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Vlad [161]
4 years ago
8

What is the length of 0.005 mm in kilometers?

Physics
1 answer:
ZanzabumX [31]4 years ago
4 0
The answer would be 5*10^9
.

If we divide \frac{5*10^9}{1*10^6}, it gives us 0.005!

Hope this helped! c:
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So we want to know how can we detect infrared rays without an instrument. Infrared rays or heat, are a part of electromagnetic spectrum. We have specialized nerve cells in our skin called thermoreceptors that can detect differences in temperature that are produced by infrared part of EM spectrum.
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3 years ago
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What is one benefit of lifelong physical activity?
kvasek [131]
The answer would be C
5 0
3 years ago
Un tubo cilindrico hueco de cobre mide 3 m de longitud tienen un diametro exterior de 4cm y un diametro interior de 2 cm¿cuanto
Irina-Kira [14]

Answer:

 W = 9.93 10² N

Explanation:

To solve this exercise we must use the concept of density

           ρ = m / V

the tabulated density of copper is rho = 8966 kg / m³

let's find the volume of the cylindrical tube

           V = A L

           V = π (R_ext  ² - R_int ²) L

let's calculate

          V = π (4² - 2²) 10⁻⁴  3

          V = 1.13 10⁻²  m³

         m = ρ V

        m = 8966 1.13 10⁻²

        m = 1.01 10² kg

the weight of the tube

        W = mg

         W = 1.01 10² 9.8

         W = 9.93 10² N

4 0
3 years ago
I don’t have a question because I posted a photoz
Scorpion4ik [409]

Part a:

Q_{1} = 56

Q_{2} = 60

Q_{3} = 63

     The quartiles are found by finding the medium of the data, and then the mediums of the two different data sets on either side of the medium. The Q_{2} is the overall medium, Q_{1} is the medium of the first half, and Q_{3} is the medium of the second half.

-> How is the medium found? When finding the medium we put the values in order least to greatest and pick the middle value.

[] See attached

Part b:

The range is 7.

The interquartile range is the range of numbers between Q_{1} and Q_{3}. In other words, it is 50% of the data, directly in the middle.

This becomes 63 - 56 = 7

Part c:

79 is an outlier.

It is an outlier because it is 1.5 above or below (in this case, above) the interquartile range.

-> 63 + (7 + \frac{7}{2}) ≤ 79

-> 63 + 10.5 ≤ 79

-> 73.5 ≤ 79

Have a nice day!

     I hope this is what you are looking for, but if not - comment! I will edit and update my answer accordingly.

- Heather

7 0
2 years ago
Find the force required to do 25 joule work when the force causes a displacement of 0.5 m​
Eduardwww [97]

Answer:

<h2>50 N</h2>

Explanation:

The force required can be found by using the formula

f =  \frac{w}{d}  \\

w is the workdone

d is the distance

From the question we have

f =  \frac{25}{0.5}  \\

We have the final answer as

<h3>50 N</h3>

Hope this helps you

6 0
3 years ago
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