Properties of matter Questions and Answers

4 The excess pressure inside a spherical drop of water is frour times that of another drop Then their respective mass ratio is 1 1 16 3 1 4 2 8 1 4 1 64
Physics
Properties of matter
4 The excess pressure inside a spherical drop of water is frour times that of another drop Then their respective mass ratio is 1 1 16 3 1 4 2 8 1 4 1 64
A heavy rope of mass m and length is placed on a smooth table It is pulled at both ends by applying force F of equal magnitude as shown Tension in the rope at point 1 A is 3F 4 l B A 1 4 1 2 F 2 B is F 2
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Properties of matter
A heavy rope of mass m and length is placed on a smooth table It is pulled at both ends by applying force F of equal magnitude as shown Tension in the rope at point 1 A is 3F 4 l B A 1 4 1 2 F 2 B is F 2
5 Two cylinders of same cross section and length L but made of two materials of densities d and d2 are connected together to form a cylinder of length 2L The combination floats in a liquid of density d with a length L 2 above the surface of the liquid If d d then 3 a d d 4 d b d 2 c d d 4 d d 4 d
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Properties of matter
5 Two cylinders of same cross section and length L but made of two materials of densities d and d2 are connected together to form a cylinder of length 2L The combination floats in a liquid of density d with a length L 2 above the surface of the liquid If d d then 3 a d d 4 d b d 2 c d d 4 d d 4 d
3 Enhancer 3 A charged spherical conductor having initial charge Q and radius R has a surface density of 0 7 cm 3 When the charge is increased by 0 44 C and the charge density by 0 14 Cm 3 If the same charge Q is uniformly distributed over the volume of a sphere of radius R having volume charge density p then auounit a Q 2 2 C p 4 2 Cm 3 b Q 2 2 C p 0 42 Cm 3 c Q 3 3 C p 4 2 Cm d Q 0 22 C p 1 4 Cm 3 3 beats mot 1570 2000 20 ep zib episrio aubuning
Physics
Properties of matter
3 Enhancer 3 A charged spherical conductor having initial charge Q and radius R has a surface density of 0 7 cm 3 When the charge is increased by 0 44 C and the charge density by 0 14 Cm 3 If the same charge Q is uniformly distributed over the volume of a sphere of radius R having volume charge density p then auounit a Q 2 2 C p 4 2 Cm 3 b Q 2 2 C p 0 42 Cm 3 c Q 3 3 C p 4 2 Cm d Q 0 22 C p 1 4 Cm 3 3 beats mot 1570 2000 20 ep zib episrio aubuning
The angle of contact does not depend upon 1 Temperature 3 Cohesive force 2 Soluble impurity 4 The inclination of surface in co
Physics
Properties of matter
The angle of contact does not depend upon 1 Temperature 3 Cohesive force 2 Soluble impurity 4 The inclination of surface in co
11 Figure shows the load extension curves for four wires A B C and D The dimensions of all the four wires are identical but materials of wire are different Which wire has largest value of Young s modulus of elasticity 1 A 3 C Load B A Extension 2 B 4 D
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Properties of matter
11 Figure shows the load extension curves for four wires A B C and D The dimensions of all the four wires are identical but materials of wire are different Which wire has largest value of Young s modulus of elasticity 1 A 3 C Load B A Extension 2 B 4 D
symmetrically by three wire each of 2 m long of mass 15 kg is supported These at each end are of copper and middle one is of steel Young s modulus of elasticity for copper and steel are 110 x 10 N m and 190 x 10 N m respectively If each wire is to have same tension ratio of their diameters will be 11 V19 30 11 2 F 19 11 11 30 The length of it is T and I length of the 3 1 T T 35 A wire ca four equ
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Properties of matter
symmetrically by three wire each of 2 m long of mass 15 kg is supported These at each end are of copper and middle one is of steel Young s modulus of elasticity for copper and steel are 110 x 10 N m and 190 x 10 N m respectively If each wire is to have same tension ratio of their diameters will be 11 V19 30 11 2 F 19 11 11 30 The length of it is T and I length of the 3 1 T T 35 A wire ca four equ
The water flows from a tap of diameter 1 25 cm with a of 5 x 10 5 m s The density and coefficient of visco S of water are 10 kg m 3 and 10 3 Pa s respectively flow of water is a steady with Reynolds number 5100 b turbulent with Reynolds number 5100 c steady with Reynolds number 3900 d turbulent with Reynolds number 3900
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Properties of matter
The water flows from a tap of diameter 1 25 cm with a of 5 x 10 5 m s The density and coefficient of visco S of water are 10 kg m 3 and 10 3 Pa s respectively flow of water is a steady with Reynolds number 5100 b turbulent with Reynolds number 5100 c steady with Reynolds number 3900 d turbulent with Reynolds number 3900
cm and 0 004 m formed of the same liquid surface tension 0 07 N m come together to form a double bubble Find the radius and the sense of curvature of the internal film surface common to both the bubbles Sol Let R be the radius of curvature of common surface when bubbles A and B of radii R and RB coalesce The excess pressure in A and B are 4T R and 4T Rg respectively A 4T PA RA and PB 4T R B
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Properties of matter
cm and 0 004 m formed of the same liquid surface tension 0 07 N m come together to form a double bubble Find the radius and the sense of curvature of the internal film surface common to both the bubbles Sol Let R be the radius of curvature of common surface when bubbles A and B of radii R and RB coalesce The excess pressure in A and B are 4T R and 4T Rg respectively A 4T PA RA and PB 4T R B
6 A piece of wax weighs 18 03 g in air A piece of metal is found to weigh 17 03 g in water It is tied to the wax and both together weigh 15 23 g in water Then the specific gravity of wax is 1 18 03 17 03 18 03 2 17 03 18 03 15 03
Physics
Properties of matter
