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Problem 2: A cylindrical cup has a mass M = 13 g and a height h = 19 cm. The inside of the cup has a radius of R = 3.1 cm. The center of mass of the empty cup is located in the center of the cup at a height of h/2. If the cup is completely full of liquid, it is trivial to show that the center of mass is still located at a height of h/2. As liquid is consumed from the cup, the center of mass moves downward because the liquid occupies the bottom part of the cup. As the cup nears empty, the center of mass again rises to height h/2. h Vem Part (a) Enter an expression for the height of the center of mass of the cup/liquid system as a function of the height of the liquid, x, in terms of the defined quantities and the density of the liquid, Q. MultipleChoice : 1) (Mh+TR? x2 e)/(M+TR2X Q) 2) 0.5 ( Mh + T R? x q )/(M+R? x Q ) 3) (Mh + 2 n R? x² e )/ (M+ R? x Q ) 4) (2 Mh+TR2 x² o ) / (M + TR²×0) 5) 0.5 (Mh+ TR2 x2 e ) / ( M + R? x Q ) Part (b) Calculate the height, in centimeters, of the center of mass of the system when the height of the liquid is x = 5.2 cm. Take o = 1.2 g/cm. Numeric : Anumeric value is expected and not an expression. Yem = 3.045 Part (c) The lower the center of mass, the more stable the system will be. Enter an expression for the height of liquid that minimizes the height of the system's center of mass, in terms of the other defined quantities. MultipleChoice : 1) (- (1+xh R2 e /M)^0.5 - 1) M/(TR2 g) 2) ((1+1x R? o/ M)^0.5 - 1) M/(AR? q) 3) (- (1+th R2e/M)^0.5 - 1) M/ 2 4) (-(1+ x R2 o /M)^0.5 - 1) M/(t R2 e) 5)((14Th R2 0/M)^0.5 - 1M/2 6) ((1+th R2 e/M)^0.5 1) M/(TR2 Q) Part (d) Calculate the height of liquid, in centimeters, that minimizes the height of the system's center of mass. Numeric : A numeric value is expected and not an expression. Xmin =

Problem 2: A cylindrical cup has a mass M = 13 g and a height h = 19 cm. The inside of the cup has a radius of R = 3.1 cm. The center of mass of the empty cup is located in the center of the cup at a height of h/2. If the cup is completely full of liquid, it is trivial to show that the center of mass is still located at a height of h/2. As liquid is consumed from the cup, the center of mass moves downward because the liquid occupies the bottom part of the cup. As the cup nears empty, the center of mass again rises to height h/2. h Vem Part (a) Enter an expression for the height of the center of mass of the cup/liquid system as a function of the height of the liquid, x, in terms of the defined quantities and the density of the liquid, Q. MultipleChoice : 1) (Mh+TR? x2 e)/(M+TR2X Q) 2) 0.5 ( Mh + T R? x q )/(M+R? x Q ) 3) (Mh + 2 n R? x² e )/ (M+ R? x Q ) 4) (2 Mh+TR2 x² o ) / (M + TR²×0) 5) 0.5 (Mh+ TR2 x2 e ) / ( M + R? x Q ) Part (b) Calculate the height, in centimeters, of the center of mass of the system when the height of the liquid is x = 5.2 cm. Take o = 1.2 g/cm. Numeric : Anumeric value is expected and not an expression. Yem = 3.045 Part (c) The lower the center of mass, the more stable the system will be. Enter an expression for the height of liquid that minimizes the height of the system's center of mass, in terms of the other defined quantities. MultipleChoice : 1) (- (1+xh R2 e /M)^0.5 - 1) M/(TR2 g) 2) ((1+1x R? o/ M)^0.5 - 1) M/(AR? q) 3) (- (1+th R2e/M)^0.5 - 1) M/ 2 4) (-(1+ x R2 o /M)^0.5 - 1) M/(t R2 e) 5)((14Th R2 0/M)^0.5 - 1M/2 6) ((1+th R2 e/M)^0.5 1) M/(TR2 Q) Part (d) Calculate the height of liquid, in centimeters, that minimizes the height of the system's center of mass. Numeric : A numeric value is expected and not an expression. Xmin =

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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7.2 please answer part d, previous parts are answered. In order to answer part d, you need to use the expression from part c.

