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physics potential gravity help needed?
This one seems pretty straight forward but i feel like i'm missing data. I'm not sure if this value is constant all the time everywhere for every planet but help would be appreciated.
Earth's average distance from the sun is 1.5 x 10 (8th power) km, the mass of Earth is 5.97 x 10 (24th power) kg, and the mass of the sun is 2.0 x 10 (30th power) kg. At this distance Earth is traveling at an average speed of 66,000 mi/h (29,500 m/s) in its orbit. (a) Calculate the average kinetic energy of Earth in its orbit; (b) calculate the average gravitational potential energy of the Earth-sun system; and (c) predict qualitatively what will happen to Earth's speed in its orbit when it is at perihelion at a distance of 1.47 x 10 (8th power)km from the sun.
i know for at least (a) we need U=(-GMm/R) and the range and two masses are given, but what value to i plug in for G?
any help is appreciated. is G constant on every planet and in every system. and am i on the right track thanks in advance.
1 AnswerPhysics9 years agohelp with physics needed please and thank you?
Students use a plastic bow-and-arrow set in a laboratory experiment to examine projectile motion. The bowstring is pulled back a distance of 16.2 cm to launch the arrow, which has a mass of .75g. When the arrow is fired horizontally from a height of 1.00 m above the floor, it lands on the floor a horizontal distance of 6.80 m from the bow. (a) Determine the launch velocity of the arrow. (b) calculate the elastic potential energy when the arrow is pulled back ready to launch. (c) Determine the elastic constant, k, of the bowstring. (d) Use energy-conservation methods to determine the speed of the arrow when it hits the floor.
converting displacement that it was pulled back to .162 m
convert mass to .075 kg.
i started with U=K mgh=.5m(v squared) v=sqrt ( (2) (9.8) (1)) v=4.43m/s
now no idea how that helps find horizontal velocity and i'm stuck thanks in advance.
1 AnswerPhysics9 years agonot sure if the book is wrong?
ok i have this problem a 5.0 kg box is being pulled at constant speed along a level sidewalk by a rope that is attached to the front of the box and inclined at an angle of 30 degrees above the sidewalk. the coefficient of friction between the box and the sidewalk surface is .70. find the tension of the rope.
so i've gone about this a couple different ways but as far as i can gather so far F-friction=ma and acceleration should be zero since its constant velocity so netforce of the x direction -friction equals zero. and friction if i remember correctly is 34.3 so that is the netforce acting in the x direction. which is Fcos30 pulling from the string. so to get the force the string is pulling at i divide 34.3 by cos30. and i think that netforce of the system -T=ma so again it should zero out. and when i do that i get around 39 for the F and T and the answer key in the book says 28N for Tension. so any help is appreciated.
3 AnswersPhysics9 years agophysics help please i'm stuck on this one for about 2 days now?
ok two cables are attached to the roof, and a piano is held up by the cables. assuming the piano weighs 350lb, find the tension in each cable.
ok there are two cables T1 which is 40 degrees above the horizon and T2 which is 55 degrees above the horizon.
1 AnswerPhysics9 years agowant to be a stand up comedian?
i'm not worried about preforming on stage and am told to have a natural ability to make people laugh from quick lines during a conversation to an experience i've had that others enjoy i've always wanted to get into comedy and would like help knowing where to begin and how to find gigs and whatnot. appreciate your help.
5 AnswersPerforming Arts10 years ago