To answer this question you need to know the definitions of work and power:

The units of work and power are Joules and watts, respectively. The passage states that the amount of time it takes to complete the task is same for both students; therefore, you can ignore the effects of time on power.

The passage also states that student Y applies less force. Since student Y applies less force, he has to expend less energy (work) than student X. This means that student Y also has to exert less power. A reduction in force will reduce work, and subsequently reduce power. We can thus conclude that student Y exerts less power than student X.

The first step is to find the force exerted by student X. The passage states that student Y exerts three times less force than student X; therefore, student X exerts three times more than student Y:

The second step is to find the amount of work done by student X.

Remember that student X directly lifts the boulder from point A to point B. This means that the distance the boulder travels is the vertical distance between point A and B ().

The question states that time for the entire process is . Since the time is the same for both students, power exerted by student X is:

The correct answer is .

Since the focal point falls behind the mirror it must be convex. The radius of curvature can be found using the focal length equation.
 

Since the mirror is concave, the focal point will be in front of the mirror. The focal length is equal to one half of the radius of curvature.



R_c is the radius of curvature. Plugging in 0.85m for R_c allows us to solve for the focal length.

0.43m is equal to 43cm.

Since the focal length is negative, the lens is diverging.

The diopter of a lens is found through the following formula:

Since the focal length of the lens is :

Water exhibits hydrogen bonding. Each hydrogen atom is bond to a highly electronegative oxygen atom, resulting in a slight positive dipole on the hydrogen and a slight negative dipole on the oxygen. Within a solution of water, these dipoles can align, causing attractive forces between the hydrogens of one water molecule and the oxygen of another. Hydrogen bonding is a specialized form of dipole moment; since water has a permanent dipole moment, dipole-dipole attractions are present. Finally, like all molecules, water exhibits London dispersion intermolecular forces as well.

Phase phenomena, such as vapor pressure, are best explained as a product of intermolecular forces. Strong intermolecular forces, like hydrogen bonds or dipole interactions, would make us expect weak vapor pressure, increasing attraction between molecules in the same phase and preventing transition. Large vapor pressures are more likely due to weak van der Waals intermolecular interactions, rather than stronger forces.

London dispersion forces are weak, temporary attracting forces caused when electrons in adjacent atoms move into asymmetrical arrangements about their nuclei, forming temporary dipoles. Since electrons are constantly moving in any atom, both polar and nonpolar substances can develop temporary dipoles and experience London dispersion forces. These forces occur randomly as electrons form spontaneous polarized distributions, before immediately dissipating as the electrons move to new positions.

The passage states that jar B has a boiling point of , the lowest of all three molecules; therefore, jar B contains molecules that have the weakest intermolecular forces. Of all the intermolecular forces, London dispersion forces (or van der Waals forces) are the weakest. These forces occur in all molecules because the movement of electrons inside molecules causes brief polarities that can be used to form intermolecular forces. Since they are so brief, London dispersion forces are very weak and easy to break.

Many nonpolar molecules, such as propane, only contain London dispersion forces and lack all forms of dipole interaction. This results in very weak intermolecular interactions for these compounds, and will affect physical properties like vapor pressure, boiling point, and surface tension. Weak intermolecular forces will increase vapor pressure, decrease boiling point, and decrease surface tension.

Remember that polar molecules, such as methanol and dichloromethane, also contain London dispersion forces. They just contain much stronger forces (hydrogen bonding and dipole-dipole interactions) that make the London dispersion forces negligible.

The effects of London dispersion forces are best demonstrated by viewing the halogen group. The lighter halogens are gaseous at room temperature, but transition to liquid and then to solid when moving down the group (bromine is a liquid, and iodine is a solid). When moving down the group, the halogens become much larger. Larger atoms experience greater London dispersion forces due to increased surface contact and greater polarizability. These increased London dispersion forces are enough to raise the boiling points of the larger atoms, causing them to exist as liquids of gases at room temperature.

While it is true that the larger atoms are heavier and will lose their valence electrons more readily, these factors do not impact the phase of the compound.