6 A piece of wax weighs 18 03 g in air A piece of metal is found to weigh 17 03 g in water It is tied to the wax and both together weigh 15 23 g in water Then the specific gravity of wax is 1 18 03 17 03 18 03 2 17 03 18 03 15 03
C and Si both have same lattice structure having 4 bonding electrons in each However C is insulator where as Si is intrinsic semiconductor This is because A B C D In case of C the valence band is not completely filled at absolute zero temperature In case of C the conduction band is partly filled even at absolute zero temperature The four bonding electrons in the case of C lie in the second orbit whereas in the case of Si they lie in the third The four bonding electrons in the case of C lie in the third orhit whereas for Si they lie in the fourth orbit
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Properties of matter
C and Si both have same lattice structure having 4 bonding electrons in each However C is insulator where as Si is intrinsic semiconductor This is because A B C D In case of C the valence band is not completely filled at absolute zero temperature In case of C the conduction band is partly filled even at absolute zero temperature The four bonding electrons in the case of C lie in the second orbit whereas in the case of Si they lie in the third The four bonding electrons in the case of C lie in the third orhit whereas for Si they lie in the fourth orbit
The pressure exerted by sea water of density 1025 kg m on a fish at a depth of g 10 m s is
Physics
Properties of matter
The pressure exerted by sea water of density 1025 kg m on a fish at a depth of g 10 m s is
6K which rain renders to vertical windshield of automobile moving with 5 constant velocity of magnitude v 12 m s Consider that raindrops fall vertically with speed u 5 m s The intensity of rainfall deposits h 2 cm of sediments in time t 1 minute p 10 kg m is the density of liquid Assume collisions are inelastic The mean pressure is 9
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Properties of matter
6K which rain renders to vertical windshield of automobile moving with 5 constant velocity of magnitude v 12 m s Consider that raindrops fall vertically with speed u 5 m s The intensity of rainfall deposits h 2 cm of sediments in time t 1 minute p 10 kg m is the density of liquid Assume collisions are inelastic The mean pressure is 9
E A material has a shearing strength of 100 mp The force necessary to punch at 10mm dia hole through a plate of this material 10 mm thick is a 314 kN c 3 14 kN b 31 4 kN d None
Physics
Properties of matter
E A material has a shearing strength of 100 mp The force necessary to punch at 10mm dia hole through a plate of this material 10 mm thick is a 314 kN c 3 14 kN b 31 4 kN d None
29 A mild steel wire of length 2 meter cross sectional area A m is fixed horizontally between two pillars A small mass m kg is suspended from the mid point of the wire If extension in wire are within elastic limit Then depression at the mid point of wire will be 3 V 1 2 M IA Mg YA Mg Mg Mg1 YA 1 3 1 3 1 3 bes syy bds Mg813 qabxA M84 23 4 bd 3 G
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Properties of matter
29 A mild steel wire of length 2 meter cross sectional area A m is fixed horizontally between two pillars A small mass m kg is suspended from the mid point of the wire If extension in wire are within elastic limit Then depression at the mid point of wire will be 3 V 1 2 M IA Mg YA Mg Mg Mg1 YA 1 3 1 3 1 3 bes syy bds Mg813 qabxA M84 23 4 bd 3 G
A wire of length 2 m and diameter 1 mm stretched so that its length increases by 6 mm the decrease in diameter of the wire is o 0 35 a 1 05 x 10 8m 1 05 x 10 6 cm b 1 05 x 10 7 m d 1 05 x 10 6m c
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Properties of matter
A wire of length 2 m and diameter 1 mm stretched so that its length increases by 6 mm the decrease in diameter of the wire is o 0 35 a 1 05 x 10 8m 1 05 x 10 6 cm b 1 05 x 10 7 m d 1 05 x 10 6m c
7 A glass flask of volume 200 cm is just filled with mercury 20 C The amount of mercury that will overflow when the temperature of the system is raised to 100 C is Yglass 1 2x10 5 C Ymercury 1 8x10 4 C 3 a 2 15 cm o pham c 2 52 cm 3 b 2 69 cm d 2 25 cm
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Properties of matter
7 A glass flask of volume 200 cm is just filled with mercury 20 C The amount of mercury that will overflow when the temperature of the system is raised to 100 C is Yglass 1 2x10 5 C Ymercury 1 8x10 4 C 3 a 2 15 cm o pham c 2 52 cm 3 b 2 69 cm d 2 25 cm
B 15m penic pressure can support a water column Olympiad 2014 stage C 5m A jet of water of cross sectional area A hits a plate normally with velocity v the plate is moving i direction of the jet with velocity V therefore the force exerted on the plate is proportional to D 25m A v Olympiad 2015 stag C v V B v PART D y Y
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Properties of matter
B 15m penic pressure can support a water column Olympiad 2014 stage C 5m A jet of water of cross sectional area A hits a plate normally with velocity v the plate is moving i direction of the jet with velocity V therefore the force exerted on the plate is proportional to D 25m A v Olympiad 2015 stag C v V B v PART D y Y
9 In an unbiased n p junction electrons diffuse from n region to p region because 1 holes in p region attract them 2 3 electrons travel across the junction due to potential difference electron concentration in n region is more as compared to that in p region 4 only electrons move from n to p region and not the vice versa
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Properties of matter
9 In an unbiased n p junction electrons diffuse from n region to p region because 1 holes in p region attract them 2 3 electrons travel across the junction due to potential difference electron concentration in n region is more as compared to that in p region 4 only electrons move from n to p region and not the vice versa
The lower surface of a cube is fixed On its uppe surface force is applied at an angle of 30 from its surface The change will be in its 1 shape 2 size LAVI
