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a) A disk-shaped grindstone of mass 3.0 kg and radius 8.0 cm is spinning at 600 rev/min. After the power is shut off, a man continues to sharpen his axe by holding it against the grindstone until it stops 10 s later. What is the average torque exerted by the axe on the grindstone? b) A rotating wheel is accelerating at 0.21 rad/s2 to a stop. What should be the initial angular speed if the wheel is to stop after exactly one revolution? c) A disk of mass M is rotating about a fixed axis with angular velocity ω. It slows down to stop with an angular displacement of θ rad. What is the angular acceleration of the disk? Answer is in terms of M, ω, and θ. d) To increase the moment of inertia of a body about an axis, you must a. increase the angular acceleration. b. increase the angular velocity. c. decrease the angular velocity. d. make the body occupy less space. e. place part of the body farther from the axis. Pls. show all necessary calculations...
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I need to produce a paragraph for my physics homework. The questions are 1) the reflecting telescope- what does it do and how does it use reflection 2) spearfishing- how does refraction make it difficult to fish using a spear 3) receiving tv signals when you live in a valley- how does diffraction allow this to happen So far, I've got these two answers Reflecting telescopes use curved mirrors instead of convex lenses to collect and focus light. A large concave mirror (the centre is thinner than the edges) collects and reflects the light to make an image. Once the image forms, the lens in the eyepiece magnifies the image. Reflecting telescopes are very helpful for viewing dim or dark objects. Large reflecting telescopes can see objects that are a millionth or a billionth the brightness of the faintest star that can be seen by the human eye alone. Light refracts as it travels from the fish in the water to the eyes of the hunter. The refraction occurs at the water-air boundary. Due to this bending of the path of light, the fish appears to be at a location where it isn't. A visual distortion occurs. Subsequently, the hunter launches the spear at the location where the fish is thought to be and misses the fish. If the hunter with the spear‘s sight is more perpendicular to the water, the amount of refraction decreases. The most successful hunters are those whose sight is perpendicular to the water. Refraction of light occurs when it crosses the boundary, visual distortions often occur. These distortions occur when light changes medium as it travels from the object to our eyes. Any better answers? And, I need help with the third question
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1. Fhoriz = 10N * cosΘ Fnormal = mg - Fhoriz * sinΘ, and therefore Ffriction = µFn = 0.345(7.7N - 10N*sinΘ) = 2.66N - 3.45N*sinΘ horizontal net F = 10cosΘ - 2.66N - 3.45sinΘ accel is maximized where net F is maximized, which is where its derivative w/r/t Θ = 0. dFnet/dΘ = 0 = -10sinΘ - 0 - 3.45cosΘ sinΘ = -0.345cosΘ tanΘ = -0.345 Θ = -19º seems plausible 2. "f" does not seem true to me. The rest do. 3. a) same on both Earth and Moon Gravity does not factor into the equation of motion for an oscillating spring.
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The way a human body reacts to hitting water is complex, but some appreciation can be found by running some numbers with appropriate approximations.Checking on youtube shows the distance a diver goes beneath the water is around twice their body length.Taking some ball-park approximate values:Human height, [math]h=1.75m[/math]Human mass, [math]m=85kg[/math]Fall height [math]s=27m[/math]Accekeration due to gravity [math]a=g=9.81ms^{-2}[/math][math]v^2=u^2+2as[/math]The speed on hitting the water is[math]v=\sqrt {u^2+2as}[/math][math]v=\sqrt {0+2\times 9.8ms^{-2}\times27m} = 23ms^{-1}[/math]That’s around 50mph.Assuming a constant deceleration under the water and rearranging the same equation:[math]a=\frac {v^2-u^2}{2\times2h}=\frac {0-(23ms^{-1})^{2}}{2\times (2\times1.75m)}=-75ms^{-2}[/math][math]F=ma=85kg\times75ms^{-2}=6375N[/math]Rearranging [math]v=u+at [/math] gives[math]t=\frac {v-u}{a}=\frac {23ms^{-1}}{75ms^{-2}}=0.25s[/math]So that would be a force of 6375N acting over 0.25s to slow down the diver.That’s a big force acting over a not insignificant time which could easily cause head injury. I guess people would rather break their legs than break their necks.
