Recent content by Newton1Law

What you know is this, in 0.03 meters of travel you achieved an additional velocity of 2 x10^5 m/s. You also know that s=0.5*a*t^2 thus 0.03 = 0.5*s*t^2 (first equation). Next you know that v=at and that the increase of velocity was 2x10^5, thus the time could be found by using 2x10^5 = a*t...

From you statement of the problem, I assumed (sorry) that the acceleration of 40 meters per second per second was the acceleration supplied by the rocket motor. Assumming up is positive then the rocket motor would apply 40 m/s^2 up while gravity would supply -9.8 m/S^2 due to gravity. From...

100 kM/hr = 27.78 m/s not 16.7 m/s. Look in your book. You should find a reference equation that looks something like v=at+vo. Since you know v0=o at time=0 the equation becomes simple. Now solve for acceleration since you have been given both the time and final velocity. Hope this helps.

Vo of the rocket is zero at time 0 and at all times for the equations used while it is accelerating to it's maximum velocity. It will reach it's maximum velocity in 2.5 seconds. Knowing this you should solvee for the velocity at time = 2.5 second(remember the total acceleration on the rocket...

f'(x) = limit [f(x+h) - f(x)]/h (as h->0) f(x) = -2x+4 f'(x) = limit [f(-2(x+h) + 4) - (-2x + 4)]/h h -> 0 f'(x) = limit [-2x - 2h+ 4 +2x - 4]/h h -> 0 f'(x) = limit [ -2 h]/h h-> 0 f'(x) = limit [-2] h -> 0 f'(x) = -2 Hope this...

Your first equation will not solve for "X". Your second equation solves for "X=50 ohms" which is the correct answer for this problem. The suggestion was for you to use two possible solution paths to get this answer. One by simple inspection of the problem recognizing that the current flow...

Consider the problem conditions given. You are told that the Galvanometer reads zero. This menas that the current flow from the 2 volt battery must be zero. In order for this to happen there must be a voltage developed across the "X" resistor to match the voltage supplied from the 2 volt...

Torque and horsepower do not equate exactly. There are 550 Foot-Pounds/Second of torque in 1 brake-horsepower. Also an accerlation of 1 foot/second*second = 0.6818166 miles/hour*second. Hope this helps to get you started.

The answer is yes, but the solution is problematic and you will spend a lot of time making up the interface between the GSM and VB. Good luck with this.

Interesting. You do not give a very good explanation of how the system was connected to the supply circuit but here is an example of how an industrial plant could get into trouble. Let's say the plant is very large and has multiple service points. (Service points meaning locations where they...

Looking at your figure you will notice you have at least one voltage supply labeled incorrectly. You show both B&C phases as 100volts at an angle of 120 degrees. Although it appears that you corrected for it in your mesh 2 equations, if mesh two is correct, then the same voltage in mesh one is...

Solution to Rate of Turn, Radius of Turn Problem Thank you for your suggestions. I was using degrees in Excel but did convert them to radians. I broke down the steps I was making towards my solution by setting solutions for each step along side of the appropriate column and was able to...

I have been working on a problem trying to solve the turn diameter of an aircraft in a steady state level turn. I have been using the standard equations for turn radius and rate of turn. The formulas I am using are: 1) Turn Radius = (Velocity*Velocity)/(11.26 *Tan (bank Angle)) , velocity...

The "real" impedance is the vector sum of the positive, negative and zero sequence components. The positive sequence impeadance is a set of balanced CCW rotating phasers, the negative sequence impedance is a set of balanced CW rotating phasers, the zero sequence impedances are a set of balanced...

I believe Averagesupernova wants you to connect the resistor to Vcc and then to pin 2. The switch then is connected between pin 2 and ground. When the switch is open, pin 2 will have a voltage on it that is close to Vcc and definitely higher than 1/3 Vcc. When you close the switch, this will...