A car negotiate a bend of radius 20m with an acceleration of 12m/s^2. What is the maximum speed the car can attain without skidding?
You have not given nearly enough information to answer the question.What kind of car? Many performance cars can pull close to a full g on a skidpad while some can’t pull 1/2 g.What suspension parts are on the car? Did it have the factory high performance handling option? Has it been upgraded with aftermarket parts?How well is it aligned? Is it aligned to factory specifications or it is compensated to the direction of the turn?What tires are on said car? Goodyear eagle racing tires will hold far more than cheaper tires.Is it raining?Is the pavement cold or warm?How warm is it outside and on the surface of the track?Are the tires warmed prior to the test to increase their grip or starting cold?How skilled is the driver? A skilled race car driver will do better than someone who rarely drives highway speeds.There are far too many undetermined variables to even consider giving a serious answer to the question.
How many times will a wheel of radius 14 cm have to rotate to cover a distance of 22m?
25 times.you can try it this following way.The distance that a wheel travels in one rotation is simply the circumference of the circle.Circumference of wheel= 2* pi* radius.2* (22/7)*14= 88 cm So wheel can travel 88 cm in 1 rotation. And it has to cover 22 metres (2200 cm)So 2200/88=25So wheel rotates 25 times to travel 22 meters.
A car is travelling on a circular road of radius 500m. At some instant its speed is 30m/s and is increasing at the rate of 2m/s. What is its acceleration?
The acceleration has 2 components : tangential ([math]a_t[/math]= 2 m/s[math]^2[/math]) and centripetal ([math]a_r=\tfrac{v^2}{r}=\tfrac{30^2}{500}=1.8[/math]m/s[math]^2[/math]).Net acceleration = [math]\sqrt{2^2+1.8^2}[/math]m/s[math]^2[/math] = 2.69 m/s[math]^2[/math].
What is the maximum speed, at which a car turns around a curve of radius 30m on a level road, if the coefficient of friction between tyres and road is 0.4?
To go around a curve, the car needs to have a centripetal force applied, and friction with the road supplies that. We will then use the centripetal acceleration below.Constants:[math]g = 10\ m/s^2[/math]Givens:[math]r = 30\ m[/math][math]\mu = 0.4[/math]Unknowns:[math]v = ?[/math]Other values to be used:[math]m[/math]It is useful to draw a force diagram on the car, which will have three forces:[math]F_g[/math] down[math]F_{normal}[/math] up[math]F_{friction}[/math] to the left or right.The relevant relationships are[1] [math]m\ \vec{a} = \sum \vec{F}[/math][2] [math]F_{friction} = \mu \ F_{normal}[/math][3] [math]F_{gravity} = m\ g[/math][4] [math]a_{centripetal} = \frac{v^2}{r}[/math]Eq [1] in the vertical direction gives[math]m (0) = F_{normal} - F_{gravity} \rightarrow F_{normal} = F_{gravity} = mg[/math]Eq [1] in the horizontal direction gives[math]m a_{centripetal} = F_{friction} \rightarrow m \frac{v^2}{r} = \mu F_{normal} = \mu m g[/math]So working with[math]m \frac{v^2}{r} = \mu m g[/math][math]\frac{v^2}{r} = \mu g[/math][math]v^2 = \mu g r[/math][math]v = \sqrt{\mu g r}[/math]It is worth explicitly noting that, while we added [math]m[/math] to solve the problem, it dropped out by the end.
How many feet per mile does the earth curve down from where you stand?
You can use pythagoream's therom. The sum of the squares of the sides of a right triangle is equal to the square of the hypotonuse.The radius of the earth is one side of the triangle. (~4000 miles)The distance away from where you are standing is the other side.The hypotonuse tells you how far that point would have to be from the earths center to stay horizontal from you.Then subtract 4000 from the hypotonuse, and that gives you how far up from the surface that point is (or how much the earth dropped).For example, calculate 1 mile out.4000^2 + 1^2 = 16000001Square root of 16000001 is 4000.000125So 1 mile out is 4000.000125 miles from the center of the earth.Subtract 4000, and you get a drop of .000125 miles5028 feet per mile times .000125 yeilds about half a foot drop. Now, the important bit here is it is NOT linear. The drop isnt so many feet per mile away. It changes because you are drawing a straight horizontal line, but the earth drops as a circle. So it wont be a constant linear feet dropped per mile away.At 1 miles, the drop is .6 foot.At 5 miles, the drop is 15 feet.At 10 miles, the drop is 63 feetAt 15 miles, the drop is 141 feet.The other thing to keep in mind is how far above the surface you are standing when you look out to the horizon. The above calculations are based on the idea that your eyeball is right on the surface. If you are 6 feet tall and standing when you look, the drop is the same, but you can see farther because you are higher.
How many images will be formed if we place two plane mirrors in front of each other?
When you are viewing an object placed in between two mirrors and you can actually see the image, in this case, the number of images would range from 1 to X (X being a very large number depending on your position relative to the mirror and object, and X not infinity). If you had two mirrors of infinite size and the surfaces perfectly reflecting, and your eyes which are capable of resolving images coming from infinite distance away, then you might see an infinite number of images as depicted below (just imagine the mirrors to extend infinitely):
A bicycle wheel has a diameter of 50 cm. How far does the bicycle travel when the wheel makes 30 turns using algebra?
HelloWheel Diameter = Wheel 2 x Radius; 50 cm / 2 = 25 cm.For one rotation of the wheel, the distance travelled = circumference of the wheel.Circumference = 2 x pi x radius = 2 x 3.1415926858 x 25 cm = 157.07 cm approx.Distance travelled with 30 turns of the wheel = 30 x 157.07 = 4712.39cm approx.Distance in meters = 4712.39cm / 100 = 47.12388 meters approx (1 cm = 1/100th of a meter).Hope this helps
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. :)