A student holds thin strip of paper below his lower lip and blows air horizontally over it. If the surface area of one side of the paper is A and the mass of the strip is m speed v, with which the air should be blown in order to keep the strip horizontal is (consider the density of air 𝜌)
(1) 𝑣= v2mg/ρA (2) 𝑣= vmg/ρA (3) 𝑣= vmg/2ρA (4) 𝑣= vmg/3ρA (5) 𝑣= v3mg/ρA
To determine the speed at which the air should be blown in order to keep the strip of paper horizontal, we can analyze the forces acting on the strip.
When the student blows air horizontally over the strip, the air exerts a force on the strip in the upward direction due to the pressure difference between the upper and lower surfaces of the strip. This upward force is counteracted by the force of gravity acting on the strip, which pulls it downward.
To find the speed at which the air should be blown to maintain the horizontal position of the strip, we need to equate the upward force exerted by the air to the downward force of gravity.
Let's break down the forces involved:
1) Force due to air pressure: The upward force exerted by the air is given by the equation:
F_air = A * ΔP
where A is the surface area of one side of the strip and ΔP is the pressure difference between the upper and lower surfaces of the strip.
2) Force due to gravity: The downward force due to gravity is given by:
F_gravity = m * g
where m is the mass of the strip and g is the acceleration due to gravity.
For equilibrium, the force due to air pressure (F_air) must be equal to the force due to gravity (F_gravity).
So we have:
A * ΔP = m * g
Now, let's express the pressure difference (ΔP) in terms of the velocity of the air (v). According to Bernoulli's principle, when the air flows over the strip of paper, the pressure difference between the upper and lower surfaces is related to the velocity of the air as follows:
ΔP = 0.5 * ρ * v^2
where ρ is the density of air and v is the velocity of the air.
Substituting this expression for ΔP into our equilibrium equation, we have:
A * (0.5 * ρ * v^2) = m * g
Now, solving for v, we get:
v^2 = 2 * (m * g) / (ρ * A)
Taking the square root of both sides, we have:
v = √(2 * (m * g) / (ρ * A))
Therefore, the correct option is (1) v = √(2 * (m * g) / (ρ * A)).
To determine the speed with which the air should be blown in order to keep the strip horizontal, we can use the principle of Bernoulli's equation.
According to Bernoulli's equation, the sum of the pressure energy, kinetic energy, and potential energy for a fluid flowing in a horizontal streamline is constant.
In this case, the pressure energy and the potential energy can be assumed to be constant. Therefore, the change in kinetic energy will be equal to zero.
The change in kinetic energy is given by:
ΔKE = (1/2)mv^2 - (1/2)m(0)^2
= (1/2)mv^2
Since the change in kinetic energy is zero, we have:
(1/2)mv^2 = 0
Simplifying the equation, we find:
mv^2 = 0
Now, let's equate the weight of the paper strip to the upward force exerted by the air blowing over it.
The weight of the paper strip is given by:
Weight = mg
The upward force exerted by the air is given by:
Upward force = Pressure x Area
Since the strip is horizontal, the pressure on the lower side of the strip is greater than the pressure on the upper side. This pressure difference exerts an upward force.
Pressure difference = (Density of air) x (Change in velocity) x (Area)
Substituting the values, we get:
Upward force = ρAΔv
Equating the weight to the upward force, we have:
mg = ρAΔv
Simplifying the equation, we find:
Δv = mg/(ρA)
Therefore, the speed with which the air should be blown in order to keep the strip horizontal is:
v = Δv = mg/(ρA)
Hence, the correct answer is (2) 𝑣= vmg/ρA.