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Okay, thank you very much. I have some more but I will post them later. Actually, here are some vector calculus questions:
Determine if the vector field f(x, y,z)= x yzi+xzj+x yk has a potential in R3
I am getting that it does not have the potential.
Hi; that's great. But unfortunately, I do not understand what you mean by CGPA and A1. I though marking was out of 500.
Phronister, is the only thing you do with your life puzzles?
Agnishom, how did your exams go? You were supposed to get your results a couple of weeks. ago, weren't you?
Oh, okay. Thank-you. The second one is a bit more complicated.
25m is 82 feet so did you simply approximate something?
Well, the units in the problem were feet, and 9.81 is approximately 32 feet. Also, "Similar to the problem in post 8. ground level will be position 0, up will be the positive direction and time t=0 will be the instant when Fred first throws the ball. The acceleration due to gravity will be a = -32."
Similar to the problem in post 8. ground level will be position 0, up will be the positive direction and time t=0 will be the instant when Fred first throws the ball. The acceleration due to gravity will be a = -32. Thus dv/dt = -32, so that v(t) = -32 + C. The problem does not give any information about the velocity of the ball at any time so I use v0 to represent initial velocity. AT time t = 0, I get v(t) = -32t + v0. So, as v(t) = dr/dt, y(t) = integral (-32t + v0)dt = -16t + v0t + C.
Yes, but assuming they start with 2, take one (half of two) and return one immediately. That probably is not the case.
I get y(t) = Integral (-32t + v subscript 0)dt = -16t^2 + v subscript 0 t + C
So y(t) = -16t^2 + v0t (o = subscript)
So 0 = -16 * 5^2 + initial velocity * 5 = -400 + 5 (initial velocity) = -400 + 5 initial velocity
Solving gives us V0 = initial velocity = 80 ft / second
Hi Stefy; for which question are you getting that?
Well yes, that seems too easy if they take one and return one.
One more: Junita is standing on the roof of a building 192 feet tall and throws a ball up in to the air with an initial speed of 64 feet / second.
(a) Find formula for velocity and position of ball at a later time. I get v(t) = -32t + 64 And for position: y(t) = -16t^2 + 64t +192
(b) How high does the ball go? I am getting 256 feet,
(c) When does the ball hit the ground? I think this is probably wrong, but I am getting 6 seconds.
Thank-you in advance.
It says "each, in turn" thus I presume that they return one chocolate bar immediately after taking the toll.
Hi; few other simple problems. Fred is standing on the ground and throws a ball up in to the air. He observes that it falls back down to the ground after 5 seconds. What was the initial velocity of the ball? I get 0 = -400 + 5V subscript 0 Thus initial velocity = 80ft / s
He is a bot....
True... Is it actually accredited? There is a possibility that the whole thing is a ploy. That, or the US education system is corrupted.
I saw some of zee's posts before. The course administrator's don't seem to be too official though.
Well, they look like grade 7-8 level questions. Maybe just for practice?
Bob, I do not understand questions 8,9,10 and 11. Oh...now I see your older post. I thought I was just being stupid.
3 is incorrect as two interior angles on the same side of the transversal are supplementary. Thus, 3 should be yes. 4 also should be yes by the same rule. Question 8 is wrongly typed as Angle 7 should equal angle 3. Thus, 9, 10 and 11 are wrong questions.
1 is right by TPT-Z rule.
I thought I saw this titled "Using her head"....
Hi; It should be around 0.82
That depends on the calculator. Usually you type 0.9176 and then press y^x and then 2.2