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Properties of matter
The lower surface of a cube is fixed On its uppe surface force is applied at an angle of 30 from its surface The change will be in its 1 shape 2 size LAVI
108 For an ideal gas a process xy is shown on P T diagram Which of the pressure temperature following is correct for the process y Doubt X 1 Internal energy decreases 2 Work done by the gas is zero 3 Volume of the gas increases
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Properties of matter
108 For an ideal gas a process xy is shown on P T diagram Which of the pressure temperature following is correct for the process y Doubt X 1 Internal energy decreases 2 Work done by the gas is zero 3 Volume of the gas increases
2 The excess pressure inside a soap bubble is equal ho to an oil column of height 3 mm The total surface area of the soap bubble in contact with air will be given density of oil 0 9 g cm surface tension of soap solution 0 03 N m and g 10 ms ACT 1 4 96 x 104 m 412 hgg 2 1 24 x 104 m 3 3 72 x 104 m 4 1 03 x 104 m 45 3 4x0 03 23x10 3x0 9x10 x 10 x R 24 44 10 4 R GTR 2 2 48 10
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Properties of matter
2 The excess pressure inside a soap bubble is equal ho to an oil column of height 3 mm The total surface area of the soap bubble in contact with air will be given density of oil 0 9 g cm surface tension of soap solution 0 03 N m and g 10 ms ACT 1 4 96 x 104 m 412 hgg 2 1 24 x 104 m 3 3 72 x 104 m 4 1 03 x 104 m 45 3 4x0 03 23x10 3x0 9x10 x 10 x R 24 44 10 4 R GTR 2 2 48 10
4 Two glass plates of width b are kept close to each other in water at a small distance d as shown in figure Density of water is p surface tension T while contact angle with glass is 0 The force with which two plates attract each other A 2T b pgd 2T d pgb C B D 4T b pgd 4T d pgb BB
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Properties of matter
4 Two glass plates of width b are kept close to each other in water at a small distance d as shown in figure Density of water is p surface tension T while contact angle with glass is 0 The force with which two plates attract each other A 2T b pgd 2T d pgb C B D 4T b pgd 4T d pgb BB
23 An ideal gas has adiabatic exponenty It contracts according to the law PV a where a is a positive constant For this process the Bulk modulus of the gas is P 1 P 3 ap 2 P a 4 1 P B
Physics
Properties of matter
23 An ideal gas has adiabatic exponenty It contracts according to the law PV a where a is a positive constant For this process the Bulk modulus of the gas is P 1 P 3 ap 2 P a 4 1 P B
Young modulus of elasticity of brass is 10 N m The increase in its energy on pressing a rod of length 0 1 m and cross sectional area 1 cm made of brass with a force of 10 kg along its length will be of Tongulial
Physics
Properties of matter
Young modulus of elasticity of brass is 10 N m The increase in its energy on pressing a rod of length 0 1 m and cross sectional area 1 cm made of brass with a force of 10 kg along its length will be of Tongulial
3 If p is density of water and B is Bulk modulus then energy density of water in a lake h metre deep is 1 3 pgh B pgh 4B 2 4 pgh 2B 2 pgh B 9 sam its w 1 Lo 2 Va 3 SH 4 La A 100 safe
Physics
Properties of matter
3 If p is density of water and B is Bulk modulus then energy density of water in a lake h metre deep is 1 3 pgh B pgh 4B 2 4 pgh 2B 2 pgh B 9 sam its w 1 Lo 2 Va 3 SH 4 La A 100 safe
A 5 m long cylindrical steel wire with radius 2 x 10 3 m is suspended vertically from a rigid sup and carries a bob of mass 100 kg at the other end If the bob gets snapped calculate the cha in temperature of the wire ignoring radiation losses Ys 2 1 x 10 Pa p 7860 kg m C 420 J kg K
Physics
Properties of matter
A 5 m long cylindrical steel wire with radius 2 x 10 3 m is suspended vertically from a rigid sup and carries a bob of mass 100 kg at the other end If the bob gets snapped calculate the cha in temperature of the wire ignoring radiation losses Ys 2 1 x 10 Pa p 7860 kg m C 420 J kg K
Adding detergents to water helps in removing dirty greasy stains This is because Read Page 153 It increases the oil water surface tension b It decreases the oil water surface tension c It increases the viscosity of the solution d Dirt is held suspended surrounded by detergen
Physics
Properties of matter
Adding detergents to water helps in removing dirty greasy stains This is because Read Page 153 It increases the oil water surface tension b It decreases the oil water surface tension c It increases the viscosity of the solution d Dirt is held suspended surrounded by detergen
The density of a metal at a normal pressure is p Its density when it is subjected to an excess pressure P is p If B is the bulk modulus of the metal the ratio p p is 1 3 1 1 P B 1 1 P B 2 1 4 P B P B unchanged when the N 24 If a sp store and 1 3 25 A bo P E ene
Physics
Properties of matter
The density of a metal at a normal pressure is p Its density when it is subjected to an excess pressure P is p If B is the bulk modulus of the metal the ratio p p is 1 3 1 1 P B 1 1 P B 2 1 4 P B P B unchanged when the N 24 If a sp store and 1 3 25 A bo P E ene
6 A 560N cyclist sits on a biycle seat and copresses the two seat springs The spring constant equals 2 2X10 4N m for each spring How much is each spring compressed because you re looking at each spring each spring takes half the weight of the cyclist thus F 280N 7 By how much does the compression increase each spring s elastic PE Use the distance x you calculated in prior question and the PE 1 2kx 2 equation
Physics
Properties of matter
6 A 560N cyclist sits on a biycle seat and copresses the two seat springs The spring constant equals 2 2X10 4N m for each spring How much is each spring compressed because you re looking at each spring each spring takes half the weight of the cyclist thus F 280N 7 By how much does the compression increase each spring s elastic PE Use the distance x you calculated in prior question and the PE 1 2kx 2 equation
