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[MansionMD]

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[CaptainSSO]

1st one is tricky -- I was trying to calculate the magnetic force w/ q = 1.6 x 10^-19 (charge on a proton), but I'm not seeing how that will lead anywhere. My gut instinct is that the speed will be greater than the initial speed after 4 seconds, b/c of the magnetic force acting on it (thus the proton is accelerating), but there are no answer choices representative of that...

2 is either A or B. Since there is negligible viscosity, the fluid can be considered ideal, and therefore follows the equation Q = AV. Flow = cross-sectional area x velocity. For ideal fluids, pressure decreases and velocity increases w/ decreasing diameter of the pipe. I think you might use Bernoulli's equation for moving fluid, P + 1/2(row)v^2 + (row)gh = K, where P is pressure, and row is density, but I'm not sure. Will keep working on it.

The last one is C, I *think*. Since the fluid splits into 3 paths, the flow is split into thirds. Q = AV, so now Q/3 = AV. (Similar to Kirchoff's laws, the fluid output at a node is equal to the fluid input, but for fluid it's flow instead of current.) Anyway, since the radius is halved, the area is divided by 4, ie: A = pi(r)^2, so pi(r/2)^2 becomes A = pi(1/4)r^2. So, the flow equation is now Q/3 = (1/4)AV, and to make the 2 sides equal you must multiply v by 4/3.

Edit: So for #2, the only thing I can think of ATM is to do a ****ty job of rounding -- I'm getting answer A by doing that. Since the new diameter is d/3, and flow is constant (definition of ideal fluid), the new velocity should be 3v. Plugging into Bernoulli's equation, the potential energy is constant, so you're left w/ P + 1/2(row)(3v)^2 = K, so P = 1/2(row)(9)(v)^2 = K, and 9/2 = 4.5 = 4 if you round incorrectly, ha ha. Obviously that doesn't seem like the correct way to approach the problem, however, so I'll keep working on it.

 

[coreankim]

Hi,

First problem:

Are you sure you typed in the correct answer choices for the magnetic field problem? The force felt by a moving charge is qVxB, which is always perpendicular to the direction of velocity. Therefore, its initial velocity shouldn't change. I'll think more about this later.

Second problem:

Use Bernoulli's equation and conservation of flow rate.

Bernoulli's equation is (v^2)/2 + gz + P/p = constant
where v = velocity, z = height, P = pressure, p = density

Conservation of flow is Av = constant
where A = area, v = velocity

A1v1 = A2v2
v2 = (A1v1)/A2 = (d/2)^2*pi*v1/((d/6)^2*pi) = 9v1

Since the pipe is horizontal and density is constant:

(v1^2)/2 + gz + P1/p = (v2^2)/2 + gz + P2/p
(v1^2)/2 + P1/p = (v2^2)/2 + P2/p, with v1 = v
(v^2)/2 + P1/p = 81(v^2)/2 + P2/p
P1 - 80(v^2)p/2 = P2

Therefore, P2 is 40(v^2)p less than P1, and choice b is correct.

Third problem:

Again, use conservation of flow rate:

A1v1 = 3(A2)(v2)
pi(R^2)v = 3pi(R^2)(v2)/4
v2 = (4/3)v

Therefore, choice c is correct.

 

[CaptainSSO]

Are you sure you typed in the correct answer choices for the magnetic field problem? The force felt by a moving charge is qVxB, which is always perpendicular to the direction of velocity. Therefore, its initial velocity shouldn't change. I'll think more about this later.

The velocity in the whatever direction it was initially headed wouldn't change, but I think its actual speed would change, would it not? Since speed is magnitude w/out direction, I'm not understanding how a force imparted on the proton would not change it's speed. I'm a chemistry man, however, so there is a good chance I'm wrong ha.

 

[coreankim]

Oh, right. I meant the initial speed wouldn't change, while the direction would change. Since the force is perpendicular to the velocity, the particle would just move in a circle with the initial speed.

 

[CaptainSSO]

Oh, right. I meant the initial speed wouldn't change, while the direction would change. Since the force is perpendicular to the velocity, the particle would just move in a circle with the initial speed.

Ah, yea I see. Like centripetal force. In that case he probably left out a "1" and the answer is C.

Edit: Ah, **** me. In an effort to avoid converting d to r (although dividing by 2 is not really an "effort", I was just trying to take shortcuts), I was using (pi)(d) as the formula for area, which is in fact the formula for circumference. But anyway, coreankim did the problem correctly.

 

[ussdfiant]

I can't believe that I used to know this stuff once!

 

[BerkReviewTeach]

Oh, right. I meant the initial speed wouldn't change, while the direction would change. Since the force is perpendicular to the velocity, the particle would just move in a circle with the initial speed.

I was thinking the exact same thing. A perpendicular force does no work, so it can't give the particle any kinetic energy, and therefore doesn't change it's speed. The change in potential energy over a proton's circular path is negligible, so the speed should be constant. The answer choices have to be a typo as first listed.