Why can't helicopters just hover and let the earth rotate beneath it?
Ok, I didn’t read 71 answers. Just 10. So maybe I’m repeating something that has already been said. However:All answers I read were wrong. They are all stating that a helicopter is, in some absolute way, stationary when it hovers over a spot on the ground, and that no forces act upon it.Actually, it is not stationary relative to anything except the Earth’s surface at that specific latitude. Relative to, say, the center of the earth, it’s traveling in excess of 1 600 km/h around it, if it’s at the equator. And yes, a lot of force is at work.And the question asks why, as I understand it.Here’s why: The atmosphere is part of Earth as such, it just happens to be a bit less dense than all the other stuff. It whirls about a lot (just like the molten stuff under the Earth’s crust, but with less effort), due to the Coriolis effect, so there is wind. But let’s leave wind out of the equation for now.So, the force at work (when there is no wind) is the rotating atmosphere (which, again, is part of the rotating Earth). The helicopter could indeed stay stationary relative to something except the surface, such as the center of the earth, but it would have to be capable of an airspeed of around 1 600 km/h.So unless you consider - like someone did in another question - the F-35B a glorified helicopter, it doesn’t “hover” in the sense meant in the question (an F-35B could, though, flying due west).(unless the helicopter hovers at the true north or south pole, in which case both meanings of the word “hover” coincide)It is, however, not possible for a helicopter (not even an F-35 or even a Falcon Heavy, also a VTOL vehicle like a helicopter) to hover relative to the sun. Much less the Universe. But that’s another story.There are no absolutes.
If an object whose mass is 20 kg is dropped from height of 10m, what is the maximum kinetic energy it will attain?
Although Kinetic Energy = 1/2 * m * v^2, you don't have to calculate the velocity directly to solve this problem. The potential energy of an item is: m*g*h m is mass in kg. g is the acceleration due to gravity in m/s^2 (e.g. 9.81) h is the height in meters This yields 10*9.81*20 = 1962 kg*m^2/sec^2 (aka N*m or joules)immediately before impact, 100 % of the potential energy has been converted to kinetic energy. 1962 Joules is my final answer.
A bear falls into a well. The depth of the well is 19.617 meters, and it takes the bear 2.0000 second to reach the bottom of the well. What is the color of the bear?
For the given valuesHeight=19.617m and time=2.0000 sec.s=ut+1/2gt2, with u=0m/sg=9.8085 Also, g= (GMe)/r2Putting the values of G, Me and above calculated gSo, r=6375.53126 Km.i.e. a place approx 4531.26 m. high above sea level.Some of which are: So, the candidate being China, Bolivia and Peru.The Andesextend from north to south through seven South American countries: Venezuela, Colombia, Ecuador, Peru, Bolivia, Chile, and Argentina.The Andes is the longest continental mountain range in the world with an average height of about 4,000 m. So, the Andean Bear can be a good choice over Chinese one.And hence, the color of the bear is Black. Thespectacled bear also known as the Andean bear or Andean short-faced bear are mostly restricted to certain areas of northern and western South America. They can range in western Venezuela, Colombia, Ecuador, Peru, western Bolivia, and northwestern Argentina. The species is found almost entirely in the Andes Mountains. **Note: · This validates till no one brought a bear of his own and pushed the beast into the well.· In such case the bear would be of any color.References: Wikipedia.
A plane flying horizontally at 100m/s at a height 1000m releases a bomb from it. Then what is the angle with which the bomb hits the target in ground with horizontal direction is in degrees?
Okay, lets make a few assumptions!Lets say the object is level with the ground( “object” is a picture drawn on the ground). There is no resistive forces in the x-axis or y-axis. The bomb is droped at the angle of 0° with the x axis.Ok first thing that comes to mind is that this is a projectile problem, so we need to find some of the unknown stuff like the time and range. Lets think about it the bomb falling vertically is the time we need, understand that the velocity in the x direction does not play a role in how long the bomb takes to fall, this means that we are able to use newtons 1-d laws of motion, specifically [delta y = Vi +1/2 gt^2] this will give you the time!For the range we must come to realise that the bomb and the plain will move at the same velocity hence the same distance, until the bomb strikes the ground( another assumption I need to make for my statement to be entirely correct both the center of mass of the plane and the bomb have to be aligned) also bear in mind none of them are accelerating(constant speed).So all you do is multiply the speed by the time to get distance, or you can do the whole [ delta x= ((Vi+Vf)/2 )t] they both give you the same answerOnce you have your range and time you can easily solve by using your 2-d law of motion equation for angle [ tanθ = (Vi sinθi - gt)/(Vi cosθi) ] θi is the angle it was initially dropped by, we assumed it to be 0.Solving the left hand side you can then just take the arctan of the left hand side, the answer you should get is 54,47°We used up as positive and down as negative!Here is a extra piece of information there is a relationship between the speeds in the x and y direction, and that the relationship is the angle, at time 0 seconds the angle was 0°, because the speed of the y direction was 0 m/s and the speed in the x-direction was 100 m/s ( this is because the y direction had no speed) when the speeds were equal the angle was 45° check and see that the Vf in the y direction is bigger than the Vf in the x direction when the angle is bigger than 45°. The angle can never be 90°, because the bomb has a velocity in the x direction, this makes sense mathematically because tan aproaches 90° but never gets to 90°.Hope this helps Good luck!
How much power is required to lift a 10 kg weight by a DC motor?
It depends upon speed of lifting.Equation of the power is:P = f*v (P is power in Watt, f is force in Newton and v is velocity in meters /sec)Force required for lifting 10 kg weight is (10*9.81=) 98.1 Newton. If you assume lifting speed to be 0.1 meter / sec, then power required is (98.1 * 0.1 =) 9.81 Watt. This much power is required without considering losses in lifting mechanism.Generally rated motor speed is 1500 rpm.Consider a simple lifting arrangement consisting of a rope and pulley. Rope tied to the weight is wound on the pulley which is rotated by the motor. Even if you use a very small pulley of say 25 mm dia directly mounted on motor shaft, it will result in lifting speed of (25*3.14*1500) mm/minute = 117750 mm / minute = 1963 mm / sec = 1.963 meter / sec, which is too high speed. Hence you should use a reduction gear, to match your speed requirement.If you use a speed reduction ratio of 19.63:1, you will get lifting speed of 0.1 meter / sec and power requirement (ignoring losses) is 9.81 Watt. Considering losses in the speed reduction gear and pulley, efficiency may be 50% and motor output power has to be 9.81 / (50/100) = 19.62 Watts.If you use seed reduction of 39.26, speed of lifting as well as power requirement will be reduced to half of above example.Hope, I have answered the question. If you need any more information related to this answer, please comment.
A car accelerates uniformly from 10m/s to 20m/s in 5 secs. How far does it travel in this time?
Uniformly suggests uniform acceleration. That acceleration is change of speed over time. (20–10)/5 = 2m/s/sDistance =initial velocity * time + (acceleration * time^2) / 2Distance = 10 *5 + (2*5^2)/2 = 75mThere are other easier (“geometrical approaches”)Simply use mean speed theorem. As the acceleation is uniform, distance travelled increases identically to its increase at average mean speed. ThereforeDistance = time * average speed = time * (initial speed + final speed) / 2Distance = 5 * (10 + 20)/2 = 5 * 15 = 75mMean speed theorem - Wikipedia