Passage I Freezing point depression occurs when a substance solute is added to a solvent decreasing the solvent s freezing point The process can be monitored over a short range depending on the solute by using a hydrometer which floats in the solution at varying depths as an indication of the mixed solution s specific gravity Freezing point is the temperature at which the solution becomes a solid at normal atmospheric pressure Three experiments were performed using distilled water mixed with precise amounts of ethylene glycol CHO propylene glycol C H O or methanol CH O in concentrations varying from 10 90 by weight All solutions were aqueous with a starting temperature of 25 C A hydrometer was used to indicate the relative change in specific gravity as the solute percentage was increased Table 1 shows the density for each of the three solutes Solute ethylene glycol propylene glycol methanol Entrance Ticket Table 1 6 4 1 Set One Density 1 110 00 kg m 1 040 00 kg m 791 30 kg m Experiment 1 Distilled water was mixed with precise concentrations of ethylene glycol in various concentrations by weight For each concentration a hydrometer was used to measure specific gravity at 25 C and the freezing point was determined using a precision thermometer Figure 1 shows results through 50 concentration freezing point C freezing point C 25 30 35 40 15 20 25 30 35 40 0 5 10 15 20 0 5 0 10 Learning Targets Trends in Data Puzzle Pieces 0 Experiment 2 Experiment 1 was repeated except propylene glycol was used instead of ethylene glycol Figure 2 shows results through 50 concentration Key freezing point specific gravity 1 10 20 30 40 50 ethylene glycol concentration Figure 1 Key freezing point specific gravity 40 10 20 30 50 propylene glycol concentration ACT Practice Figure 2 1 09 1 08 1 07 1 06 1 05 Sum It Up 1 04 1 03 1 02 1 01 1 06 1 05 1 04 1 03 1 02 specific gravity 1 01 1 00 specific gravity Experiment 3 Experiment I was repeated except methanol was used instead of ethylene glycol and specific gravity was not recorded because of methanol s low evaporation point Figure 3 shows results for all concentrations of the three solutes freezing point C 0 10 20 30 40 50 60 70 80 90 100 0 Key ethylene glycol propylene glycol methanol 20 F G H J 40 60 80 solute concentration Figure 3 1 In Experiment 1 if 60 ethylene glycol had been added to the sample the freezing point would most likely have been A greater than 15 C B between 15 C and 25 C C between 25 C and 35 C D less than 35 C 100 2 Consider two of the solutes temporarily named Solute A and Solute B At increasing concentrations both solutes initially provided similar levels of freeze protection At 70 concentration however Solute B began to lose effectiveness while Solute A continued to provide an increasing level of freeze protection as concentration increased What are the actual names for Solutes A and B Solute A ethylene glycol propylene glycol methanol propylene glycol Solute B propylene glycol methanol ethylene glycol ethylene glycol 3 Suppose another concentration of propylene glycol had been tested in the same way described in Experiment 2 The specific gravity for a 60 propylene glycol solution would most likely have been Entrance Ticket Learning Targets A greater than 1 04 B between 1 02 and 1 04 C between 1 01 and 1 02 D less than 1 01 4 Based on Table 1 the mass of 0 5 m of propylene glycol would be closest to which of the following values F 395 65 kg G 520 00 kg H 555 00 kg J 791 30 kg 5 Based on the results of the three experiments which of the following combinations of solute solute weight and solution weight will provide the lowest possible freezing point Trends in Data A B C D Solute Weight Solution Weight Solute 80 lbs 200 lbs 70 lbs 280 lbs 90 lbs 150 lbs 180 lbs 90 lbs END OF SET STOP DO NOT GO ON TO THE NEXT F UNTIL TOLD TO D Puzzle Pieces methanol ethylene glycol ethylene glycol propylene glyco ACT Practice Sum It Up
Physics
Properties of matter
Passage I Freezing point depression occurs when a substance solute is added to a solvent decreasing the solvent s freezing point The process can be monitored over a short range depending on the solute by using a hydrometer which floats in the solution at varying depths as an indication of the mixed solution s specific gravity Freezing point is the temperature at which the solution becomes a solid at normal atmospheric pressure Three experiments were performed using distilled water mixed with precise amounts of ethylene glycol CHO propylene glycol C H O or methanol CH O in concentrations varying from 10 90 by weight All solutions were aqueous with a starting temperature of 25 C A hydrometer was used to indicate the relative change in specific gravity as the solute percentage was increased Table 1 shows the density for each of the three solutes Solute ethylene glycol propylene glycol methanol Entrance Ticket Table 1 6 4 1 Set One Density 1 110 00 kg m 1 040 00 kg m 791 30 kg m Experiment 1 Distilled water was mixed with precise concentrations of ethylene glycol in various concentrations by weight For each concentration a hydrometer was used to measure specific gravity at 25 C and the freezing point was determined using a precision thermometer Figure 1 shows results through 50 concentration freezing point C freezing point C 25 30 35 40 15 20 25 30 35 40 0 5 10 15 20 0 5 0 10 Learning Targets Trends in Data Puzzle Pieces 0 Experiment 2 Experiment 1 was repeated except propylene glycol was used instead of ethylene glycol Figure 2 shows results through 50 concentration Key freezing point specific gravity 1 10 20 30 40 50 ethylene glycol concentration Figure 1 Key freezing point specific gravity 40 10 20 30 50 propylene glycol concentration ACT Practice Figure 2 1 09 1 08 1 07 1 06 1 05 Sum It Up 1 04 1 03 1 02 1 01 1 06 1 05 1 04 1 03 1 02 specific gravity 1 01 1 00 specific gravity Experiment 3 Experiment I was repeated except methanol was used instead of ethylene glycol and specific gravity was not recorded because of methanol s low evaporation point Figure 3 shows results for all concentrations of the three solutes freezing point C 0 10 20 30 