The nickel's image in Figure P23.35 has four times the diameter of the nickel when the lens is 2.88 cm from the nickel. Determine the focal length of the lens.
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the magnifying of the lens: M = 4 and by definition \nM = -di/do\ndo is the distance from lens to object\ndi is the distance from lens to image\nso -di/do = 4 ---> di = -4do\nthe thin lenses equation:\n1/f = 1/do + 1/di\n1/f = 1/do + 1/(-4do)\n1/f = 1/do - 1/(4do) = 3/(4do)\nf = 4do/3 = (4*2.88 cm)/3\nf = 3.84 cm
45313
change in diameter=0.10xcoef.of thermal elongation(350-20)=y\nnew dim.=0.1-(change in diam.)=z\n%age change in area=1-(z/0.1)**2
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You are measuring the distances from the mirror.\n\nThe focus is behind the mirror. Therefore take that as - 53cm.\n\nThe object is at + 25cm.\n\nWe have to find v.\n\n1/f = 1/u + 1/v. \n\n-1/53 = 1/25 + + 1/v..\n\n+ 1/v = -1/53 - 1/25 = - 17 cm.\n\n17cm to the back of the mirror.\n\nA convex mirror focus distant objects at the focus.  If the object is brought nearer and nearer, the image moves from the focus toward the mirror.\n In your problem the object is very near to the mirror.  The image must be nearer to the mirror. \n Your answer -47 cm implies it is very near to the focus (-53cm) and it cannot be the correct answer.
263607
0.40(35) + 0.60(37) = 36.2
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I think the last 2 questions may refer to breakage problems. Since objectives are different lengths, switching from a shorter to a longer one could damage the slide and any part of the lens that protrudes. I guess you'd want to be off the microscope and looking at the slide and objective from the side when you change, to ensure clearance. When you stow the microscope, I'd have the shortest lens in place and the tube cranked all the way up.
127663
Both are used to magnify
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Convert all this to MKS units and find the cross-section of that wire.\n\nE (stainless-steel) is given to be 18 x 10^10 N/m^2.\nCross-section is from pi(r^2).  This must be in m^2, not mm^2.\n3.14159(0.125*10^-3 m)^2 = 4.909 x 10^-8 m^2 cross-section.\n\nYoung's modulus (E) is given as,\n\nE = (delta F / cross-section) / (delta length / original length)\n\nDelta F is the tension in the wire, thats the first part of solving this problem.  So rearrange the equation for delta F, renaming delta F as "tension" and substitute the value given in place of E:\n\nTension = (18 x 10^10 N/m^2) * (8 x 10^-5 m / 0.028 m) * (4.909 x 10^-8 m^2)\n\nTension in the wire is therefore 25 N.\n\nThe magnitude of force on the tooth is 25 sin 30 deg or 12.5 N\n\nWarning!:  While the reasoning is sound, double-check all of my math.  Also, note that I rounded the intermediate answer 25 N.
16334
20 cm unless you are near-sighted, in which case it is closer
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The mass of a nickel is a 5.02 g. Its density is 8.90 g/mL!\n\nI hope I'm first!
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17857.142857142857142857142857143 wavelengths fit into the gap\n\ntime taken= dist/vel= 3.33* 10^(-11)
310881
1) 1 inch =25.4 mm\n2) F = 9/5 * C + 32     where F is the temperature in fahrenheit and C is the temperature in celcius.
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Your equation was partially cut off. This is what it looks like in my browser:\nInside Diameter=Outside Diameter + [2(Height-Tolerance)(1-Compres...\n\nIf this is the final part of the equation:\n...(1-Compression)]\nthen the equation, solved for Height, is:\n\nLet:\nI = Inside Diameter\n0 = Outside Diameter\nH = Height\nT = Tolerance\nC = Compression\n\nI=O+[2(H-T)(1-C)]\nI-O=2(H-T)(1-C)\nI-O=2(1-C)(H-T)\nI-O=(2-2C)(H-T)\n(I-O)/(2-2C)=H-T\n[(I-O)/(2-2C)]+T=H\n\nAnswer:\nH=T+[(I-O)/(2-2C)]
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A radius is half of a diameter. So the diameter is twice the radius. Multiply 15x by 2 and you get:\n\n30x
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It has nothing to do with the two halves of the lens anyway. The way I can explain it is spherical abberation, i.e due to the difference in the foci for the edge of the lens and the middle of the lens. But that would form a blurred image and not two images.
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(x-3)² / 25 + (y+2)² / 169 = 1\n\nie (x-3)² / 5² + (y+2)² / 13² = 1\n\nThus centre ≡ (3, -2)\n\nLength of semi-major axis (a) = 13 on vertical axis\nLength of semi-minor axis (b) = 5 on horizontal axis\n\nEccentricity (e) = √(1 - (b/a)²)\n= √(1 - (5/13)²)\n= 12/13\n\nFocal length = ea\n= 12/13 * 13\n= 12\nSo distance between the foci = 24\n\nThus focus 1 ≡(3, -2+ 12) ≡(3, 10)\nand focus 2 ≡(3, -2 - 12) ≡(3, -14)
536218
0.07896 km\n153 mm\n0.578  km
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Since the foci are along the y-axis, the major axis of the ellipse will be along the y-axis.  Also, the center of the two foci is the center of the ellipse, which will be (0,0) or the origin.\n\nSo the only thing left to figure out is the length of the minor axis.\n\nThe major axis, minor axis, and focal length (distance from the foci to the center) are related by the Pythagorean theorem:\n\na² + b² = c²\n\nYou know that a = 26/2 = 13, c = 12, so b = 5.\n\nTherefore the equation is:\n\nx²/25 + y²/169 = 1