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2 years ago

8

A cylinder is filled with a liquid of density d upto a height h. If the beaker is at rest ,then the mean pressure on the wall is

?

1 answer:

tangare [24]2 years ago

7 0

Answer:

h over 2 dg

Explanation:

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A rocket in deep space has an empty mass of 220 kg and exhausts the hot gases of burned fuel at 2500 m/s. What mass of fuel is n

Scilla [17]

Answer:

Explanation:

Let fuel is released at the rate of dm / dt where m is mass of the fuel

thrust created on rocket

= d ( mv ) / dt

= v dm / dt

this is equal to force created on the rocket

= 220 dv / dt

so applying newton's law

v dm / dt = 220 dv / dt

v dm = 220 dv

dv / v = dm / 220

integrating on both sides

∫ dv / v = ∫ dm / 220

lnv = ( m₂ - m₁ ) / 220

ln4000 - ln 2500 = ( m₂ - m₁ ) / 220

( m₂ - m₁ ) = 220 x ( ln4000 - ln 2500 )

( m₂ - m₁ ) = 220 x ( 8.29 - 7.82 )

= 103.4 kg .

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3 years ago

sergeinik [125]

Answer:

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3 years ago

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NISA [10]

Answer:

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Explanation:

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3 years ago

Cuál fue la primera derrota que sufrieron los fascistas

valentinak56 [21]

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2 years ago

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

<u>Answer:</u> The average atomic mass of the given element is 20.169 amu.

<u>Explanation:</u>

Average atomic mass of an element is defined as the sum of masses of the isotopes each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

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3 years ago