value

   T_average = ∑ T_{i} / n

   T_average = (99.8 + 99.7 + 99.7 + 99.8 + 99.9) / 5

   T_average = 99.78ºC

the absolute error or uncertainty of the measurement is half the appreciation of the instrument

       ΔT = ± 0.05ºC

so the result must be written

        T = (99.78 ±0.05) ºC

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3 years ago
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