Read the passage "The Movement of Water" in the box below. The Movement of Water

Osmosis is the movement of water from low concentrations of solutes, like pure water, to high concentrations of solutes, like 30% salt water. Plants use osmosis to take in and hold onto water because they have a higher solute concentration in their cells than the freshwater that surrounds them.

Using the passage, a student can conclude that a celery stalk placed in concentrated salt water would wilt, but a celery stalk placed in pure water would stand up tall because —
A.
water is rushing into the celery stalk cells and applying a pushing force on the cell walls to make the plant firm.
B.
salt is rushing into the celery stalk cells and applying a pushing force on the cell walls to make the plant firm.
C.
water is rushing out of the celery stalk cells and applying a pushing force on the cell walls to make the plant firm.
D.
salt is rushing out of the celery stalk cells and applying a pulling force on the cell walls to make the plant firm.

Part f - example: finding two forces (part i) two dimensional dynamics often involves solving for two unknown quantities in two separate equations describing the total force. the block in (figure 1) has a mass m=10kg and is being pulled by a force f on a table with coefficient of static friction îľs=0.3. four forces act on it: the applied force f (directed î¸=30â above the horizontal). the force of gravity fg=mg (directly down, where g=9.8m/s2). the normal force n (directly up). the force of static friction fs (directly left, opposing any potential motion). if we want to find the size of the force necessary to just barely overcome static friction (in which case fs=îľsn), we use the condition that the sum of the forces in both directions must be 0. using some basic trigonometry, we can write this condition out for the forces in both the horizontal and vertical directions, respectively, as: fcosî¸â’îľsn=0 fsinî¸+nâ’mg=0 in order to find the magnitude of force f, we have to solve a system of two equations with both f and the normal force n unknown. use the methods we have learned to find an expression for f in terms of m, g, î¸, and îľs (no n).

The frequency of vibrations of a vibrating violin string is given by f = 1 2l t ρ where l is the length of the string, t is its tension, and ρ is its linear density.† (a) find the rate of change of the frequency with respect to the following. (i) the length (when t and ρ are constant) (ii) the tension (when l and ρ are constant) (iii) the linear density (when l and t are constant) (b) the pitch of a note (how high or low the note sounds) is determined by the frequency f. (the higher the frequency, the higher the pitch.) use the signs of the derivatives in part (a) to determine what happens to the pitch of a note for the following. (i) when the effective length of a string is decreased by placing a finger on the string so a shorter portion of the string vibrates df dl 0 and l is ⇒ f is ⇒ (ii) when the tension is increased by turning a tuning peg df dt 0 and t is ⇒ f is ⇒ (iii) when the linear density is increased by switching to another string df dρ 0 and ρ is ⇒ f is ⇒

For which pair of objects would adding the same a