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...
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.
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.
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
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...
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...
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...
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...
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?