The water potential in one plant cell was calculated to be -4 bar. The water potential in another cell was found to be -11 bar. What was the difference in the water potential?

The difference in the water potential ∆P is 7 bar.

∆P = 7 bar

Step-by-step explanation:

Given;

The water potential in one plant cell was calculated to be -4 bar

P1 = -4 bar

The water potential in another cell was found to be -11 bar.

P2 = -11 bar

The difference in the water potential ∆P is;

∆P = P1 - P2

Substituting the given values;

∆P = -4 - (-11) bar

∆P = -4 +11 bar

∆P = 7 bar

∠cbe and ∠deb are same-side interior angles.

i took the test

0.02

step-by-step explanation:

time of 9 192 631 770 periods of the radiation corresponding to the transition between the 2 hyperfine levels of the ground state of the caesium 133 atom at 0 k (but other seconds are sometimes used in astronomy)