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Mounted on a low-mass rod of length 0.52 m are four balls. Two balls (shown in red on the diagram), each of mass 0.59 kg, are mounted at opposite ends of the rod. Two other balls, each of mass 0.29 kg (shown in blue on the diagram), are each mounted a distance 0.13 m from the center of the rod. The rod rotates on an axle through the center of the rod (indicated by the "x" in the diagram), perpendicular to the rod, and it takes 1.4 seconds to make one full rotation. (a) What is the moment of inertia of the device about its center? I = kg·m2 (b) What is the angular speed of the rotating device? = radians/s (c) What is the magnitude of the angular momentum of the rotating device? abs(L^^->) = kg·m2/s Here are the picture http://www.webassign.net/userimages/stick_4balls.gif?db=v4net&id=32722
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When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.30 cm -tall object is 55.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm . I1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses
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Weight of object = 0.249 * 9.8 = 2.4402 N Since the whole block is under water, its volume is equal to the volume of displaced water. Buoyant force = 1000 * 9.8 * V = 9800 * V The weight of the block is equal to the buoyant force minus the weight of the object. Density = Mass of wood ÷ volume of block
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(a) Flux = [math]N*B*A*\cos\theta[/math]Area [math]A = 12cm^2 = 12*10^{-4}m^2[/math]Since Area vector is parallel to the magnetic fieldTherefore, [math]\cos{0} = 1[/math]=> [math]200*12*10^{-4}[/math]=> [math]0.24 Wb[/math](b) Flux = [math]N*B*A*\cos\theta[/math]Area vector becomes perpendicular to the Magnetic FieldTherefore, [math]\cos{90} = 0[/math]Hence, Flux [math]= 0[/math].(c) EMF = Change in Flux/Time=> [math]\frac{0.24 T - 0T }{0.04}[/math]=> [math]6 V[/math]Cheers!
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a. Assuming the motion is vertical the forces acting on the block are: its weight acting vertically down ... so that's 800 N and the opposing force acting vertically up ... so that's 900 N b. Net force = 900 - 800 = 100 N vertically up c. If the force exerted by the cord on the block was only 700 N then the net force on the block would be 800 - 700 = 100 N vertically down so the resultant force is downwards and the block would continue to fall (and accelerate downwards) and if the cord is stretched beyond its limit it'll probably snap ... not good if there's a person on the end of it particularly since the mass of a person with a weight of 800 N is only about 80 kg ... maybe it's just for kids
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(1) Let x be the distance from moon's center to the spacecraft. The earth's mass is about 81 moon-masses, so you have 81/(240000 mi - x)^2 = 1/x^2 => 81x^2 = 240000^2 - 480000x + x^2 => 80 x^2 + 480000x - 240000^2 = 0 => x^2 + 6000x - 3000*240000 = 0 => x = -3000 + (1/2)*sqrt(6000^2 + 12000*240000) = -3000 + (1/2)*54000 = 24000. About 24000 miles from the moon's center and 216000 miles from the earth's center. (2) v^2 = GM/r where M is the mass of the earth and v the speed of the satellite. Radius of the orbit (relative to earth's center) is 2.0 x 10^7 m + 6.37 x 10^6 m v = sqrt((6.67 x 10^(-11) m^3/(kg s^2))*5.97 x 10^24 kg/(2.64 x 10^7 m)) = 3884 m/s. The circumference of the orbit is 2*pi*2.64 x 10^7 m = 1.659 x 10^8 m. Dividing this by 3884 m/s you get the period, which is 42700 s = 11.9 hours, which seems to be close to the "Google" value for GPS period.