Problem 2: A cylindrical cup has a mass M = 13 g and a height h = 19 cm. The inside of the cup has a radius of R = 3.1 cm. The center of mass of the empty cup is located in the center of the cup at a height of h/2. If the cup is completely full of liquid, it is trivial to show that the center of mass is still located at a height of h/2. As liquid is consumed from the cup, the center of mass moves downward because the liquid occupies the bottom part of the cup. As the cup nears empty, the center of mass again rises to height h/2. h cm Part (a) Enter an expression for the height of the center of mass of the cup/liquid system as a function of the height of the liquid, x, in terms of the defined quantities and the density of the liquid, Q. MultipleChoice : 1) ( Mh + aR? x? g )/ (M + TR? x g) 2) 0.5 ( Mh + 1R? x g )/(M + R? x ) 3) ( Mh + 2 tR? x²e)/(M+tR? x q ) 4) ( 2 M h + t R² x² o ) / ( M + T R² x o ) 5) 0.5 ( M h + AR² x² e ) / ( M + t R² x ) Part (b) Calculate the height, in centimeters, of the center of mass of the system when the height of the liquid is x = 5.2 cm. Take o = 1.2 g/cm³. Numeric : A numeric value is expected and not an expression. Ycm = 3.045 Part (c) The lower the center of mass, the more stable the system will be. Enter an expression for the height of liquid that minimizes the height of the system's center of mass, in terms of the other defined quantities. MultipleChoice : 1) ( - (1+ th R2 Q / M ) ^ 0.5 - 1 ) M / (1R² Q ) 2) ((1 + Tx R? Q / M) ^ 0.5 - 1 ) M / ( IR² ) 3) (- (1+ Th R2 e /M ) ^ 0.5 - 1 ) M / 2 4) (- (1 +1x R2 Q / M ) ^ 0.5 - 1 ) M / ( IR² e ) 5) ((11Th R² 0 / M ) ^ 0.5 - 1 ) M / 2 6) ((1+th R? Q/ M) ^ 0.5 - 1 ) M / (IR² Q ) Part (d) Calculate the height of liquid, in centimeters, that minimizes the height of the system's center of mass. Numeric : A numeric value is expected and not an expression. Xmin =
Transcribed Image Text:Problem 2: A cylindrical cup has a mass M = 13 g and a height h = 19 cm. The inside of the cup has a radius of R = 3.1 cm. The center of mass of the empty cup is located in the center of the cup at a height of h/2. If the cup is completely full of liquid, it is trivial to show that the center of mass is still located at a height of h/2. As liquid is consumed from the cup, the center of mass moves downward because the liquid occupies the bottom part of the cup. As the cup nears empty, the center of mass again rises to height h/2. h cm Part (a) Enter an expression for the height of the center of mass of the cup/liquid system as a function of the height of the liquid, x, in terms of the defined quantities and the density of the liquid, Q. MultipleChoice : 1) ( Mh + aR? x? g )/ (M + TR? x g) 2) 0.5 ( Mh + 1R? x g )/(M + R? x ) 3) ( Mh + 2 tR? x²e)/(M+tR? x q ) 4) ( 2 M h + t R² x² o ) / ( M + T R² x o ) 5) 0.5 ( M h + AR² x² e ) / ( M + t R² x ) Part (b) Calculate the height, in centimeters, of the center of mass of the system when the height of the liquid is x = 5.2 cm. Take o = 1.2 g/cm³. Numeric : A numeric value is expected and not an expression. Ycm = 3.045 Part (c) The lower the center of mass, the more stable the system will be. Enter an expression for the height of liquid that minimizes the height of the system's center of mass, in terms of the other defined quantities. MultipleChoice : 1) ( - (1+ th R2 Q / M ) ^ 0.5 - 1 ) M / (1R² Q ) 2) ((1 + Tx R? Q / M) ^ 0.5 - 1 ) M / ( IR² ) 3) (- (1+ Th R2 e /M ) ^ 0.5 - 1 ) M / 2 4) (- (1 +1x R2 Q / M ) ^ 0.5 - 1 ) M / ( IR² e ) 5) ((11Th R² 0 / M ) ^ 0.5 - 1 ) M / 2 6) ((1+th R? Q/ M) ^ 0.5 - 1 ) M / (IR² Q ) Part (d) Calculate the height of liquid, in centimeters, that minimizes the height of the system's center of mass. Numeric : A numeric value is expected and not an expression. Xmin =
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