Holding a glass of water filled to a depth of 7.8 cm you step into an elevator. The elevator moves upward with constant acceleration, increasing its speed from zero to 2.2 m/s in 3.1 s.

Find the change in pressure exerted on the bottom of the glass during the time the elevator accelerates.

I understand that you use the P_2_=P_at_+ p_water_(g+a)h!

I find P_2_ and I attained a difference of pressures to be 820. I keep rechecking my answers and submiting new ones, I just cant seem to understand it!

PLEASE HELP!

The pressure at the bottom of the glass changes from

rho(water)*g*h to rho(water)*(g+a)*g
when the elevator is accelerating

I have neglected the atmospheric pressure term, which does not change, and is the same on both sides of the bottom of the glass, anyway.

The pressure increase is rho(water)*a*h

h = 0.078 m
a = (2.2 m/s)/3.1s = 0.71 m/s^2
rho(water) = density = 1000 kg/m^2

THANK YOU!

I meant to write

<<The pressure at the bottom of the glass changes from
rho(water)*g*h to rho(water)*(g+a)*h>>
but the last letter came out g.

The final answer remains correct. The pressure will be in Pascals, ths standard S.I. (Systeme International) unit.

To find the change in pressure exerted on the bottom of the glass during the time the elevator accelerates, you can use the equation:

∆P = P2 - P1

Where:
∆P is the change in pressure
P2 is the pressure exerted at the bottom of the glass after the elevator accelerates
P1 is the pressure exerted at the bottom of the glass before the elevator accelerates

Now let's break down the steps to find P2 and P1:

Step 1: Calculate P2
We can use the equation:
P_2 = P_atm + ρ_water * (g + a) * h

Where:
P_atm is the atmospheric pressure (typically around 101,325 Pa)
ρ_water is the density of water (typically around 1000 kg/m^3)
g is the acceleration due to gravity (typically around 9.8 m/s^2)
a is the acceleration of the elevator (given as 2.2 m/s^2)
h is the depth of water in the glass (given as 7.8 cm = 0.078 m)

Substituting the values into the equation, we get:
P_2 = 101325 Pa + 1000 kg/m^3 * (9.8 m/s^2 + 2.2 m/s^2) * 0.078 m

Calculating this expression, we find:
P_2 ≈ 103557.84 Pa

Step 2: Calculate P1
Since P1 is the pressure exerted at the bottom of the glass before the elevator accelerates, the glass is at rest. In this case, the acceleration of the elevator is 0 (zero).

Using the same equation as above but with a = 0, we get:
P_1 = 101325 Pa + 1000 kg/m^3 * (9.8 m/s^2 + 0 m/s^2) * 0.078 m

Calculating this expression, we find:
P_1 ≈ 102270.6 Pa

Step 3: Calculate ∆P:
Substituting the values we calculated into the equation ∆P = P_2 - P_1, we get:
∆P = 103557.84 Pa - 102270.6 Pa

Calculating this expression, we find:
∆P ≈ 1287.24 Pa

Therefore, the change in pressure exerted on the bottom of the glass during the time the elevator accelerates is approximately 1287.24 Pa.