A Blowhard Tests of lung capacity show that adults are able to exhale 1.5 liters of air through their mouths in as little as 1.0 second.
A. If a person blows air at this rate through a drinking straw with a diameter of 0.49cm , what is the speed of air in the straw?
v = ___ m/s?
B. If the air from the straw in part A is directed horizontally across the upper end of a second straw that is vertical to what height does water rise in the vertical straw?
h = ____ m?
To calculate the speed of air in the straw, we can use the equation for the flow rate of a fluid through a pipe:
Q = A * v
Where Q is the flow rate, A is the cross-sectional area of the pipe, and v is the speed of the fluid.
A. To find the speed of air in the straw:
Step 1: Convert the diameter of the straw to meters
d = 0.49 cm = 0.49 * 10^-2 m
Step 2: Calculate the cross-sectional area of the straw using the formula for the area of a circle
A = π * (d/2)^2
Step 3: Substitute the values into the equation for the flow rate and solve for v
Q = 1.5 L = 1.5 * 10^-3 m^3
v = Q / A
Now, let's perform the calculations:
Step 1: Convert the diameter to meters
d = 0.49 * 10^-2 m
Step 2: Calculate the cross-sectional area of the straw
A = π * (d/2)^2
= 3.14 * (0.49 * 10^-2 / 2)^2
Step 3: Calculate the speed of air in the straw
v = Q / A
= 1.5 * 10^-3 m^3 / [3.14 * (0.49 * 10^-2 / 2)^2]
After evaluating the expression, the final answer will give you the speed of air in the straw.
B. To find the height to which water rises in the vertical straw:
Knowing the speed of air from part A, we can apply Bernoulli's equation, assuming no losses or friction, to calculate the height to which water rises.
P + (1/2) * ρ * v^2 = P0 + ρ * g * h
Where P is the pressure of the air, ρ is the density of water, v is the speed of air, P0 is the atmospheric pressure, g is the acceleration due to gravity, and h is the height to which water rises.
Step 1: Convert the speed of air from part A to m/s
v = ___ m/s (from part A)
Step 2: Find the density of water
ρ = 1000 kg/m^3
Step 3: Substitute the values into Bernoulli's equation and solve for h
P + (1/2) * ρ * v^2 = P0 + ρ * g * h
Now, let's perform the calculations:
Step 1: Convert the speed of air to m/s
v = ___ m/s
Step 2: Find the density of water
ρ = 1000 kg/m^3
Step 3: Solve Bernoulli's equation for h
P + (1/2) * ρ * v^2 = P0 + ρ * g * h
After evaluating the expression, the final answer will give you the height to which water rises in the vertical straw.
A. To find the speed of air in the straw, we can use the equation v = Q/A, where v is the velocity of air, Q is the volume flow rate, and A is the cross-sectional area of the straw.
Given:
Volume flow rate (Q) = 1.5 liters = 0.0015 m^3 (1 liter = 0.001 m^3)
Diameter of the straw (d) = 0.49 cm = 0.0049 m (since 1 cm = 0.01 m)
First, we need to find the cross-sectional area (A) of the straw using the formula A = π * (d/2)^2, where π is approximately 3.14159.
A = 3.14159 * (0.0049/2)^2
A = 3.14159 * 0.0024^2
A = 3.14159 * 0.0000057601
A = 0.000018065 m^2
Now, we can substitute the values into the equation v = Q/A:
v = 0.0015 m^3 / 0.000018065 m^2
v ≈ 83 m/s
Therefore, the speed of air in the straw is approximately 83 m/s.
B. To find the height to which the water rises in the vertical straw, we can apply Bernoulli's principle, which states that the pressure exerted by a fluid decreases as its velocity increases.
Assuming the air from the horizontal straw pushes the water upward, the pressure at the top of the vertical straw can be considered atmospheric pressure (P1). At the bottom of the vertical straw, the pressure (P2) is the sum of atmospheric pressure and the pressure due to the height of the water column (ρgh).
Given:
Density of water (ρ) = 1000 kg/m^3
Acceleration due to gravity (g) = 9.8 m/s^2
Let's assume H is the height of the water column.
Using Bernoulli's principle, P1 + (1/2)ρv1^2 = P2 + (1/2)ρv2^2, and since the air in the vertical straw is at rest (v2 = 0), the equation simplifies to:
P1 = P2 + (1/2)ρv1^2
At the top of the vertical straw, the pressure is atmospheric pressure, which can be approximated as 101325 Pa.
101325 Pa = P2 + (1/2) * 1000 kg/m^3 * (83 m/s)^2
Now, let's solve for P2:
P2 = 101325 Pa - 0.5 * 1000 kg/m^3 * (83 m/s)^2
P2 = 101325 Pa - 0.5 * 1000 kg/m^3 * 6889 m^2/s^2
P2 ≈ 101325 Pa - 3,444,500 Pa
P2 ≈ -3,343,175 Pa
Since pressure cannot be negative, we disregard the negative sign and consider P2 = 0 Pa (atmospheric pressure).
Now, we can equate the pressure at the bottom of the vertical straw (P2) to the sum of atmospheric pressure and the pressure due to the height of the water column:
0 Pa = 101325 Pa + 1000 kg/m^3 * 9.8 m/s^2 * H
Solving for H:
1000 kg/m^3 * 9.8 m/s^2 * H = -101325 Pa
H ≈ -101325 Pa / (1000 kg/m^3 * 9.8 m/s^2)
H ≈ -10.33 m
Again, we disregard the negative sign, so the height to which the water rises in the vertical straw is approximately 10.33 meters.
(1500cm^3)/(1.0s) / (pi*(.49/2)^2 cm^2) = 7954 cm/s or 7.9 m/s
not sure on B.