What is wrong with the Indian education system?
Thinking.Science for intelligent ones.Commerce for “good in maths” students.Arts for weak students.Question papers.Bihar boards’ open cheating.Repeating the same questions from past papers. Leading to proliferation of “guidebooks” business.Clash of views.If your views don’t match with the teachers’, you won’t get desired marks.Long hauls.11000 UPSC aspirants on an average don’t make it in the interview. 2000 of them don’t find their place in the merit list. They have to go through the process again. They could have had easily filled any group B job.Mugging over understanding.“Sun rises in the east and sets in the west.”People writing XII boards mostly remember the solution of the question asked in their exam.Stickiness in thought process.Reforms are envisaged in 10 years. It takes 10 years to implement them.K. Kasturirangan to head panel on National Education Policy. He will suggest some fresh reforms. But alas, all those reforms will bite the dust.TSR Subramanian’s recommendations on Higher education are yet to be implemented.Last day hauls.“Ma’am told us these important topics. Learn these by heart. You will get good marks.”Sad part is that these persons really do top the exams.It leads to “Yes men” around teachers and professors.Vicious cycles.Intelligent—- Science—-Send to Kota—-Depressed—-Suicide.Commerce — Prepare for CA — Feels like not interested — ends up wasting precious time.Arts --Prepare for UPSC—not interested—- ends up wasting many years.Extraordinary delays.CBSE did some serious mistakes in this year’s results. Now, many students' marks are increasing. But college cut off won’t count those increased marks. These students miss the cutoffs of their favourite colleges.UPSC exams takes an year to complete the process. Many young years are drained into it.Things are beaten into the head.My friend’s sibling had started writing sentences. I asked him to write something. I asked him to write something in 5 different ways.He wrote the word “something” in five different fonts!When after few months, the teacher asked him to write something, he did the same. He was scolded by the teacher. He came to home crying.AImage: Don't Let Perfect Get In the Way of Progress.
A rocket, rising vertically, is tracked by a radar station that is on the ground 12 miles from the launchpad..?
A rocket, rising vertically, is tracked by a radar station that is on the ground 12 miles from the launchpad. How fast is the rocket rising when it is 16 miles high and the rocket's distance from the radar station is increasing at a rate of 1600 mph?
Calculus: A rocket rising vertically is tracked by a radar station that is on the ground 5 miles from the launchpad?
At time t let x(t) be the rocket's vertical distance and l(t) the distance to the radar station. So l² = 5² + x² and 2ldl/dt = 2xdx/dt→ldl/dt = xdx/dt. Now when x = 12, dl/dt = 2400, l = √(5² + 12²) = 13→ 13 x 2400 = 12dxdt→dx/dt = 2600 mph.
A rocket, rising vertically, is tracked by a radar station that is on the ground 5 mi from the launch pad...?
If s is the distance from the radar station, and h miles is the vertical height of the rocket: s^2 = h^2 + 5^2 2s ds/dt = 2h dh/dt dh/dt = (s / h)(ds/dt) When h = 4 mi, and ds/dt = 2000 mi/sec: s = sqrt(16 + 25) = sqrt(41) mi. dh/dt = (sqrt(41) / 4) * 2000 mi/sec. = 500 sqrt(41) mi/sec.
Classical Physics: A bomb is dropped from a plane flying horizontal with speed of 720km/hr, at an altitude of 980m. After what time the bomb will hit the ground?
Using the equation y=(v*t) + (1/2)a(t)^2 with 980m for y, 0 for v as the initial velocity was entirely horizontal, and 9.8m/s^2 for a as it is very close to the acceleration near the surface of Earth. We come to 980 meters = 4.9t^2 meters per second^2 which when further simplified yields 200 s^2 = t^2. T then must equal 10*2^(-2) seconds or about 14.14 seconds.Using this time we can also calculate how far horizontally the bomb traveled by taking 720 kilometers per hour, multiplying it by a 1000 to convert it to meters per hour and dividing it by 3600 to convert it to meters per second. This yields the value of 200 m/s which when multiplied with the time we found earlier results in a horizontal distance of around 2826.427 meters (if one multiplies by the exact value of time found earlier). So in summary the bomb fell 980m in around 14.14 seconds and traveled horizontally at a velocity of 200 m/s to a total horizontal distance of 2828.427m.Edit: This assumes no air resistance which is not entirely realistic but in the case of a bomb that would likely amount to at most a 2 second drop increase and maybe a reduction in horizontal distance of at the very greatest around 2000m without taking wind speeds into account. Wind speeds might reduce the distance by a maximum value of an additional 200 to 300m. Keep in mind this is a fairly critical evaluation of the factors at play and is a bit of an overestimate of what the actual values might be.Hope this was helpful! Edit 2: By the way, if you are dropping a fission or fission/part fusion weapon you would want to be much higher than that and also turning away your plane (if you are using a plane) so as to escape before it detonates, besides the fact that most fission bombs are designed to be slowly dropped by a parachute and detonated in air to allow for the greatest destructive potential and the safe escape of the bomber and crew.
Calculus word problem?
A rocket rising vertically, is tracked by a radar station that is on the ground 5 miles from the launchpad. How fast is the rocket rising when it is 4 miles high and its distance from the radar station is increasing at a rate of 2000 mi/hr?
Can someone help me solve this word problem for my Calculus class?
r² = a² + d² r = slant range to the rocket from the radar station = √(4² + 5²) = √(16+25) = √41 a = altitude of the rocket above the launch pad = 4 miles d = distance from the radar station to the launch pad = constant = 5 miles 2rdr/dt = 2ada/dt da/dt = (r/a)dr/dt = (√41)*2000/4 = 500√41 mph ≈ 3202 mph ...
Calculus problem #2. Need help a.s.a.p?
The diagonal distance of the rocket from the radar station we shall call D, and the vertical distance of the rocket from the ground we shall call H. The horizontal distance of the station from the rocket is 5 miles. Then, by the Pythagorean Theorem: D² = (5 mi)² + H². So, when H = 4 mi, then D = √(25 mi² + 16 mi²) = √41 mi. Differentiating D² = (5 mi)² + H² with respect to time, we get this: 2D (dD/dt) = 0 + 2H (dH/dt), where dD/dt is the rate at which the diagonal distance of the rocket from the station is increasing, and dH/dt is the vertical velocity the rocket is rising. Solving for dH/dt symbolically, we get this: 2D(dD/dt)/2H = (dH/dt) (D/H)(dD/dt) = (dH/dt). Plugging in our calculated value for D and our given value for H, we get this: [(√41) mi/4 mi](2000 mi/hr) = dH/dt (√41) ∙ (500 mi/hr) = dH/dt 3201.56 mi/hr ≈ dH/dt. The rocket is rising about 3201.56 mi/hr when it is 4 miles high.
Calculus help! !!?
draw a right triangle. h = height z = distance between the rocket and the radar 5^2 + h^2 = z^2 2h dh/dt = 2z dz/dt h dh/dt = z dz/dt they gave you dz/dt = 2000 mi/hr and h = 4 mi 4 dh/dt = z (2000) dh/dt = 500z so in order to find dh/dt, you need know z recall that h^2 + 5^2 = z^2 and h = 4 4^2 + 5^2 = z^2 z = root(41) so dh/dt = 500 root(41) mi/hr Yep! you did right! good job!