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anonymous
 one year ago
Physics,Kinemaics=HELLPP!!!
anonymous
 one year ago
Physics,Kinemaics=HELLPP!!!

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0David travels in an airplane a distance of 3180 km. For onehalf of the distance, the airplane flies at a speed of 624 km/h, and for the rest of the distance, it flies at a speed of 720 km/h. How long does the trip take? 3) Sarah roller skates for 3/4 hours with a constant speed of 24 km/h and then for another 2 hours 30 minutes with a constant speed of 12 km/h. What is her average speed for the total trip? 4) With the sun as a reference point, what is the velocity of the Earth in km/h? 5) With the sun as a reference point, what is the speed of the Earth in km/h? 6) You jump off a bridge that is only 34 feet high and takes you 1.46 seconds to hit the water. What velocity do you hit the water at? the acceleration due to gravity is 32.2 ft/s/s

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Please give the formula for how you solved the questions as well,this would be greatly appreciated.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1for david they give you the velocity and the distance traveled and now you just have to find the time. now here's the equation for velocity\[v=\frac{ d }{ t }\]

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1whoops. messed up that. ok so for half the distance he travels at the speed of 624 km/hr so for 1590 km he is going 624 km/hr (1590km)*(1hr/624km)=2.55hrs then the remaining distance he travels at the speed of 720km/hr (1590)*(1hr/720km)=?

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1number three you are using the same equation.\[v=\frac{ d }{ t }\] this time it's asking you to find the distance traveled by giving you the duration the person was traveling at a certain speed. so just find the distance for one segment and then the distance for the other segment and add them together. then you divide the combined distance by the total time to find the average speed.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1for the first segment d1=(0.75hr)*(24km/hr)=18 km then the same for the second part but it's tricky with using 2 hours and 3o0 minutes but just remember that an hour is 60 minutes so half of an hour is 30 minutes so really you can used 2.5 hours there. d2=(2.5hr)*(12km/hr)=? also crucial is total time so to find that is just adding all the times so t=(0.75hr)+(2.5hr)=? and then for the next part to find the average speed (or average velocity) use \[v_{avg}=\frac{ (d1+d2) }{ t_{tot} }\]

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1cripes. we'll have to look bigger. ok. same distance but we'll be looking at months. one full year around the sun is 12 months. so divide by 4 we'll get 3 months. using the same equation Arc Length= (theta)*(radius) then divide that by 3 months. \[ArcLength=(90)*(149,600,000km)=13464000000km\] now take that and divide by the time to find the speed.\[speed=\frac{ 13,464,000,000km }{ time }\] lets assume we started measuring in January so we'll stop at March and that would be 31 days in january, 28 days in february, and 31 days in march and add that up to get 90 days then in one day we have 24 hours so multiply that 90 by 24 and get 2160. so for time we'll use 2160 hours. \[speed=\frac{ 13,464,000,000km }{ 2160hr }\]

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1now speed is a scalar because it doesn't have a direction so for velocity i guess you would say the direction it's traveling around the sun counter clockwise i guess. not sure about that one.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1so for the last one you have distance you'll travel, time it takes to travel that distance, and the acceleration that is affecting you in that fall. you can use three of the four kinematic equations.\[v^{2}v^{2}_0+2*a*d\]\[v=v_0+a*t\]\[d=\frac{ v_0+v }{ 2 }*t\] now you're starting velocity is at 0 sinnce you'll be jumping off the bridge and the question doesn't specify that you jumped at an upward angle so we'll assume you jumped out horizontally so really your initial velocity is 0 so that clears up all the v_(0) in the equations.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1whoops that first equations should have an equals sign there and not a minus/dash sign.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1that last one you'll have to do some cross multiplication to isolate your v final so just use one of the first two.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1just use the second one. it's so much easier.

zephyr141
 one year ago
Best ResponseYou've already chosen the best response.1i think most of that is right except for the sun questions. can you check @IrishBoy123 ?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1the OP should put each question in a separate thread dumping it all in one makes it hard to follow.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.1sun orbit stuff.... \(\dfrac{2 \pi \cdot 150,000,000(km)}{365 \cdot 24}= 108,000 \; kph\) the distinction between velocity and speed in this context seems meaningless.
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