40 50 60 70 80 90 100 0 Key ethylene glycol propylene glycol methanol 20 F G H J 40 60 80 solute concentration Figure 3 1 In Experiment 1 if 60 ethylene glycol had been added to the sample the freezing point would most likely have been A greater than 15 C B between 15 C and 25 C C between 25 C and 35 C D less than 35 C 100 2 Consider two of the solutes temporarily named Solute A and Solute B At increasing concentrations both solutes initially provided similar levels of freeze protection At 70 concentration however Solute B began to lose effectiveness while Solute A continued to provide an increasing level of freeze protection as concentration increased What are the actual names for Solutes A and B Solute A ethylene glycol propylene glycol methanol propylene glycol Solute B propylene glycol methanol ethylene glycol ethylene glycol 3 Suppose another concentration of propylene glycol had been tested in the same way described in Experiment 2 The specific gravity for a 60 propylene glycol solution would most likely have been Entrance Ticket Learning Targets A greater than 1 04 B between 1 02 and 1 04 C between 1 01 and 1 02 D less than 1 01 4 Based on Table 1 the mass of 0 5 m of propylene glycol would be closest to which of the following values F 395 65 kg G 520 00 kg H 555 00 kg J 791 30 kg 5 Based on the results of the three experiments which of the following combinations of solute solute weight and solution weight will provide the lowest possible freezing point Trends in Data A B C D Solute Weight Solution Weight Solute 80 lbs 200 lbs 70 lbs 280 lbs 90 lbs 150 lbs 180 lbs 90 lbs END OF SET STOP DO NOT GO ON TO THE NEXT F UNTIL TOLD TO D Puzzle Pieces methanol ethylene glycol ethylene glycol propylene glyco ACT Practice Sum It Up
Passage III One way to use the principle of density to isolate macromolecules from cell extracts is by separating out the various parts of a cellular extract using a density gradient 4 A Experiment 1 To make the density gradient 6 mL aliquots for each step a layer of liquid that has a different density than the other layers of the density gradient were made by mixing together pure water with different amounts of 60 wt vol iodixanol solution The density of the 60 iodixanol solution is known 1 32 g mL See Table 1 for measurements of the various steps of the density gradient Step 1 2 3 4 5 Mass of water g 0 00 0 91 1 69 2 48 3 00 Table 1 4 4 3 Set Three Mass of iodixanol Total mass Density g mL g solution g 8 10 6 87 5 81 4 73 4 05 8 10 7 78 7 50 7 21 7 05 1 35 1 30 1 25 1 20 1 18 Deriment 2 To make steps in the density gradient that are lower sity than Step 5 a Falcon tube was tared on an electronic nce Water and acetonitrile were added and weighed in Falcon tube until the total volume in the Falcon tube was Several more steps of the density gradient were made dding different masses of water and acetonitrile the urements for which are noted in Table 2 Step 6 7 8 9 10 Condition 6 00 5 00 4 00 3 00 2 00 virus caspase 1 virus caspase 5 Mass of Mass of Total mass Density water g acetonitrile g g g mL virus only virus trypsin virus proteinase K virus MMP 9 Table 2 11 0 00 0 78 1 56 2 35 3 13 Experiment 3 To determine the various densities of HPV virus fragments that are processed by the cell and by extracellular proteases 293TT cells were infected with HPV and then harvested after 18 hours To some of the cellular extracts proteases such as trypsin proteinase K MMP 9 caspase 1 and caspase 5 were added and incubated with the extracts prior to loading them onto the density gradient To locate where the HPV virus was on the gradient it was tagged with GFP prior to infection the virus will show up as a green band in one or more of the steps of the gradient The locations of the HPV virus on the gradient after the various conditions are listed in Table 3 means that GFP was detected in that step of the gradient and means GFP was not detected in that step of the density gradient Table 3 11 1 Step of the density gradient 1 2 3 4 5 6 7 8 9 10 1111 6 00 5 78 5 56 5 35 5 13 TING 1 00 0 96 0 93 0 89 111 11 11 II 1 1 III 0 85 E of Hypotheses Part 2 4 A A A A A A A A A 4 11 The data from Table 3 was interpreted to mean that MMP 9 does not affect the location of the HPV virus on the gradient Is that interpretation correct A No the GFP signal from the virus MMP 9 condition is vastly different from that of virus alone B Yes the GFP signals from the virus alone and virus are same MMP 9 C No the GFP signal from the virus MMP 9 condition was seen in the steps of the gradient with the least density D Yes the GFP signal from virus MMP 9 was similar to that of the virus caspase 1 condition 12 A student predicted that if a solid with a known density of 2 33 g mL were mixed with the solution from Step 5 of Experiment 1 the solid would float to the top Based on the information in Table 1 is the student s prediction correct F Yes because the mass of the solid 2 33 g is less than mass of the Step 5 iodixanol solution 7 05 g G Yes because the density of the solid is less than the of the solutions in the table density of H No because the solid would dissolve and create a homogenous solution J No because the density of the solid is greater than the density of the solution from Step 5 13 Do the results of Experiment 2 indicate that the density of acetonitrile is less than the density of water 14 A student stated that the location of the HPV virus on the gradient is affected by each protease in different ways Do the results of Experiment 3 verify this statement A Yes the table indicates that only 3 13 g of acetonitrile can be added to the solution while a maximum of 6 g of water can be added to the same solution B Yes as more acetonitrile and less water is added to the same volume of solution the density goes down indicating acetonitrile has a lower density C No the density of acetonitrile solutions goes down the more water is added indicating that water has a lower density D No the mass of water added in each step goes down while the density goes up showing that acetonitrile has a higher density than