redbeardthegiant
redbeardthegiant
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1] Separate the horizontal & vertical components. 2] d = .5 at^2 ---> t = SQR[2d/a] 3] Once you know t, then use f=ma --> a = f/m to find horizontal a, then use d = .5at^t 4] To get impact speed you have to add quadratically the Vx and…
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1] Separate the horizontal & vertical components. 2] d = .5 at^2 ---> t = SQR[2d/a] 3] Once you know t, then use f=ma --> a = f/m to find horizontal a, then use d = .5at^t 4] To get impact speed you have to add quadratically the Vx and…
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Set it equal to zero and use synthetic substitution
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Set it equal to zero and use synthetic substitution
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t + u = 14 t - u = 2 OR -2; you don't specify which is larger rearrange t + u = 14 to t = 14 - u insert into the second equation to get 14 - u - u = 2 or -2 14 - 2u = 2 or -2 12 = 2u or 16 = 2u u = 6 or 8 t = 8 or 6 So it is either 68 or…
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Well, leaving aside that the physics is WRONG... You ave 1/4 the power, so you must have 4 times the resistance
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Well, leaving aside that the physics is WRONG... You ave 1/4 the power, so you must have 4 times the resistance
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J = 2P P = A + 5 J+5 = 3A Combine 1&3 2P + 5 = 3A Substitute in 2nd eq 2[A+5] +5 = 3A 2A + 15 = 3A A = 15 P = 20 J = 40
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1] Doctors are trained to be cool about teen males getting wood. Do not let that stop you. If your bad experience included doctor misbehavior, report it. If it was wood and an unsympathetic doc, well, some docs, like some people in general, are @ssh…
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1] Doctors are trained to be cool about teen males getting wood. Do not let that stop you. If your bad experience included doctor misbehavior, report it. If it was wood and an unsympathetic doc, well, some docs, like some people in general, are @ssh…
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J = 2P P = A + 5 J+5 = 3A Combine 1&3 2P + 5 = 3A Substitute in 2nd eq 2[A+5] +5 = 3A 2A + 15 = 3A A = 15 P = 20 J = 40
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Well, leaving aside that the physics is WRONG... You ave 1/4 the power, so you must have 4 times the resistance
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Set it equal to zero and use synthetic substitution