Recent content by MJC8719

Thanks guys...i was being sloppy with the units...long day of work, class, then volunteering, plus an exam to study for tonight...much appreciated

I am trying to solve for the velocity correct...so I can use it in my equation for objects rotating around a circle... So (1/2)mv^2 = KE = 45keV I would then convert this all to joules...then I would have the velocity which could then be plugged into the equation: r = mv/(qB) so I could...

1 keV = 1.783 x 10-33J of energy...i am pretty sure i did that part right Our professor also gave us the fact that the mass of e = 511kev/C2 So that would mean (1/2)mv^2 = 45 v^2 = (45/0.5x511) v = 0.41967 Then B = (511KeV)(0.41967)/(1.60X10-19)(200) Which means B = 6.7016e18 T...

High above the surface of the Earth, charged particles (such as electrons and protons) can become trapped in the Earth's magnetic field in regions known as Van Allen belts. A typical electron in a Van Allen belt has an energy of 45 keV and travels in a roughly circular orbit with an average...

The rotor in a certain electric motor is a flat, rectangular coil with 90 turns of wire and dimensions 2.50 cm by 4.00 cm. The rotor rotates in a uniform magnetic field of 0.800 T. When the plane of the rotor is perpendicular to the direction of the magnetic field, it carries a current of 9.1...

Hi, We have been working on a lab that dealt with oscillations and simple harmonic motion. We attached two springs to an air cart, placed it on an air track, and released a distance of 10cm from the equilibirum point. To investigate the relationship between mass and the period, we added 50...

The pendulum consists of a uniform disk with radius r = 10 cm and mass 900 g attached to a uniform rod with length L = 500 mm and mass 100 g. (a) Calculate the rotational inertia of the pendulum about the pivot. kgm2 (b) What is the distance between the pivot and the center of mass of...

At t = 0, a 730 g mass at rest on the end of a horizontal spring (k = 128 N/m) is struck by a hammer, which gives it an initial speed of 2.74 m/s. (a) Determine the period of the motion. s Determine the frequency of the motion. Hz (b) Determine the amplitude. m I have found parts A...

yes..i do write out the units on my paper...its just quicker to not type them out on the computer... thanks for all the help, greatly appreciated.

So, if i understand you correctly, then my 3.33 g/cm^3 = (3.33 x 10^6)/1000 (i can't do the fancy stuff like you did lol) which gives me a value of 3330 kg/m^3 So then the equation would read: m = (3330)(4/3pi(1000^3)) so m = 1.3949 x 10^13 which is them multipled by 0.5v^2 And in part A, i...

In the early morning hours of June 14, 2002, the Earth had a remarkably close encounter with an asteroid the size of a small city. The previously unknown asteroid, now designated 2002 MN, remained undetected until three days after it had passed the Earth. Suppose that at its closest approach...

A disk-shaped merry-go-round of radius 2.63 m and mass 155 kg rotates freely with an angular speed of 0.615 rev/s. A 59.4 kg person running tangential to the rim of the merry-go-round at 3.99 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the...

So, when I tried to solve this i got: (2.74)(51.7)(2.86) + (1/2(168(2.74^2)(0.737) = [(.5(168)(2.74^2) + (51.7 x 2.74^2)] w then i solved for w and got 0.8539 which is not correct what am i missing?

Actually, ignore all of this...the kinetic energy is not relevant...i need to focus on angular momentum therefore the equation should be: rmv(child) + Iw = (i + mr^2)(w) is this correct?

A disk-shaped merry-go-round of radius 2.74 m and mass 168 kg rotates freely with an angular speed of 0.737 rev/s. A 51.7 kg person running tangential to the rim of the merry-go-round at 2.86 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the...