water F Yes because the results of Table 3 show that the location of the virus was detected in different steps for each protease G Yes because the results of Table 3 show that the location of the virus was detected in the same steps for each protease H No because the results of Table 3 show that the location of the virus was detected in different steps for each protease J No because the results of Table 3 show that the location of the virus was detected in the same steps for each protease 15 Is the statement caspase 5 is denser than caspase 1 supported by the information in Table 3 A Yes because caspase 5 reacts with the HPV virus at densities of 5 and 6 while caspase 1 reacts with the HPV virus at densities of 4 and 5 B Yes because out of all of the proteases listed in the table caspase 5 is the densest as shown by its placement in Table 3 and its reaction with the HPV virus C No because caspase 5 reacts with the HPV viru at later steps of the gradient indicating that it has lower density than caspase 1 D No because the table does not compare the densit of the various proteases but instead shows h the relative densities of HPV virus fragments produced by the different proteases END OF SET STOPI DO NOT GO ON TO THE NE
Physics
Properties of matter
Passage III One way to use the principle of density to isolate macromolecules from cell extracts is by separating out the various parts of a cellular extract using a density gradient 4 A Experiment 1 To make the density gradient 6 mL aliquots for each step a layer of liquid that has a different density than the other layers of the density gradient were made by mixing together pure water with different amounts of 60 wt vol iodixanol solution The density of the 60 iodixanol solution is known 1 32 g mL See Table 1 for measurements of the various steps of the density gradient Step 1 2 3 4 5 Mass of water g 0 00 0 91 1 69 2 48 3 00 Table 1 4 4 3 Set Three Mass of iodixanol Total mass Density g mL g solution g 8 10 6 87 5 81 4 73 4 05 8 10 7 78 7 50 7 21 7 05 1 35 1 30 1 25 1 20 1 18 Deriment 2 To make steps in the density gradient that are lower sity than Step 5 a Falcon tube was tared on an electronic nce Water and acetonitrile were added and weighed in Falcon tube until the total volume in the Falcon tube was Several more steps of the density gradient were made dding different masses of water and acetonitrile the urements for which are noted in Table 2 Step 6 7 8 9 10 Condition 6 00 5 00 4 00 3 00 2 00 virus caspase 1 virus caspase 5 Mass of Mass of Total mass Density water g acetonitrile g g g mL virus only virus trypsin virus proteinase K virus MMP 9 Table 2 11 0 00 0 78 1 56 2 35 3 13 Experiment 3 To determine the various densities of HPV virus fragments that are processed by the cell and by extracellular proteases 293TT cells were infected with HPV and then harvested after 18 hours To some of the cellular extracts proteases such as trypsin proteinase K MMP 9 caspase 1 and caspase 5 were added and incubated with the extracts prior to loading them onto the density gradient To locate where the HPV virus was on the gradient it was tagged with GFP prior to infection the virus will show up as a green band in one or more of the steps of the gradient The locations of the HPV virus on the gradient after the various conditions are listed in Table 3 means that GFP was detected in that step of the gradient and means GFP was not detected in that step of the density gradient Table 3 11 1 Step of the density gradient 1 2 3 4 5 6 7 8 9 10 1111 6 00 5 78 5 56 5 35 5 13 TING 1 00 0 96 0 93 0 89 111 11 11 II 1 1 III 0 85 E of Hypotheses Part 2 4 A A A A A A A A A 4 11 The data from Table 3 was interpreted to mean that MMP 9 does not affect the location of the HPV virus on the gradient Is that interpretation correct A No the GFP signal from the virus MMP 9 condition is vastly different from that of virus alone B Yes the GFP signals from the virus alone and virus are same MMP 9 C No the GFP signal from the virus MMP 9 condition was seen in the steps of the gradient with the least density D Yes the GFP signal from virus MMP 9 was similar to that of the virus caspase 1 condition 12 A student predicted that if a solid with a known density of 2 33 g mL were mixed with the solution from Step 5 of Experiment 1 the solid would float to the top Based on the information in Table 1 is the student s prediction correct F Yes because the mass of the solid 2 33 g is less than mass of the Step 5 iodixanol solution 7 05 g G Yes because the density of the solid is less than the of the solutions in the table density of H No because the solid would dissolve and create a homogenous solution J No because the density of the solid is greater than the density of the solution from Step 5 13 Do the results of Experiment 2 indicate that the density of acetonitrile is less than the density of water 14 A student stated that the location of the HPV virus on the gradient is affected by each protease in different ways Do the results of Experiment 3 verify this statement A Yes the table indicates that only 3 13 g of acetonitrile can be added to the solution while a maximum of 6 g of water can be added to the same solution B Yes as more acetonitrile and less water is added to the same volume of solution the density goes down indicating acetonitrile has a lower density C No the density of acetonitrile solutions goes down the more water is added indicating that water has a lower density D No the mass of water added in each step goes down while the density goes up showing that acetonitrile has a higher density than water F Yes because the results of Table 3 show that the location of the virus was detected in different steps for each protease G Yes because the results of Table 3 show that the location of the virus was detected in the same steps for each protease H No because the results of Table 3 show that the location of the virus was detected in different steps for each protease J No because the results of Table 3 show that the location of the virus was detected in the same steps for each protease 15 Is the statement caspase 5 is denser than caspase 1 supported by the information in Table 3 A Yes because caspase 5 reacts with the HPV virus at densities of 5 and 6 while caspase 1 reacts with the HPV virus at densities of 4 and 5 B Yes because out of all of the proteases listed in the table caspase 5 is the densest as shown by its placement in Table 3 and its reaction with the HPV virus C No because caspase 5 reacts with the HPV viru at later steps of the gradient indicating that it has lower density than caspase 1 D No because the table does not compare the densit of the various proteases but instead shows h the relative densities of HPV virus fragments produced by the different proteases END OF SET STOPI DO NOT GO ON TO THE NE
