physicS IMP PLS

Consider a rocket in space that ejects burned fuel at a speed of Vex = 1.5 km/s with respect to the rocket. The rocket burns 5% of its mass in 280 seconds.
(a) What is the speed V of the rocket after a burn time of 140 seconds (in m/s)?
(b) What is the instantaneous acceleration a of the rocket at time 140.0 seconds after start of the engines (in m/s^20)? I have tried to solve, but not been successful. Thanks for your help.

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  1. I have tried to solve this problem the following way:
    5% mass is burned in 280s.
    1.5 km/s = 1500 m/s
    u=fuel speed m/s
    Solution:
    V= u*ln(1/(1-p))
    1500*ln(1/0.950) = 76.9399416 m/s
    a= V/t = 76.9399416/280 = 0.274785506 m/s^2
    What is confusing to me what is the speed after 140s and instantaneous acceleration a 140 s after the start of the engines. Thanks for your help.

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  2. THE instantaenous acc is just the acceleration of the rocket

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  3. Hi Anonymus, So my question is V= 76.939416 m/s and a=0.2747 m/s^2 correct. Is the other information just trying to through us off? Thanks.

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  4. do you know how to solve the ruler question?

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  5. no it is wrong!!! he ask you "What is the speed V of the rocket after a burn time of 140 seconds (in m/s)? " that is the half of time and half of % . so ..

    1500*ln(1/0.750) = 37.91 m/s
    a=v/t=37.91/140=0.27

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  6. sorr that waw

    1500*ln(1/0.9750) = 37.91 m/s

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  7. Hi Anonymous, I tried that problem and was unsuccessful. My answer was 8 radians. I knew it was wrong. I put that off to the side. I am just trying to get a passing grade. Thanks for your help. If I learn any thing I will let you know. Also Thanks Greco and everyone who was kind enough to help me out.

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  8. always here Hill...if you can help with ruler problem it will pe magnificent

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  9. Hi, Greco. To be honest Questions 1 and the ruler are by far the most difficult for me. when I started the test I knew I was in trouble. I saw this approach:
    Apply the conservation of energy
    U= m*g*h cm
    Ek = m*g* (L/2)+0
    l=1/3*m*L
    Einitial= m*g*(L/2)+0
    Efinal= m*g*(2/L)cos30+1/2*L*w^2
    Solve for Einitial = Efinal -> w =
    Maybe you saw this. I hope it helps.

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  10. i know this ,i already done it.thx for the efford!

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  11. I=1/3*m*L^2
    Eini= mg(L/2) + 0
    Efin= mg(L/2)cos30 + 1/2*I*w^2
    Eini=Efin ->
    w=sqrt(3*g (1-cos(theta))/L)

    b)
    alpha=3*mg/2*sin(theta)/L
    ax = -L/2*sin(theta) w^2 + L/2*cos(theta) alpha
    Fx=m*ax
    Fy=m*g*cos(theta)-m*w^2*(L/2)

    c)
    cos(theta)=2/3
    theta=48.19

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  12. I=1/3*m*L*L?

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  13. @Greco I tried it and it workrd for a and c and for Fy but Fx was wrong!why did you use alpha? please help with this question!

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