The diagonal of a unit square has length sqrt(2), but taking a large number of tiny horizontal/vertical steps along the diagonal yields a path of length 2. How can we reconcile this contradiction?
It never goes to sqrt(2).It can appear to you that it is getting closer to being a diagonal but it isn't.Imagine that you start with a number of steps and then you double them. At the same time you zoom in into the image and you notice that nothing changed. You still have steps horizontal/vertical that still form a 45º angle with the diagonal, so each 2 steps that form a triangle with the diagonal create a path that is always 2/sqrt(2) bigger than the corresponding piece of diagonal. It doesn't matter how many times you double the number of steps, you will always see the same.The error you're making when visualizing the increase in steps is that you keep looking at them from a "far" perspective which makes it seem like they're turning into a line, but it clearly fades away when you imagine yourself zooming in.
Walking diagonally please help?
I've always been known for walking diagonally. I don't walk straight unless I'm directly focusing on doing so. For a couple years after I was born I walk/ran with my feet pointing in towards each other. People are always complaining that I walk into them but I really can't help. Please explain what this could be and/or why this is happening. Don't just say to go to a doctor, that is not an option that I have. Thanks! I'm a 15 year old girl.
How much shorter is to walk diagonally across a rectangular field 80 m long and 60 m wide than along two of its adjacent sides?
The rectangular field is 80 b x 60 m.The diagonal = [80^2+60^2]^0.5 = 100 m.It is 40 m shorter walking along the diagonal of the field than walking along the length and breadth of the field, to reach the opposite corner.
There is a square with each side 1 km long. A person walks with a speed of 60 km/hr on the first side, 30 km/hr on the second side, and 60 km/hr on the third side. With what speed should he travel on the fourth side of the square to average out to a speed of 60 km/hr?
Average speed to that point: 45 kph over 4 minutes. To reach 60 kph, we must travel at N kph (where N > 60) for 60/(N-60) minutes over a total distance of N/(N-60) km. Critically, this is possible because speed is calculated without reference to direction. (For that matter, if this was a question of velocity the problem would be impossible no matter the speed of travel, since the net velocity would be 0.) So, simply walk forwards and backwards along the fourth side so that we arrive at the far corner having walked the desired time at the desired speed.For example: walking at 120 kph, I walk halfway along the fourth side, turn around and walk back to the corner, turn again and walk all the way across. I have walked across the fourth side, taking 1 minute, going 120 kph, so my average speed is (45*4 + 120*1)/(4+1) = 60 kph.A trick question deserves a trick answer.
I have the four sides of an irregular quadrilateral and none of its angles. How can I calculate its area?
Brahmagupta (c.598 - 665) was an Indian mathematician who gave his formula for finding the area of a quadrilateral, which does not require the angle measurements. Only the 4 lengths are sufficient to find the area. We should give him due credit!Brahmagupta's FormulaBrahmagupta's formula finds the area of a cyclic quadrilateral. The formula for the area of a cyclic quadrilateral with sides a, b, c, d is given bywhere s = [a+b+c+d]/2and a, b, c and d are the four sides.Example 1: Let ABCD be a kite. DA = AB = 3, DC = CB = 5. Let us find the area of the kite.s = (3+3+4+4)/2 = 14/2 = 7Area = [(7–3)(7–4)(7–4)(7–3)]^0.5 = (4x3x4x3)^0.5 = 12.As you can see DAC is a RAT as also ABC, with AC as the hypotenuse (5). Therefore area of DAC and ABC = 3*4/2 = 6 each, and so area of ABCD = area DAC + area ABC = 6+6 = 12, same as what we got from Brahmagupta’s formula.Example 2: Let PQRS be a kite. SP = PQ = 5, QR = RS = 12. Let us find the area of the kite.s = (5+5+12+12)/2 = 34/2 = 17Area = [(17–5)(17–12)(17–12)(17–5)]^0.5 = (12x5x5x12)^0.5 = 60.As you can see SPR is a RAT as also PQR, with PR as the hypotenuse (13). Therefore area of SPR and PQR = 5*12/2 = 30 each, and so area of PQRS =area SPR + area PQR = 30 + 30 = 2*30 = 60, same as what we got from Brahmagupta’s formula.
Homework Question: Joe walked 4 miles north, 9 miles east, 8 miles north, and then 7 miles east. If Joe now decides to walk straight back to where he started, how far must he walk?
The answer provided by User-9597828242429937573 is great, but there are couple of omissions that make its precision unacceptable for practical purposes:1. Our planet is not spherical: its height is less than width due to its rotation. We should take it into account;2. Joe is not a point, but rather a material object. We have to make calculations probably for his geometrical center or, maybe, his center of mass. It should be discussed and decided. If agreed upon geometrical center, 1/2 of Joe's height should be added to the radius;3. Since the journey is long, the result depends on the time he starts and his speed - because of thermal expansion of the Earth's surface heated by the Sun. If Joe is walking while the surface is expanding, his trip back may be longer if the temperature is still high;4. Of course, we should consider the Earth deformation due to the Moon's gravity;5. The answer has to be expressed in closest integer number of Plank lengths. Otherwise, it would be absolutely unrealistic.Other than that, it is pretty accurate and will probably satisfy Joe. :)
Why do some people walk diagonally?
My right leg is just a bit longer than my left. No one would ever know. It gives me a "wiggle to my walk" I have been complimented on. Just as often though, my husband, family and friends will just steer me straight. If we are walking with say, ice cream cones, hot coffee, etc. No one wants to walk anywhere near me though. My poor big dog has no idea how to walk a straight line either since I am the only one that has ever walked him. We are quite a pair when we are out walking as we are tripping all over each other. LOL
If a hole is made right through Earth so it reaches the other side, what will happen when one jumps into it?
Earth’s center of mass is pretty much at its core. It is therefore clear that the attractive force exerted by the Earth towards all bodies is directed towards its center.Why is this important? Say the particular hole looks like thisSay the center is at B (it’s not exactly B, but slightly below B). When the person is in the first leg of the journey, i.e. not yet reached the center, he is falling to the center, or basically towards the point of attraction. So, his speed naturally increases.Once he passes B, though, he is now falling away from the center, which is kind of equivalent to jumping up. The force he now experiences is against the direction of his motion, so he now decelerates. Once he completely stops, he falls back towards to the center and the entire cycle repeats.If you’ve noticed, this is basically an SHM (simple harmonic motion). He will keep oscillating about the center with a time period of about 43 minutes (if I’m not wrong).This, however, is assuming that you initially simply dropped into the hole or had low start speed. If you were actively launched into the hole at a high speed, it is possible you may not decelerate fast enough to stop within the hole. You may fly right out the other side. If the speed is high enough, you could even reach space.
A rectangular ground is 40 m long and 30 m wide. A 3 m wide path, parallel to its length and width, is in the middle of the ground. What is the area of the path?
The rectangular ground is 40 m x 30 m. A 3 m wide path is parallel to the length and the width of the ground. What is the area of the path.The area of the path = [40+30]*3 - 3*3 = 70*3–9 = 210–9 = 201 sq m.