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distance degrees questions and answers

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Q: What is it's angle of trajectory when it is at a horizontal distance of 1/3 of it's range in degrees?
A ball is kicked from level ground at a speed of 90 m/s at an angle of 10 degrees. It travels a certain range and returns to the ground.

A: t = flight time u = initial speed g = gravity t1 = time to cover 1 / 3 of range x' = horizontal velocity y' = vertical velocity a = angle of projection b = angle required. t = 2u sin(a) / g t1 = 2u sin(a) / (3g) x' = u cos(a) y' = u sin(a) - gt1 = u sin(a) - 2u sin(a) / 3 = u sin(a) / 3 tan(b) = y' / x' = tan(a) / 3 tan(b) = tan(10) / 3 b = 3.36 deg.

Q: Distance for degrees, minutes, seconds on a map?
Can anyone define the distance at the equator (such as miles or feet, etc.) for a degree, minute, second and an arcsecond is on a map or globe? I know it will vary the further you go away from the Equator, but give me your best estimate. If you have a website, that will also help. I'll check back in 1 hour for answer. It's 1:25 pm PST right now. Best answer will definitely get the points. Promise... Thanks Apparently, I have to wait 4 hours to award the points. Will come back then to give 'em out. Thanks!

A: You can figure this out. Find the radius of Earth and multiply it by 2*pi (or find the diameter and multiply by pi) to find the Earth's circumference. Then divide that number by 360 (number of degrees in a circle) to find how "long" a degree is. Divide that by 60 (number of minutes in a degree) to find how long a minute is. And finally divide that by 60 (number of seconds in a minute) to find how long a second is.

Q: Are there any accredited Christian seminaries that provide distance learning for graduate degrees?


A: I would avoid them. Your work load is greater and they often times will require you to travel long distances just to take exams at specified testing places. I did it for a few semesters and hated it. God bless!

Q: a plane flies from base camp to lake A, a distance of 280km at a direction of 20.0 degrees north of east. ?
After dropping off supplies, the plane flies to lake B, which 190km and 30.0 degrees west of north from lake A. Determine the distance from lake B to the base camp. Answer: 310 km 57 degrees south of west im looking for a detailed explanation if possible.

A: Hint: attach the tail of "190" to the head of "280" in an xy coordinate system: Need more help let me know. 190km and 30.0 degrees west of north means 190@120 degrees CCW Sum the x components: 280* Cos( 20)+ 190*Cos(120) =168.114 i Km Sum the y components: 280* Sin(20)+ 190* Sin(120)=260.310 j Km Angle with x-axis: =57.14 degrees vector magnitude 309.877 Km

Q: what us the horizontal distance traveled by a soccer ball kicked into the air at 32 degrees above the?
what us the horizontal distance traveled by a soccer ball kicked into the air at 32 degrees above the horizontal with an initial velocity of +26.0 m/s? a. 2.4 m b. 31 m c. 62 m d. 39 m

A: First you need to convert the angled direction into horizontal and vertical velocities... In the y direction (the upward velocity), the speed is 26sin(32) = 13.778m/s = Viy (initial V in the y-direction) In the x direction (horizontal velocity), the speed is 26cos(32) = 22.049m/s = Vix (initial V in the x-direction) With the Vy value (velocity in Y direction) you can figured out how long it takes for the ball to reach the maximum height... time = (Vfinal - Vinitial)y / acceleration in the y-direction = (0m/s - 13.778m/s)/9.8m/s^2 = 1.406s So it takes 1.406 seconds for the ball to reach its max height...that means it takes 2.812 seconds for the ball to hit the ground again (1.406 seconds going up, and 1.406 seconds coming down). Plug in the value t=2.812s into the displacement formula given below for x-values Dist = velocity(x-direction)*time + 0.5acc(x-direction)*time^2 = 22.049m/s*2.812s + .5 * 0m/s * 2.418^2 = 22.049 x 2.812 = 62m So your horizontal distance travelled is 62 meters

Q: why most of the universities offering quality graduation and post graduation degrees through distance learning
mode are located in tamilnadu. can you name few universities offering career oriented courses through distance education located in north india except IGNOU.

A: Try in Google, you can get every information Cheers Prince

Q: Where can i find a wrist watch that has both a digital compass (degrees) and a distance indicator (meters)?
ive been looking for a watch with a compass and a sort of pedometer where i can measure my own walking or running distance. I can find one or the other, but not both on the same watch

A: I found my husbands on ebay. We got the "fully loaded" one with gps, pedometer, and a hundred other items on it. Just be willing to spend money on a good one.

Q: a distance measued in degrees east or west of the prime meridian is called?


A: Distances east or west of the Prime Meridian are longitude.

Q: Any brick and mortar colleges that off distance/online CS degrees?
I am interested in earning a Computer science degree (info security and/or software engineering), but I am a construction worker, so I work all over the state/region, and work 10-12 hrs a day, so i really have not time to go to the actual school, so i need an online degree program that is also FULLY accredited/respectable. I have read too much about the degree mills, so I feel that i need to go through a college that has an actual campus. Any suggestions/advice/links would be really helpful.

A: I am an adjunct at Franklin University in Columbus, OH, teaching technical communications to MIS/ITEC students online. Franklin is accredited, has a brick and mortar campus, and does a nice job online. Check them out at Franklin.edu. We specialized in "non-traditional" students such as you.

Q: why does throwing a ball 45 degrees yield the greatest distance?
at 45 degrees the ball is neither in the air the longest nor does it have the greates intial velocity. why?

