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A log-log graph of the area of glaciers and their volume is shown in the figure below. The linearized equation is shown on top of the graph. One glacier has a volume V and area A. If another glacier has an area of 0.538A, what is the volume of this other glacier? log10 V = -1.52 + 1.36log10 A 1000 100 10 0.1 0.01 0.001 0.01 0.1 10 100 1000 10000 log10 A (km²) O 1.86 O 0.430 0.538 O 2.32 log10 V (km³) 1,

A log-log graph of the area of glaciers and their volume is shown in the figure below. The linearized equation is shown on top of the graph. One glacier has a volume V and area A. If another glacier has an area of 0.538A, what is the volume of this other glacier? log10 V = -1.52 + 1.36log10 A 1000 100 10 0.1 0.01 0.001 0.01 0.1 10 100 1000 10000 log10 A (km²) O 1.86 O 0.430 0.538 O 2.32 log10 V (km³) 1,

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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Physics 114 Equation Sheet Constants and Conversions Kinematics Continued g = 9.80 m/s Free-fall acceleration Δν Instantaneous ainst. = lim At-o At Acceleration 1N = 1 kg m/s? Newton Uniform motion (v) = (v); = constant Position in uniform X = x + (v)At Mathematics, Scaling and Vectors b = a* + loga (b) = x motion Logarithm Constant (v); = (v,); + azAt acceleration: 1 log(ab) = log(a) + log (b) x, = x, + (v,),At +a, (at)? log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere V = Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = mã %3D Volume of a cylinder V = arl Newton's second law Fnetx = EF = ma, %3D Surface area of a A = 2ar? + 2rl Component form Fnety = ER, = may cylinder Mass density p = m/V Newton's Third Law FA en =- ton A A, = A cos e (rel. to x-axis) Weight w = mg x -component of a vector Å Apparent weight Wapp = magnitude of supporting forces y -component of a Ay = A sin 8 (rel to x-axis) vector Å Kinetic friction fk = Han Magnitude of vector Ả Static friction A = JA + A, Reynolds number Re = pvl/n Direction of A relative 8 = tan-(Ay/A,) 1 Drag (high Reynolds number) =CopAv? to x-axis Addition of two vectors If = Å + B, then C, = A, + B, D = 6nyrv Drag (low Reynolds number) Cy = Ay + By Circular Motion Kinematics Centripetal acceleration a = Displacement Ax = x - X Average Velocity Ax Frequency 1 Varg T 2ar At Ax Vinst. = lim Instantaneous Velocity At+0 At Av Average Acceleration davg Δε
Transcribed Image Text:Physics 114 Equation Sheet Constants and Conversions Kinematics Continued g = 9.80 m/s Free-fall acceleration Δν Instantaneous ainst. = lim At-o At Acceleration 1N = 1 kg m/s? Newton Uniform motion (v) = (v); = constant Position in uniform X = x + (v)At Mathematics, Scaling and Vectors b = a* + loga (b) = x motion Logarithm Constant (v); = (v,); + azAt acceleration: 1 log(ab) = log(a) + log (b) x, = x, + (v,),At +a, (at)? log Ax" = n log x + log A (v,); = (v,)} + 2a,Ax Volume of a sphere V = Surface area of a A = 4ar? Forces sphere Newton's second law Fnet = EF = mã %3D Volume of a cylinder V = arl Newton's second law Fnetx = EF = ma, %3D Surface area of a A = 2ar? + 2rl Component form Fnety = ER, = may cylinder Mass density p = m/V Newton's Third Law FA en =- ton A A, = A cos e (rel. to x-axis) Weight w = mg x -component of a vector Å Apparent weight Wapp = magnitude of supporting forces y -component of a Ay = A sin 8 (rel to x-axis) vector Å Kinetic friction fk = Han Magnitude of vector Ả Static friction A = JA + A, Reynolds number Re = pvl/n Direction of A relative 8 = tan-(Ay/A,) 1 Drag (high Reynolds number) =CopAv? to x-axis Addition of two vectors If = Å + B, then C, = A, + B, D = 6nyrv Drag (low Reynolds number) Cy = Ay + By Circular Motion Kinematics Centripetal acceleration a = Displacement Ax = x - X Average Velocity Ax Frequency 1 Varg T 2ar At Ax Vinst. = lim Instantaneous Velocity At+0 At Av Average Acceleration davg Δε
A log-log graph of the area of glaciers and their volume is shown in the figure below. The linearized equation is shown on top of the graph. One glacier has a volume V and area A. If another glacier has an area of 0.538A, what is the volume of this other glacier? log10 V = -1.52 + 1.36log10 A 1000 100 10 0.1 0.01 0.001 0.01 0.1 10 100 1000 10000 log10 A (km²) O 1.86 O 0.430 0.538 O 2.32 O 0.289 log10 V (km³) 1,
Transcribed Image Text:A log-log graph of the area of glaciers and their volume is shown in the figure below. The linearized equation is shown on top of the graph. One glacier has a volume V and area A. If another glacier has an area of 0.538A, what is the volume of this other glacier? log10 V = -1.52 + 1.36log10 A 1000 100 10 0.1 0.01 0.001 0.01 0.1 10 100 1000 10000 log10 A (km²) O 1.86 O 0.430 0.538 O 2.32 O 0.289 log10 V (km³) 1,
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