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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.
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.
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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.
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.
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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.
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.
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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.
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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.