Recent content by Bingo1915

How can I calculate the following? Given A=\hat{x}(x+2y)-\hat{y}(y+3z)+\hat{z}(3x-y) Determine a unit vector parallel to A at point P(1,-1,2).

I hope you can see the pic. I'm not sure what you would consider equilibrium in the pic. I think it would be:\sqrt{0.45^2 + 0.25^2}. Can you advise?

Mass m=1kg Spring constant k=200 N/m Unstretched spring length 0.1m When the system as show in the attatachment is realeased from rest the spring contracts. Use conservation of energy to determine the magnitude of velocity of the mass when the...

1. A projectile is launched at 10 m/s from a sloping surface. The angle \alpha=80 deg. Determine the range R. 2. Attached is the drawing. 3. Treat as 2 equations. x-direction Initial time t=0 Initial V_{x}=V_{0}Cos\theta a_{x}dv_{}x/dt = 0 V_{x}=Initial VCos\theta =...

I see where I was incorrect. Thanks for the help.

Using pi/6 to pi/3 I get -2pi/3 + pi/3 = lnw-ln10 -pi/3 + ln10 = lnw From here I think it is 1/[e^(-pi/3+ln10)] = w Can you check me on this part?

[SOLVED] Angular Velocity 1. The angular acceleration is given \alpha=-2w^2 rad/s^2 where \omega is the angular velocity in rad/s. When \theta=30 deg the angular velocity is 10 rad/s. What is the angular velocity when \theta=60deg? [b]2. Used \alpha=d\varpi/d\theta * \varpi...

Thats what I was forgetting. Therefore s=(10t^3)/3+ct+d. Where t=3 c=-10 d=40 Thanks for the help.

For velocity I got v=10t^2 + c, and pluged 3 in for t and -10 in for c. This gave me v=80 m/s. If i integrate once more to get position I get s=(10t^3)/3+ct. Am I correct with the constant? Plug 3 in for t, and -10 in for c again. I think I am getting confused with the constant.

Finding position I have run across the same type of problem, but I am being asked what the position is. Below is what was given. Acceleration of a point is a=20t m/s^2. When t=0, s=40 meters and v=-10 m/s. What is the position at t=3 sec? I've used a=dv/dt to get the velocity, v=80...