A car misses a turn and sinks into a shallow lake to a depth of 8.9 m. If the area of the car door is 0.51 m2, what is the force exerted on the outside of the door by the water? Note: 1 atm = 101.325 kPa.

What is the force exerted on the inside of the door by the air, assuming the inside of the car is at atmospheric pressure? Think about what the occupant should do to get the door open.

p(h)=p0+D•g•h,

where D is the density of water.
p(8.9 )=101325+1000• (9.8) • (8.9)=188545 Pa
Force is pressure multiplied by area.
F(out)=p•A=188220 • (0.51)=96158 N
F(in)=p0•A=101325• (0.51)=51676 N
The door may be opened when p (in ) = p (out).
Wait the moment when the water gets into the car

Well, it seems like the driver really took a wrong turn and ended up in quite a pickle! Let's see if we can help them out.

To calculate the force exerted on the outside of the door by the water, we can use the equation F = P x A, where F is the force, P is the pressure, and A is the area. The pressure exerted by the water depends on the depth.

Using the equation P = ρgh, where P is the pressure, ρ is the density of water, g is the acceleration due to gravity, and h is the depth, we can find the pressure exerted by the water.

With the given depth of 8.9 m, the density of water (ρ) is approximately 1000 kg/m³, and the acceleration due to gravity (g) is 9.8 m/s², we can plug in these values and calculate the pressure exerted by the water.

Now, to find the force exerted on the outside of the door, we multiply the pressure by the area of the door. So, F = P x A.

Calculating this gives us the force exerted on the outside of the door by the water. Keep in mind that the direction of the force is towards the inside of the car.

As for the force exerted on the inside of the door by the air, assuming the inside of the car is at atmospheric pressure, well, the air inside the car is at the same pressure as the outside atmosphere. So, we don't have to worry about any additional force being exerted by the air on the inside of the door.

But hey, I have a suggestion for the occupant to open the door - just roll down the window and swim to safety! It's probably easier than trying to tackle the force of the water and air. Stay safe out there!

To find the force exerted on the outside of the car door by the water, we can use the concept of pressure.

Step 1: Calculate the pressure exerted by the water:
The pressure exerted by a fluid is given by the formula P = ρgh, where P is the pressure, ρ is the density of the fluid, g is the acceleration due to gravity, and h is the depth.

Given:
Density of water (ρ) = 1000 kg/m^3 (assumed)
Depth (h) = 8.9 m

Using the formula, we can calculate the pressure exerted by the water:
P = ρgh
= (1000 kg/m^3)(9.8 m/s^2)(8.9 m)
= 88,220 Pa

Step 2: Convert the pressure to kilopascals (kPa):
We know that 1 Pa = 1 N/m^2 and 1 kPa = 1000 Pa. Therefore, the pressure exerted by the water can be written as:
P = 88,220 N/m^2 = 88,220 Pa
= 88.22 kPa

Step 3: Calculate the force exerted on the outside of the door:
The force is given by the formula F = PA, where F is the force, P is the pressure, and A is the area.

Given:
Area of the car door (A) = 0.51 m^2
Pressure (P) = 88.22 kPa

Using the formula, we can calculate the force exerted on the outside of the door:
F = PA
= (88.22 kPa)(0.51 m^2)
= 44.9722 kN

Therefore, the force exerted on the outside of the door by the water is approximately 44.9722 kilonewtons (kN).

Now, let's move on to the force exerted on the inside of the door by the air. Considering the inside of the car is at atmospheric pressure, the force exerted on the inside of the door is simply the atmospheric pressure multiplied by the area of the car door.

Step 4: Calculate the force exerted on the inside of the door by the air:
Given:
Area of the car door (A) = 0.51 m^2
Atmospheric pressure = 1 atm = 101.325 kPa

Using the formula, we can calculate the force exerted on the inside of the door:
F = PA
= (101.325 kPa)(0.51 m^2)
= 51.69975 kN

Therefore, the force exerted on the inside of the door by the air is approximately 51.69975 kilonewtons (kN).

To open the door, the occupant should push against the door with a force greater than the force exerted by the air (51.69975 kN) to overcome the atmospheric pressure and create a pressure imbalance, allowing the door to open.

To find the force exerted on the outside of the car door by the water, we can use the concept of pressure. The force exerted by a fluid on an object is equal to the pressure multiplied by the area over which the force is distributed.

First, let's convert the depth of the water from meters to kilopascals (kPa). 1 meter of water exerts a pressure of approximately 9.8 kPa (assuming water density is 1000 kg/m³ and the acceleration due to gravity is 9.8 m/s²).

So, the pressure exerted by the water on the car door can be calculated as follows:
Pressure = depth * fluid density * acceleration due to gravity

Since the car is submerged to a depth of 8.9 meters, we can calculate the pressure as:
Pressure = 8.9 m * 1000 kg/m³ * 9.8 m/s²
Pressure = 86920 Pa or 86.92 kPa

Next, we can calculate the force exerted on the outside of the car door by multiplying the pressure by the area of the door:
Force = Pressure * Area

Given that the area of the car door is 0.51 m², we can calculate the force as:
Force = 86.92 kPa * 0.51 m²
Force = 44.3792 kN (kilonewtons)

Therefore, the force exerted on the outside of the door by the water is approximately 44.3792 kilonewtons.

Now, to find the force exerted on the inside of the door by the air, assuming the inside of the car is at atmospheric pressure, we need to consider the equilibrium of forces.

Since the car is submerged in water, the pressure inside the car will be equal to the atmospheric pressure. The force exerted by the air on the inside of the door is equal to the atmospheric pressure multiplied by the area of the door:
Force = Pressure * Area

Given that the atmospheric pressure is approximately 101.325 kPa and the area of the door is 0.51 m², we can calculate the force as:
Force = 101.325 kPa * 0.51 m²
Force = 51.704175 kN (kilonewtons)

Therefore, the force exerted on the inside of the door by the air, assuming the inside of the car is at atmospheric pressure, is approximately 51.704175 kilonewtons.

To open the door, the occupant should exert a force greater than the force exerted by the air. They can do this by rolling down the window or opening another door to equalize the pressure inside and outside the car, making it easier to open the door.