Recent content by chazgurl4life

thanx so i guess my prof was wrong thanx

Homework Statement 2. Consider an electrochemical cell as shown, with Zn in ZnCl2(aq) and Cu in Cu(NO3)2(aq), and a salt bridge containing KNO3(aq). The overall chemical reaction is Zn(s) + Cu2+(aq)  Zn2+(aq) + Cu(s). a. How many moles of electrons are transferred in this...

The relationship between free energy and the equilibrium constant is Go = -RT ln K By measuring the pH at various temperatures, the Kb for NH3(aq) was found to be temperature dependent, yielding the following values: temperature (K) Kb 283 1.34 X 10-5 293 1.42 X 10-5 303 1.50 X...

The relationship between free energy and the equilibrium constant is Go = -RT ln K By measuring the pH at various temperatures, the Kb for NH3(aq) was found to be temperature dependent, yielding the following values: temperature (K) Kb 283 1.34 X 10-5 293 1.42 X 10-5 303 1.50 X...

ok i think i got it...so ...66*9.8=-(132.3)x X in this case is -4.8 ms.. so if i add that to the 28 meters the unstretched cord should be in length equal to 23.1 meters ...that should be rite

so how do i figure out the unstretched length...now I am confused

ok so...if i use v= (T/m/l)^1/2 it should be v=(.07154/4.4x10^4 kg/m/.150m)^1/2 which would turn out to be...32.7 m/s...im confused

ok i know that Force = - kx could x=length of the bungee cord... and which would lead me to believe that M*g=force also equals -Kx so...should i be using 66kg (9.8)=-132.3(x) and solve for x?

A bungee jumper with mass 66.0 kg jumps from a high bridge. After reaching his lowest point, he oscillates up and down, hitting a low point eight more times in 35.5 s. He finally comes to rest 28.0 m below the level of the bridge. Calculate the spring stiffness constant of the bungee cord...

frequency= 1/2pi(g/L)^1/2= so...-->4.02 Hz vo=2*pi*frequency*amplitude= v=+/-Vo ( 1-x^2/A^2)^1/2 i don't know the wave equation

One end of a horizontal string is attached to a small-amplitude mechanical 57 Hz vibrator. The string's mass per unit length is 4.4 10-4 kg/m. The string passes over a pulley, a distance L = 1.50 m away, and weights are hung from this end Assume the string at the vibrator is a node, which is...

Ok so i know that K= (2pi*freq.)^2* mass and then once i have K i can uses vmax= 2pi Af and solve for A?

An object with mass 3.5 kg is attached to a spring with spring stiffness constant k = 300 N/m and is executing simple harmonic motion. When the object is 0.020 m from its equilibrium position, it is moving with a speed of 0.55 m/s. b) Calculate the maximum velocity attained by the object...

A 0.55 kg mass at the end of a spring vibrates 2.0 times per second with an amplitude of 0.16 m. (a) Determine the velocity when it passes the equilibrium point. I have no idea how to figure this out i mean to figure out velocity ...we use this equstion vmax= 2pi*Amplitude* Frequency I...

The tires of a car make 85 revolutions as the car reduces its speed uniformly from 90 km/h to 30 km/h. The tires have a diameter of 0.80 m. (a) What was the angular acceleration of the tires? rad/s2 angular acceleration = Omega final- Omega intial / Delta Time does that mean that in...