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lyudmila [28]
2 years ago
15

How do I do this? Help please!

Physics
1 answer:
chubhunter [2.5K]2 years ago
6 0

Answer:

17.5g/cm3

Explanation:

find the slope of the line

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Explain any two areas affected by migration​
mel-nik [20]

Answer:

<h2>Migration is affected by various factors like age, sex, marital status, education, occupation, employment etc. Age and sex are main demographic factors that affect the migration. Men, generally, migrate to other places quite often though there are more women who migrate to husbands' places after marriage.</h2>
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An airplane starts from rest and reaches a takeoff velocity of 60.0m/s due north in 4.0 s. How far has it traveled before takeof
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A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai
Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

Pb' - Pat = d \times g \times  (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000

Pb' = 122 KPa

3 0
3 years ago
What is the main difference between an internal cumbustion engine and a external combination engine
aleksandrvk [35]
Internal and external combustion engines are two types of heat engines: they convert thermal energy into mechanical energy. The main difference between internal and external combustion engine is that in internal combustion engines, the working fluid burns inside the cylinder, whereas in external combustion engines, combustion takes place outside the cylinder and heat is then transferred to the working fluid.
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New 5G networks utilize millimeter-wave radiation. Millimeter-wave radiation refers to electromagnetic waves with frequencies in
seraphim [82]

Answer:

It corresponds to 1mm-10 mm range.

Explanation:

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  • As in any wave, there exists a fixed relationship between speed, frequency and wavelength, as follows:

        v = \lambda * f  (1)

  • Replacing v= c=3*10⁸ m/s, and the extreme values of f (which are givens), in (1) and solving for λ, we can get the free-space wavelengths that correspond to the 30-300 GHz range, as follows:

       \lambda_{low} = \frac{c}{f_{high}}  = \frac{3e8m/s}{300e9Hz} = 1 mm (2)

      \lambda_{high} = \frac{c}{f_{low}}  = \frac{3e8m/s}{30e9Hz} = 10 mm (3)

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