The 16 kg mass has a weight of M*g = 156.8 N. Presumably, the cable that lifts the weight is wound around the axle and the cable that one pulls on is wrapped around the wheel outer rim. Input torque = 26 N*(2.3/2) m = 29.9 N-m Output torque = 156.8 N * (0.092/2 m) = 7.213 N-m ...
Yes, seems like I did.
The height of the cylindrical column of hot chocolate is H = 12.6 - 2.5 = 10.1 cm. The volume of hot chocolate is V = pi*(d^2/4)*H where d = 9.2 cm is the diameter. V = 67.1 cm^3 That is less than 3 fluid ounces.
I do not see a question here. Whether or not the bar fails will depend upon where the axial load is placed. This may represebnt your loading situation: http://www.engineersedge.com/beam_bending/beam_bending19.htm
I have no idea what the question is. We do not share your text or test material.
Are you asking for the value of the principal after 4.25 years? How often is interest compounded? Is 29,200 the initial principal? If compounding is quarterly and the initial principal is 29,200, the final principal is 29,200(1.015)^17 = 37,610.19 You need to specify currency ...
Let the image distance be Di 1/45 + 1/Di = 1/15 1/Di = 2/45 Di = 22.5 cm You do not need to know that the lens is double convex. All you need is the focal length.
Let N be the number of rotations. Each rotation advances the tire and the unicycle 2*pi*R The distance traveled and the radius must use the same dimensions when using the formula. N*2*pi*R = 1000 (meters) R = 1.85 meters N = (1000/1.85)*[1/(2pi)] = 86
If the length L is cut in half, the cross sectional area A must double. Since the resistance is proportional to L/A, it must decrease by a factor of 4. That makes r = 4 ohms.
The fundamental (first harmonic) wavelength is 17.0 meters. Second harmonic: 8.5 m Third harmonic: 5.67 m Sixth harmonic: 2.83 m Seventh harmonic: 2.429 m Frequency*wavelength = constant = 1923 m/s Seventh harmonic frequency = 792 Hz
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