1)A car misses a turn and sinks into a shallow lake to a depth of 11.4 m. If the area of the car door is 0.56 m2, what is the force exerted on the outside of the door by the water? Note: 1 atm = 101 kPa.
Assume the whole door is at 11.4 m
pressure = rho g h + 1 atm
rho = 10^3 kg/m^3
pressure = 10^3 (9.8)*11.4 + 1 atm
= 112*10^3 + 101*10^3
213*10^3 N/m^2
213*10^3 * .56 = 119*10^3 N
NOTE: You still actually have at least one atm on the inside of the door that you brought down from the surface. That will increase as water leaks into the car and eventually the pressure will equalize inside and out as the car fills with water. You may not wish to wait that long to open the door easily, but right from the start the problem is not quite as bad as implied here.
PS
If you wait until the pressure equalizes inside and out, you will have 1.13*10^5 Pa of pressure above atmospheric in your lungs.
If you then exit the car and swim up to the surface without expelling air from your lungs you will die as your lungs try to expand from the excess of air in them. Divers and submariners know about this and practice expelling air during free ascents using air under pressure (scuba or hard hat). It is called "air embolism".
Well, if I were the car, I'd probably be saying, "Water you doing to me?! I'm sinking here!" But let's do the math.
To find the force exerted on the outside of the door, we need to calculate the pressure of the water at that depth. Pressure is defined as force per unit area, so we can use the formula:
Pressure = Density x Gravity x Depth
The density of water is roughly 1000 kg/m3, and gravity is approximately 9.8 m/s2. Plugging in the values, we get:
Pressure = 1000 kg/m3 x 9.8 m/s2 x 11.4 m = 113,040 Pa
Now, we need to convert from pascals to kilopascals, because the note says 1 atm = 101 kPa. Divide the pressure by 1000 to get:
Pressure = 113,040 Pa / 1000 = 113.04 kPa
Finally, to find the force, we multiply the pressure by the area of the door:
Force = Pressure x Area = 113.04 kPa x 0.56 m2 = 63.22 kN
So, the force exerted on the outside of the door by the water is approximately 63.22 kilonewtons. That's a lot of water pressure!
To find the force exerted on the outside of the car door by the water, we can use the concept of pressure.
The pressure exerted by a fluid at a certain depth is given by the equation:
P = ρgh
Where:
P is the pressure,
ρ (rho) is the density of the fluid,
g is the acceleration due to gravity, and
h is the depth of the fluid.
In this case, we are given the depth of the water (h = 11.4 m) and the area of the car door (A = 0.56 m2), but we need to determine the density of water (ρ).
The equation to calculate the density of water is:
ρ = m/V
Where:
m is the mass of the water, and
V is the volume of the water.
Since we know the density of water is 1000 kg/m3, we can use this information to find the mass of the water.
To calculate the mass of the water, we need to determine the volume of the water. The volume can be found by multiplying the area of the car door (A) by the depth of the water (h).
V = A * h
Now we can substitute the known values into the equations to find the force exerted on the outside of the car door by the water.
First, calculate the volume of the water:
V = 0.56 m^2 * 11.4 m
Next, calculate the mass of the water:
m = ρ * V
Now that we have the mass of the water, we can calculate the force exerted on the outside of the car door using the equation:
F = m * g
Substitute the known values into the equation to find the force exerted on the outside of the car door by the water.