Figure 1 shows a lightweight plank supported at its right end by a 7 0 mm diameter rope with a tensile strength of 6 0x 107 N m Figure 1 of 1 Part A What is the maximum force that the rope can support Express your answer with the appropriate units Tmax Submit Part B O 0 Value H 1 03 A Request Answer A What is the greatest distance measured from the pivot that the center of graw machinery can be placed without snapping the rope Express your answer with the appropriate units Units m BE Submit Previous Answers Request Answer
Physics
Properties of matter
Figure 1 shows a lightweight plank supported at its right end by a 7 0 mm diameter rope with a tensile strength of 6 0x 107 N m Figure 1 of 1 Part A What is the maximum force that the rope can support Express your answer with the appropriate units Tmax Submit Part B O 0 Value H 1 03 A Request Answer A What is the greatest distance measured from the pivot that the center of graw machinery can be placed without snapping the rope Express your answer with the appropriate units Units m BE Submit Previous Answers Request Answer
Given the following reaction CuO s H g Cu s H O g If 357 L of hydrogen gas are used to reduce copper II oxide at STP what mass of copper is to be expected
Physics
Properties of matter
Given the following reaction CuO s H g Cu s H O g If 357 L of hydrogen gas are used to reduce copper II oxide at STP what mass of copper is to be expected
Calculate the mass of 3 7 L of CH4 at STP
Physics
Properties of matter
Calculate the mass of 3 7 L of CH4 at STP
A concrete bridge is built of 345 cm long concrete slabs with an expansion joint between them The slabs just touch on a 115 F day the hottest day for which the bridge is desig Part A What is the gap between the slabs when the temperature is 0 F Express your answer with the appropriate units 100 HA CHIC
Physics
Properties of matter
A concrete bridge is built of 345 cm long concrete slabs with an expansion joint between them The slabs just touch on a 115 F day the hottest day for which the bridge is desig Part A What is the gap between the slabs when the temperature is 0 F Express your answer with the appropriate units 100 HA CHIC
A disk between vertebrae in the spine is subjected to a shearing force of 530 N Find its shear deformation using the shear modulus of 1 0 x 10 N m The disk is equivalent to a solid cylinder 0 77 cm high and 6 7 cm in diameter Round your answer to 9 decimal places
Physics
Properties of matter
A disk between vertebrae in the spine is subjected to a shearing force of 530 N Find its shear deformation using the shear modulus of 1 0 x 10 N m The disk is equivalent to a solid cylinder 0 77 cm high and 6 7 cm in diameter Round your answer to 9 decimal places
The hanging wall of a fault has been displaced downward by 24 cm relative to the footwall and there has been a total horizontal displacement of 35 cm What is the slope of the fault and what type of fault is it Slope cm cm It is a select fault
Physics
Properties of matter
The hanging wall of a fault has been displaced downward by 24 cm relative to the footwall and there has been a total horizontal displacement of 35 cm What is the slope of the fault and what type of fault is it Slope cm cm It is a select fault
acceleration of the mass after it falls 4 1 cm A 5 2 kg mass hung from a spring vibrates with a period of 1 2 s Calculate the spring constant
Physics
Properties of matter
acceleration of the mass after it falls 4 1 cm A 5 2 kg mass hung from a spring vibrates with a period of 1 2 s Calculate the spring constant
A horst is bounded on the east and west by two normal faults The fault plane of the west fault has a slope of 0 88 m m to the west The plane of the east fault slopes 0 76 m m to the west The sedimentary layers cut by the faults have been displaced 65 meters How much extension has occurred Extension m
Physics
Properties of matter
A horst is bounded on the east and west by two normal faults The fault plane of the west fault has a slope of 0 88 m m to the west The plane of the east fault slopes 0 76 m m to the west The sedimentary layers cut by the faults have been displaced 65 meters How much extension has occurred Extension m
Express the answer in scientific notation An ore deposit consisting of chert and hematite occurs in a deposit measuring 5 900 m long by 1 300 m wide by 57 m high The deposit has an average specific gravity of 3 2 How many tonnes of iron can potentially be obtained from the mine Specific gravity can be converted to density by multiplying by the density of water Assume SGhematite 5 3 SGchert 2 5 and g cm tonne m The mass of iron that can potentially be obtained from the mine is x 10 tonnes
Physics
Properties of matter
Express the answer in scientific notation An ore deposit consisting of chert and hematite occurs in a deposit measuring 5 900 m long by 1 300 m wide by 57 m high The deposit has an average specific gravity of 3 2 How many tonnes of iron can potentially be obtained from the mine Specific gravity can be converted to density by multiplying by the density of water Assume SGhematite 5 3 SGchert 2 5 and g cm tonne m The mass of iron that can potentially be obtained from the mine is x 10 tonnes
A vertical solid steel post 27 cm in diameter and 2 8 m long is required to support a load of 8618 kg You can ignore the weight of the post What is the stress in the post Round your answer to 2 decimal places
