Basic Physics Homework Question Help!? Topic: Basic Physics Homework Question Help!?
June 18, 2019 / By Abagale
Question: A 5900-kg satellite is in a circular earth orbit that has a radius of 3.2*10^7 m. A net external force must act on the satellite to make it change to a circular orbit that has a radius of 8.9*10^6 m. What work must the net external force do? I have asked this question before but the answer was wrong. Thank you for your help! Best Answers: Basic Physics Homework Question Help!? Sparrow | 5 days ago
the work is equal to the variation of potential gravitational energy W=-GmM/(r2-r1) W=GmM/(r1-r2) G=6.67x10^-11 m^3/(kg s^2) gravitational constant M=5.97x10^24 kg Earth mass m=5.9x10^3 kg satellite mass r1-r2=2.31x10^7 m delta radius W=6.67x10^-11*5.9x10^3*5.97x10^24/2.31x... W=101.7x10^9 J
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Did you like the answer? Basic Physics Homework Question Help!? Share with your friends Originally Answered: Spanish Homework question. Am I translating these basic phrases correctly?
Ella quiere ir AL baño. (when a is followed by el, it forms a contraction al. Note: this is not the case with la, los, las, or él) No deben/deberían beber cerveza. (Deben is more of a subtle command. Deberían is more of a suggestion. Also cerveza is spelled with a z) No debes/deberías ir a la fiesta. (Depends on what your saying see above.) Correct Correct. Originally Answered: Spanish Homework question. Am I translating these basic phrases correctly?
(1) She wants to go to the bathroom: Ella quiere ir al baño. (2) They should not drink beer: Ellos no deben beber cerveza. (3) You shouldn’t go to the party: Tú no debes ir a la fiesta. (4) I am not going to do the homework: Yo no voy a hacer la tarea. (5) Are you going to go to the party: ¿Vas a ir a la fiesta? Osman
you have been given height or vertical distance h = 2.5t^3 you have been given t1 = 2.10s you have been asked t2 =? locate an equation that appropriate all those. I provide you some tricks. •An merchandise in loose fall stories an acceleration of -9.8 m/s/s. (The - sign exhibits a downward acceleration.) whether explicitly pronounced or no longer, the fee of the acceleration interior the kinematic equations is -9.8 m/s/s for any freely falling merchandise. •If an merchandise is basically dropped (as against being thrown) from an greater effective height, then the preliminary velocity of the article is 0 m/s. •If an merchandise is projected upwards in a splendidly vertical path, then it fairly is going to decelerate because it rises upward. the on the spot at which it reaches the peak of its trajectory, its velocity is 0 m/s. This fee may be utilized as between the action parameters interior the kinematic equations; as an occasion, the main suitable velocity (vf) after vacationing to the peak may be assigned a value of 0 m/s. •If an merchandise is projected upwards in a splendidly vertical path, then the rate at which that's projected is equivalent in magnitude and opposite in sign to the rate that it has while it returns to the comparable height. it fairly is, a ball projected vertically with an upward velocity of +30 m/s might have a downward velocity of -30 m/s while it returns to the comparable height. i might purely use this or a mixture or manipulation of acceleration equations. purely becareful on the subject of the indicators. h = vt1 + a million/2at^2
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Hey, here's a PDF that I found to be really helpful. It shows you how to get the answer. It should be on the fourth page. Good luck! http://wc.pima.edu/~solson/phy121/notes/notes19.pdf
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When the rock is at the top of its path, the velocity will equal 0. So, v_f = 15 - 9.8t = 0 Solve for t, and when you get that answer plug that time into the position function: x_f = x_0 + v_0*t + .5at^2 "_" is subscript.

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