Physics,Kinemaics=HELLPP!!!

At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

Physics,Kinemaics=HELLPP!!!

Physics
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

David travels in an airplane a distance of 3180 km. For one-half 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
Please give the formula for how you solved the questions as well,this would be greatly appreciated.
for 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\[v=\frac{ d }{ t }\]

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

whoops. 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)=?
number 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.
for 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 \[v_{avg}=\frac{ (d1+d2) }{ t_{tot} }\]
cripes. 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. \[speed=\frac{ 13,464,000,000km }{ 2160hr }\]
now 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.
so 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.
whoops that first equations should have an equals sign there and not a minus/dash sign.
that last one you'll have to do some cross multiplication to isolate your v final so just use one of the first two.
just use the second one. it's so much easier.
i think most of that is right except for the sun questions. can you check @IrishBoy123 ?
the OP should put each question in a separate thread dumping it all in one makes it hard to follow.
sun 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.

Not the answer you are looking for?

Search for more explanations.

Ask your own question