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I didn't think that Magnesium would combine with oxygen... thnx stmoe.

The actual question is Mg reacts with CO2 (Carbon dioxide) to give a white solid and a black solid. Provide the balanced equation for this reaction.

well... the problem only gave me the first part of the equation which was the CO2 + Mg. This is the chemical process that occurs when dry ice is mixed with burning Magnesium pellets, which creates this fire work type of reaction. I figured that the magnesium and carbon will combine and release...

I don't know if I solved this problem correctly. Balance the equation: CO2+ Mg -> MgC + O2 Is this the right end result and can someone help me figure out the state of matter for each.

Can some one help me with this Milliken oil drop problem. Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data: Voltage needed to suspend drop (V) Drop diameter (microns) 103------ 1.1 44.7----- 1.2 127------ 1.6...

No one wants to try?

come on anyone please help me or at least tell me if I'm in the right direction.

Can some one help me with this Milliken oil drop problem. Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data: Voltage needed to suspend drop (V) Drop diameter (microns) 103------ 1.1 44.7----- 1.2 127------ 1.6...

Milliken Oil Drop Problem anyone? Help please. Can some one help me with this Milliken oil drop problem. Data from a Milliken oil drop experiment are given below. Calculate the value of the charge on an electron from this data: Voltage needed to suspend drop (V) Drop diameter (microns)...

1. A ball A is thrown vertically upwards from the ground at a velocity of 20 m/s. A second identical ball is thrown vertically upwards, at the same velocity, one second later from a platform 10 meters above the ground. The two balls will be at the same height above ground when ... a. A is...

for 1. I don't know what Lorentz force is 2. The direction of the magnetic field should be towards the south-pole end.

1. A charged particle moves in a straight line through a certain region of space. The magnetic field in that region a) has a magnitude of zero b) has a zero component perpendicular to the particle's velocity c) has a zero component parallel to the particle's velocity 2. The north-pole end...

umm... 3.78 x 10^-7 V

so i multiply my emf by 2 which gives me -7.56 x 10 ^ -6 V

yay... so I use the emf equation and get -3.78 x 10 ^-6 V. does that sound good?

ohhh ok. here's what i got so far 3.15 x 10 ^ -4 T for the magnetic field.

well... i got a magnetic field of 1.14 x 10^-6 T

how do i figure out what n is if I don't have length. Should I use the area to find the length which is equal to 2r?

can i use this equation to solve for the magnetic field? B=MIn

Here's the problem: Can someone help me start it off? A long straight solenoid of cross-sectional area 6 cm^2 is wound with ten turns of wire per centimeter, and the windings carry a current of 0.25 A. A secondary winding of two turns encircles the solenoid. When the primary circuit is...

Anyways, thank you very much again for all of your help SpaceTiger.

WOW ur a genius! Will u be my homework buddy? LOL Thanks a lot for the help even though it's like uber late there in NJ. I might be bugging you for more physics stuff. Hope you don't mind.

Wahh... 1) it's watts not joules ... sorry. :cry: 2) wow.. I^2 R --> (5.00Amps)^2 (0.200 ohms) = 5.00 watts

lol.. ur funny. Well... here's what I have now: a) emf =1.00 V b) I=5.00 Amps, Fmag=1.25N c) Fv=5.00j wait.. u got me.. crap.. I keep messing up. Thnx for dealing with my overwhelming stupidity.

Everything moves 1 decimal point to the right.

OOps... sry.. its only 0.500 m = to 50 cm. Sorry bout that. need to change all my data now...

Ok. a) emf= Bvl --> (4.00m/s)(0.500T)(0.0500m) = .100 V b) I=emf/R --> (0.100V)/(0.200 ohms) = 0.500 A Fmag= IBl --> (.500Amps)(0.500T)(0.0500m) = 0.0125N

Ok, I got 0.0500 j for the work. And I got .500 Amps for the current. Now what?

yup. I got 0.0125 N.

Ummm... here's the entire problem: A conducting rod AB makes contact with metal rails CA and DB. The apparatus is in a uniform magnetic field 0.5 T, perpendicular to the plane of the diagram. The rod is 50 cm long. a)find the magnitude and direction of the emf induced in the rod when it is...

Ok, the other formulas for power are P=V^2/R and P=VI. I don't know anything about the force.

I don't quite understand what I'm suppose to do in this question: --> Compare the rate at which mechanical work is done by the force (Fv) with the rate of development of heat in the circuit (i^2 R).

Can anyone help me with those other problems that I couldn't get the answers to, especially the potential difference problems.

Here are my homework problems. I did most of them, but I don't understand how to do the others. 1. A rod of length 0.40 m moves in a magnetic field of magnitude B=1.2 T. The emf induced in the moving rod is found to be 2.40 V. a)what is the speed of the rod? b)if the total circuit...

Here is a problem that I don't quite understand. P: Two hikers are reading a compass under an overhead transmission line that is 5.0 meters above the ground and carries a current of 400 Amps in a direction from south to north. a. Find the magnitude and direction of the magnetic field at...

So is my second solution the right one? or are they both wrong?

Well... since I knew how much volts there is, I figured that it is faster to just use that formula where k is the Coulomb constant (8.99x10^9 Nm^2/C^2), then just solve for the radius. Is it possible to use this equation like that?

I'm confused if I am using the right equations to solve this problem. P: A singly charged Li^7 ion has a mass of 1.16x10^-26kg. It is accelerated through a potential difference of 500 Volts and then enters a magnetic field of 0.4 Teslas, moving perpendicular to the field. What is the radius...

So what you're trying to say is that since I know the voltage, I should multiply it by the charge of an electron (1.60 x 10^-19 C). By doing so, I will get the total energy, which I can use in the kinetic energy equation to solve for the velocity. That sounds great. So there's no need for me to...

Sorry, I don't know the answer. I wasn't sure about the velocity part either. Because I used U=1/2 mv^2, which is for springs, i think. How did u solve it? And the V= U/q is for volts.

I am really trying hard to pass my IB Physics class, but the homework is just too hard. Please help me with this problem. In a TV picture tube, an electron in the beam is accelerated by a potential difference of 20,000 V. Then it passes through a region of transverse magnetic field, where it...

I am really trying hard to pass my IB Physics class, but the homework is just too hard. Please help me with this problem. In a TV picture tube, an electron in the beam is accelerated by a potential difference of 20,000 V. Then it passes through a region of transverse magnetic field, where it...