Recent content by VSCCEGR

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    Re-Learning AutoCAD 2004: What Are the Best Books?

    I have had a couple of CAD classes, but that was 3 yr. ago using AutoCAD 14. I have recently borrowed a copy of AutoCAD 2004 and i have noticed that, from what i can remember, a lot of the menues and where they lead have changed. My question is what are the better books for re-learning AutoCAD...
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    Solving the Bullet-Block Impulse Problem

    Found (a): V=Vo+at 0=Vo+2.54*1.2 Vo=3.04m/s mv_1=mv_2 .02[V_1/cos(30)]=(4+.02)3.04 V_1=707m/s Sorry ramollari, but you do using this method 707m/s is correct.
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    Solving the Bullet-Block Impulse Problem

    by the way would this problem be done in the same way? 4oz ball @ 9ft/s strikes a 10oz plate on springs with no loss of energy. find Vel. of ball after hitting plate, force exerted by ball on plate.
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    Solving the Bullet-Block Impulse Problem

    Found (a): V=Vo+at 0=Vo+2.54*1.2 Vo=3.04m/s mv_1=mv_2 .02[V_1/cos(30)]=(4+.02)3.04 V_1=707m/s as for (b) i think they mean: Imp_1,2=F(delta)t so I am guessing they want to find F(delta)t as a whole using m_1*v_1+F(delta)t=m_2*v_2
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    Solving the Bullet-Block Impulse Problem

    Thats what I thought. So: V_0=? V=0 a=9.81m/s verticly or 2.54m/s down incline m=4.02kg t=1.2s cos(30)V_o=velocity of the bullet slong incline
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    Solving the Bullet-Block Impulse Problem

    V_o along the incline is (V_o/2) Initial Vel. Of block is 0 But how does this apply? Impulse & Momentum: sum(mv_1)+sum(Imp_1,2)=sum(mv_2) Right, So: sum(mv_1)=mass bullet *v_o sum(Imp_1,2)=.5F-sin(15)4kg*9.81m/s sum(mv_2)=(mass bullet+mass block)*v_2 F being the force exerted by the...
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    Solving the Bullet-Block Impulse Problem

    Problem: A 20-g bullet is fired into a 4-kg block and becomes embeded in it. knowing that the bullet and block then move up the incline for 1.2 seconds before coming to a stop Determine; (a) the Magnitude Of the bullets initial velocity, (b) the Magnitude of the Impulsive force by the bullet...
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    Solve Impulsive Motion Problem: Initial Velocity 9m/s, uk=.3, Theta=20 Deg

    My bad, 3.06s
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    Solve Impulsive Motion Problem: Initial Velocity 9m/s, uk=.3, Theta=20 Deg

    Yes, I did place the positive x-axis in the direction of movement, but it really should not matter because I am using the Principle of Impulse and Momentum. (a) part was finding the time taken if (theta)=0deg. (3.06s I got this one right) So I know my method works. There is just something I am...
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    Solve Impulsive Motion Problem: Initial Velocity 9m/s, uk=.3, Theta=20 Deg

    Problem: The initial velocity of block A is 9m/s. Knowing that uk=.3 Determine the time for the block to reach zero velocity. When (theta)=20 deg. Here is my work so far. [FONT=Times New Roman]mv_1+Imp_1,2=mv_2 Imp_1,2=Ff(t)+sin(theta)W(t) Ff=Cos(theta)uk(mg) W=mg...
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    Calculating Spring Compression: Finding Height and Max Velocity

    Never Mind. I Solved It With Out You All.
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    Calculating Spring Compression: Finding Height and Max Velocity

    k=2626.8N/m x=.1524m weight collar=71.04N g=9.81m/s^2
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    Calculating Spring Compression: Finding Height and Max Velocity

    You guys going to help or not? :mad:
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    Calculating Spring Compression: Finding Height and Max Velocity

    (15lb/in.*6in)/2=16lb*h h=16.87in Actual answer is 45.2in We are both missing something.
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    Calculating Spring Compression: Finding Height and Max Velocity

    T1+V1=T2+V2 T1=.5mv^2=.5(.497)0=0 T2=.5(.497)*? V1=mgh=.497(32.2)0=0 V2=.497(32.2)h This is what you are saying right? If so what goes in the "?" spot? If not, Why not?