Physics
Properties of matter
A vertical solid steel post 27 cm in diameter and 2 8 m long is required to support a load of 8618 kg You can ignore the weight of the post What is the stress in the post Round your answer to 2 decimal places
Two students each build a piece of scientific equipment that uses a 855 mm long metal rod One student uses a brass rod the other an invar rod Part A If the temperature increases by 4 0 C how much more does the brass rod expand than the invar rod Express your answer with the appropriate units A
Physics
Properties of matter
Two students each build a piece of scientific equipment that uses a 855 mm long metal rod One student uses a brass rod the other an invar rod Part A If the temperature increases by 4 0 C how much more does the brass rod expand than the invar rod Express your answer with the appropriate units A
By how much does a 65 kg mountain climber stretch her 0 545 cm diameter nylon rope when she hangs 46 m below a rock Dutcropping For nylon Y 1 35 x 10 Pa Round your answer to 4 decimal places
Physics
Properties of matter
By how much does a 65 kg mountain climber stretch her 0 545 cm diameter nylon rope when she hangs 46 m below a rock Dutcropping For nylon Y 1 35 x 10 Pa Round your answer to 4 decimal places
A vertebra is subjected to a shearing force of 458 N Find the shear deformation taking the vertebra to be a cylinder 9 66 cm high and 4 16 cm in diameter Round your answer to 8 decimal places
Physics
Properties of matter
A vertebra is subjected to a shearing force of 458 N Find the shear deformation taking the vertebra to be a cylinder 9 66 cm high and 4 16 cm in diameter Round your answer to 8 decimal places
For the demand function q D p 433 p find the following a The elasticity b The elasticity at p 124 stating whether the demand is elastic inelastic or has unit elasticity c The value s of p for which total revenue is a maximum assume that p is in dollars a Find the equation for elasticity E p b Find the elasticity at the given price stating whether the demand is elastic inelastic or has unit elasticity E 124 Simplify your answer Type an integer or a fraction Is the demand elastic inelastic or does it have unit elasticity elastic unit elasticity inelastic c Find the value s of p for which total revenue is a maximum assume that p is in dollars Round to the nearest cent Use a comma to separate answers as needed
Physics
Properties of matter
For the demand function q D p 433 p find the following a The elasticity b The elasticity at p 124 stating whether the demand is elastic inelastic or has unit elasticity c The value s of p for which total revenue is a maximum assume that p is in dollars a Find the equation for elasticity E p b Find the elasticity at the given price stating whether the demand is elastic inelastic or has unit elasticity E 124 Simplify your answer Type an integer or a fraction Is the demand elastic inelastic or does it have unit elasticity elastic unit elasticity inelastic c Find the value s of p for which total revenue is a maximum assume that p is in dollars Round to the nearest cent Use a comma to separate answers as needed
Bonnie owns a cheese shop She just got 3 different cylindrical wheels of cheese The ta below shows the mass diameter and height of each wheel of cheese Cheese Manchego Smoked Cheddar Port Salut Mass lb 7 6 5 4 5 Diameter in Height in 7 4 8 5 3 8 5 2 Put the cheeses in order from the least dense to the most dense Manchego Smoked Cheddar Port Salut What is the mean of the densities of the three types of cheese If necessary round your answer to the nearest thousandth
Physics
Properties of matter
Bonnie owns a cheese shop She just got 3 different cylindrical wheels of cheese The ta below shows the mass diameter and height of each wheel of cheese Cheese Manchego Smoked Cheddar Port Salut Mass lb 7 6 5 4 5 Diameter in Height in 7 4 8 5 3 8 5 2 Put the cheeses in order from the least dense to the most dense Manchego Smoked Cheddar Port Salut What is the mean of the densities of the three types of cheese If necessary round your answer to the nearest thousandth
tangular beam of breadth 100 mm and depth 300 mm is simply supported over a span of 4 m beam is loaded with a uniformly distributed load of 5 kN m Which of the given options will be the value of the maximum bending stress in N mm Revisit La
Physics
Properties of matter
tangular beam of breadth 100 mm and depth 300 mm is simply supported over a span of 4 m beam is loaded with a uniformly distributed load of 5 kN m Which of the given options will be the value of the maximum bending stress in N mm Revisit La
The temperature of a iron bar rises by 10.0°C when it absorbs 2.35 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of iron from these data.
kJ/kg. °C
Physics
Properties of matter
The temperature of a iron bar rises by 10.0°C when it absorbs 2.35 kJ of energy by heat. The mass of the bar is 525 g. Determine the specific heat of iron from these data. kJ/kg. °C
In the above structure use the following values: s= 1.012 m; u = 3.473 m; v = 4.375 m; M = 9.817 N*m; F = 1.199 N; w= 4.408 N/m. There is a fixed support at A, the left end of the beam. M is a couple moment. F is a concentrated force. w is a distributed load. 
Find the resultant moment at point A (the left end) from the loading. Enter the resultant moment at A in the box below in units of N*m. Don't include the unit with your answer. Make sure to use the correct sign (- is clockwise / + is counterclockwise).
Physics
Properties of matter
In the above structure use the following values: s= 1.012 m; u = 3.473 m; v = 4.375 m; M = 9.817 N*m; F = 1.199 N; w= 4.408 N/m. There is a fixed support at A, the left end of the beam. M is a couple moment. F is a concentrated force. w is a distributed load. Find the resultant moment at point A (the left end) from the loading. Enter the resultant moment at A in the box below in units of N*m. Don't include the unit with your answer. Make sure to use the correct sign (- is clockwise / + is counterclockwise).