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rosijanka [135]
2 years ago
14

Bonus: (It's not that hard, you just have to pay attention to units.) The Saturn V rocket first stage

Physics
1 answer:
agasfer [191]2 years ago
6 0

v = 2.45×10^3\:\text{m/s}

Explanation:

Newton's 2nd Law can be expressed in terms of the object's momentum, in this case the expelled exhaust gases, as

F = \dfrac{d{p}}{d{t}} (1)

Assuming that the velocity remains constant then

F = \dfrac{d}{dt}(mv) = v\dfrac{dm}{dt}

Solving for v, we get

v = \dfrac{F}{\left(\frac{dm}{dt}\right)}\;\;\;\;\;\;\;(2)

Before we plug in the given values, we need to convert them first to their appropriate units:

The thrust <em>F</em><em> </em> is

F = 7.5×10^6\:\text{lbs}×\dfrac{4.45\:\text{N}}{1\:\text{lb}} = 3.34×10^7\:\text{N}

The exhaust rate dm/dt is

\dfrac{dm}{dt} = 15\dfrac{T}{s}×\dfrac{2000\:\text{lbs}}{1\:\text{T}}×\dfrac{1\:\text{kg}}{2.2\:\text{lbs}}

\;\;\;\;\;= 1.36×10^4\:\text{kg/s}

Therefore, the velocity at which the exhaust gases exit the engines is

v = \dfrac{F}{\left(\frac{dm}{dt}\right)} = \dfrac{3.34×10^7\:\text{N}}{1.36×10^4\:\text{kg/s}}

\;\;\;= 2.45×10^3\:\text{m/s}

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The area of an equilaterla triangle is increasing at a rate of 5 m^2/hr. find the rate at witch the height is chganging when the
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The rate at which the height is changing is ( 5 / x ) m / hr

We know that,

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h = \sqrt{3} x / 2

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Given that,

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h = \sqrt{3} x / 2

Differentiate both sides with respect to t

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Differentiate both sides with respect to t

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Rate of change of height is defined as the rate at which height of an object changes with respect to time. It is represented as dh / dt

Therefore, the rate at which the height is changing is ( 5 / x ) m / hr

To know more about Rate of change of height

brainly.com/question/13283964

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