A: Hi jset - There's a powerful way to prove this using calculus, but let's do it with simple reason in a "thought" experiment. (Einstein loved thought experiments!) First, let's start by throwing the ball at a ninety degree angle. We observe it goes very high, but not very far. In fact, it goes nowhere - lands at our feet. Okay, now throw the ball at a zero degree angle - we observe it doesn't go very high (in fact, it just goes down) and it also goes nowhere, landing at our feet. Now let's think about throwing the ball at ever increasing angles, starting at zero and going on up to 45 degrees. We observe the ball goes farther and farther until we reach 45 degrees, then as we continue to increase the angle of throw beyond 45 degrees, the distance it goes decreases until it again becomes zero at ninety degrees. Q.E.D. as they say in geometry class - Quo Erat Demonstadium. So what's the deal? Why??? The answer is a trade-off between height and distance in a force field - the field caused by gravity. The energy imparted to the ball when it is thrown goes into doing two things - resisting the vertical pull of gravity and translating the ball horizonally. At 45 degrees, the optimum energy trade-off between vertical height and horizonal translation is achieved. By the way, in order to conduct a valid experiment, the ball must be thrown with the same inital velocity, which equates to equal energy, in each case. That means that the thowing force must be modulated for each throw to account for the force of gravity. For example, imagine the ball weighs 100 lbs (a cannon ball?) and to conduct the experiment we decide to throw the ball each time at 100 ft/sec. Then when trowing it at ninety degrees, the thrower must exert enough force to accelerate the ball to that speed against the force of gravity (which is 100lbs in opposition to the desired direction of travel) . However, when throwing the ball at zero degrees, the gravitational force of 100lbs is acting in the same direction of travel, so the force required of the thrower is less by that much. And of course, at 45 degrees, the throwing force will be half that required at ninety degrees and twice that required at zero degrees. You see, if the ball were thrown at equal FORCE in each case, the resulting initial velocity, and thus the ball's energy, would be different in each case. Fun - huh?? Hope that helps your insight! CQ

Q: A toy shoots off the floor. If Velocity is 4.5 m/s at an angle of 37 degrees, what distance is covered?
If a toy is able to shoot off the ground at velocity of 4.5 m/s at an angle of 37 degrees, how much distance will it travel? Mass is not known. Please show your work...

A: Mass is not needed for this question... This is vector analysis... For the 37 degree V vector it's horizontal component will be: 4.5m/s * cos37 = 4.5m/s * .7986 = 3.59 m/s The vertical vector will be: 4.5m/s * sin37 = 4.5m/s * .6018 = 2.71 m/s Gravity will overcome this vertical vector so: v*t = g*t^2 v/g = t 2.71m/s / 9.8m/s^2 = t t = .2765 since this is the time to reach the apex, the highest trgectory, the time is exactly the same to fall back down so we multiply that time by 2: t = .5530 seconds Last part horizontal distance: d = v*t d = 3.59m/s * .5530 s d = 1.99 m or 2.0 meters (sigfig)

Q: Postgraduate degrees by distance learning?
Does anyone know of a UK university that offers masters and higher degrees by distance learning (other than the OU)?

A: I think that several universities now offer distance learning higher degrees. I'm studying an MSc at the University of Derby at the moment (via distance learning). My subject is not available at this university by any other method, but I know other subjects do not offer distance learning at Derby. You need to be very self-directed to manage distance learning at this level. I am really enjoying it, and find that it's really flexible. I work full-time as well, and would therefore not be able to manage to study any other way. I think your best bet would be to look at university websites for the subject you are interested in, and see what methods are available. Good luck with your studies.

Q: What is the distance around the earth at 33 degrees North latitude?


A: distance = 2.pi.radius_at_equator.cos33 =33.6x10^3 km edit: there will be some error as I haven't taken into account the ellipticity of the Earth, but it'll be fairly minor.

Q: Online/distance learning degrees?
Is there anyway I can do an arts degree online/or by distance which isn't very expensive? I don't need a library or tuition or anything All I want is a syllabus, deadlines, marking of my assignments and an accredited degree certificate.

A: Not all distance degrees are trash. They've all earned a pretty bad rep since the University of Phoenix came around, and all the various scandals associated with that school. I live in Canada and I know a highly accredited distance learning school. Their MBA program is one of the top 75 internationally, and it's done COMPLETELY online. It's Athabasca University (http://www.athabascau.ca). It's based in Canada but is done by students worldwide. They offer whole programs and even some graduate programs. Another popular one is Open University based in the UK, which I think was one of the first distance universities out there (http://www.open.ac.uk/about/ou/). Overall in the UK, they're VERY highly ranked out of all the universities, regardless of the fact that they're a distance based university. BOTH these schools were created by the government, rather than a private company (like University of Phoenix) so I would say in a sense, they are more reliable and I personally would prefer attending one of those because they don't operate on a "for-profit" basis. As for the expense factor, all universities, online or not, have rising tuition fees. Your best bet is to contact that schools Financial Aid department. If you live in Ontario, OSAP (Ontario Student Assistance Program) really helps you out, and I know for a fact that if you attend Athabasca, they'll still give you funding, even though it's based in another province. But before I can answer the rest of your question, what do you mean by you don't need a library or tuition? What exactly are you talking about when you're referring to tuition?

Q: Do companies nowadays consider distance and online degrees on par with traditional college degrees?


A: It depends on who you plan on working for. If it's an accredited school you should be alright. (a lot of universities offer pretty good on line MBA programs)