An elevator is accelerating downwards with an acceleration of 4.9 ms`""^(-2)`. A barometer placed in this elevator reads 75 cm of Hg. Calculate the air pressure inside the elevator.

Answer

To solve the problem of calculating the air pressure inside an elevator that is accelerating downwards, we will follow these steps:

### Step 1: Understand the effective acceleration due to gravity
The elevator is accelerating downwards with an acceleration of \( a = 4.9 \, \text{m/s}^2 \). The acceleration due to gravity is \( g = 9.8 \, \text{m/s}^2 \). When the elevator accelerates downwards, the effective acceleration due to gravity (\( g_{\text{eff}} \)) that acts on the fluid inside the barometer will be:
\[ g_{\text{eff}} = g - a = 9.8 \, \text{m/s}^2 - 4.9 \, \text{m/s}^2 = 4.9 \, \text{m/s}^2 \]

### Step 2: Convert the height of the mercury column
The barometer reads \( 75 \, \text{cm} \) of mercury. We need to convert this height into meters:
\[ H = 75 \, \text{cm} = 0.75 \, \text{m} \]

### Step 3: Use the formula for pressure
The pressure exerted by a column of liquid is given by the formula:
\[ P = \rho g_{\text{eff}} H \]
where:
- \( P \) is the pressure,
- \( \rho \) is the density of mercury,
- \( g_{\text{eff}} \) is the effective acceleration due to gravity,
- \( H \) is the height of the mercury column.

### Step 4: Substitute the values
The density of mercury (\( \rho \)) is approximately \( 13.6 \times 10^3 \, \text{kg/m}^3 \). Now we can substitute the values into the pressure formula:
\[ P = (13.6 \times 10^3 \, \text{kg/m}^3) \times (4.9 \, \text{m/s}^2) \times (0.75 \, \text{m}) \]

### Step 5: Calculate the pressure
Now we perform the multiplication:
\[ P = 13.6 \times 10^3 \times 4.9 \times 0.75 \]
Calculating this step-by-step:
1. \( 4.9 \times 0.75 = 3.675 \)
2. \( 13.6 \times 10^3 \times 3.675 = 49995 \, \text{Pa} \) (approximately)

Thus, the air pressure inside the elevator is approximately:
\[ P \approx 49995 \, \text{Pa} \text{ or } 49995 \, \text{N/m}^2 \]

### Final Answer
The air pressure inside the elevator is approximately \( 49995 \, \text{Pa} \).

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