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vodomira [7]
4 years ago
6

A rocket has initail mass M begins to move from space , with an exhaust constant speed . Find the mass of the rocket while has t

he maxmum momentum.
Physics
1 answer:
irina1246 [14]4 years ago
3 0

Answer:

m=\frac{m_{0}}{e}

Explanation:

Equation of the rocket is,

m\frac{dv}{dt} =F-v'\frac{dm}{dt}

Here, v' is the relative velocity of rocket.

In space F is zero.

So,

m\frac{dv}{dt} =-v'\frac{dm}{dt}\\dv=-v'\frac{dm}{m} \\v=-v'ln\frac{m}{m_{0} }

Now the momentum can be obtained by multiply by m on both sides.

P=-v'mln\frac{m}{m_{0} }

Now for maxima, \frac{dP}{dm}=0

-v'ln\frac{m}{m_{0} }-v'm\frac{m_{0}}{m }m_{0=0

Now,

ln(\frac{m}{m_{0} } )=-1\\\frac{m}{m_{0} }=\frac{1}{e} \\m=\frac{m_{0}}{e}

Therefore, the mass of the rocket while having maximum momentum is \frac{m_{0}}{e}

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3 years ago
the car starts from a stop to travel 1100 meters in 14 seconds. it is clocked at 65 m/s at point k. find its average speed and a
inysia [295]

Answer:

The average velocity of the car is, V = 74.04 m/s

Explanation:

Given data,

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The time period of travel, t = 14 s

The velocity of the car at point k, v = 65 m/s

Using the II equation of motion,

                      S = ut + ½  at²

Substituting the given values,

                      1100 = 0 + ½ x a x 14²

                          a = 11.22 m/s²

Using the III equation of motion

                         v² = u² + 2 as

                          v = √(2as)              (∵ u = 0)

Substituting,

                           v = √(2 x 11.22 x 1100)

                              = 157.11 m/s

The average speed of the car,

                        V=\frac{0+65+157.11}{3}

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4 0
3 years ago
Find the moments of inertia Ix, Iy, I0 for a lamina that occupies the part of the disk x2 y2 ≤ 36 in the first quadrant if the d
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I(x)  = 1444×k ×{\pi}

I(y)  = 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}  

Explanation:

Given data

function =  x^2 + y^2 ≤ 36

function =  x^2 + y^2 ≤ 6^2

to find out

the moments of inertia Ix, Iy, Io

solution

first we consider the polar coordinate (a,θ)

and polar is directly proportional to a²

so p = k × a²

so that

x = a cosθ

y = a sinθ

dA = adθda

so

I(x) = ∫y²pdA

take limit 0 to 6 for a and o to \pi /2 for θ

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} y²p dA

I(x) = \int_{0}^{6}\int_{0}^{\pi/2} (a sinθ)²(k × a²) adθda

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (sin²θ)dθ

I(x) = k  \int_{0}^{6}a^(5)  da ×  \int_{0}^{\pi/2}  (1-cos2θ)/2 dθ

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I(x)  = k ×  ({6}^{5}) ×   {\pi /4}

I(x)  = 1444×k ×{\pi}    .....................1

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and here I(o) will be  I(x) + I(y) i.e

I(o) = 2 × 1444×k ×{\pi}

I(o) = 3888×k ×{\pi}   ......................2

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