Recent content by antoman

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    Linear Drag Force Homework: Acceleration & Constant Speed Calculations

    If you mean in colledge probably yes, but me personaly don't know that equation.
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    Linear Drag Force Homework: Acceleration & Constant Speed Calculations

    Homework Statement We sink ball with density 1450 kg/m^3 in some liquid with density 1400 kg/m^3 a) What's acceleration in moment we drop the ball? b) With what constant speed will the ball sink in balace? Assuming that the drag force is linear. Radius of ball is 3mm and viscosity of...
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    Electric motor(calc. angular velocity

    Wrot...KE Wk1.. KE at t=0 Wk2...KE when t=7 A.. work
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    Electric motor(calc. angular velocity

    Problem is i don't know if M is constant, but guessing from what you wrote its not, so my equations are all wrong. So... Wrot=2*ω^(2)/J ω^2=Wrot*2/J edited: A=Wk2-Wk1 --> A=Wrot :smile: ω=sqrt((A*2*η)/J)=8,366 rad/s Thanks :)
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    Electric motor(calc. angular velocity

    First of all I am sorry for my awful english. Homework Statement Electric motor starts turning wheel with a constant power of P=25W and J=3 kg*m^2. With what angular velocity the wheel spins after 7s(from start), if the efficiency is 60%Homework Equations P=M*ω M=J*αThe Attempt at a Solution...
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    Bypass time of satelite arround Mars

    Now i know what was the problem :D Correct formula is: t=(2∏*(...))/sqrt(...) + g0 to km/s^2, and it all works out perfect, iven makes more sense :)
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    Bypass time of satelite arround Mars

    No, i don't think the formula is right, that's why i asked for help.
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    Bypass time of satelite arround Mars

    I used 3.6*10^-3 km/s^2 for g0 when i calculated, so its (km*km^(3/2))/sqrt(km*s^(-2)*km^2). Then km^(3/2)/km^(3/2)=1, so all it stays is km*s. Repair me if I am wrong :)
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    Bypass time of satelite arround Mars

    Ok i made some changes, so now t= (2*r(R+h)^(3/2))/(sqrt(g0*R^2)) r=R+h R=3400km h=150km g0=3,7 m/s^2... assuming i should of know g0 from mars. I calculated and it came out t=7261412,315 km/s...:confused:
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    Bypass time of satelite arround Mars

    g0=9,81m/s^2 hmm … don't you need to know either the mass of Mars, or the gravitational acceleration (g) at the surface? i could calculate g at surface using g=g0*R^(2)/(R+h)^2, but then again i have no idea what to use it for .. nevermind g0 is different on Mars then on Earth so..
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    Bypass time of satelite arround Mars

    Homework Statement On high 150km above the surface of Mars, there is satelite That is ciculating arround Mars. What's his bypass time? Radius of Mars is 3400km. Homework Equations The Attempt at a Solution I found equation like this : t= (2*r(R+h)^(3/2))/(sqrt(g0*R^2)) R=r+h r=3400km h=150km...
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