The drag force on a falling coffee filter can be modeled as a linear function with respect to velocity, F⃗ D =−bv, where b is a
positive constant related to the surface area and shape of the coffee filter. The teacher demonstrates how the value of constant b can be determined by releasing coffee filters from rest and having them fall toward a motion detector, as shown. The students repeat the activity and use the motion detector to measure the position, velocity, and acceleration of different coffee filters as they fall. Which of the following would provide the data needed to determine the value of b ?
1 answer:
Newton's second law and graphical analysis allow us to find the correct answer for the measure of the constant b is:
D) Stack several filters to change the mass measure the acceleration of the system.
Newton's second law establishes a relationship between the net force, the mass and the acceleration of the bodies.
∑ F = m a
Where the bold letters indicate vectors, F is the force, m the mass and the acceleration of the body.
They indicate that the drag force on the filters is
F = - b v
Where b is a positive constant that depends on the shape and area of the filter and v is the speed of the filter.
Let's write Newton's second law.
W - fr = ma
W -bv = ma
In the experiment the students indicate that they can measure the position, velocity and acceleration of the body.
Based on the above, if we place several filter weights and measure their speed and acceleration in each case, we can make a graph of velocity versus acceleration, we can take the value of the constant b from the slope.
Let's analyze the different answers:
A) False. The constant b depends on the shape of the filter therefore it must be kept constant.
B) False. The constant depends on the area, so it must be kept constant.
C) May be. In this case, since we have the terminal velocity, the acceleration is zero, Newton's second law remains.
B v - W = 0
b =
The problem with the method is the difficulty of measuring the compression terminal velocity.
D) True. According to the discussion of the velocity versus acceleration. graph, the constant b is equal to the slope of the graph.
E) May be. The problem with this method is finding a reliable value for the terminal velocity.
In conclusion, using Newton's second law and graphical analysis we can find the correct answer for the measure of the constant b is:
D) Stack several filters to change the mass measure the acceleration of the system.
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Answer:
The ball is in the air for 3.5 seconds
The initial horizontal component of velocity is 28.6 m/s
The vertical component of the final velocity is 34.3 m/s downward
The final velocity is 44.7 m/s in the direction 50.2° below the horizontal
Explanation:
A ball is thrown horizontally
That means the vertical component of the initial velocity
The initial velocity is the horizontal component
The ball is thrown from the top of a 60 m
That means the vertical displacement component y = 60 m
→ y = t +
gt²
where g is the acceleration of gravity and t is the time
y = -60 m , g = -9.8 m/s² ,
Substitute these values in the rule
→ -60 = 0 + (-9.8)t²
→ -60 = -4.9t²
Divide both sides by -4.9
→ 12.2449 = t²
Take √ for both sides
∴ t = 3.5 seconds
* <em>The ball is in the air for 3.5 seconds </em>
The initial velocity is the horizontal component
The ball lands 100 meter from the base of the building
That means the horizontal displacement x = 100 m
→ x = t
→ t = 3.5 s , x = 100 m
Substitute these values in the rule
→ 100 = (3.5)
Divide both sides by 3.5
→ = 28.57 m/s
<em>The initial horizontal component of velocity is 28.6 m/s</em>
The vertical component of the final velocity is
→ =
+ gt
→ = 0 , g = -9.8 m/s² , t = 3.5 s
Substitute these values in the rule
→ = 0 + (-9.8)(3.5)
→ = -34.3 m/s
<em>The vertical component of the final velocity is 34.3 m/s downward</em>
The final velocity v is the resultant vector of and
→ Its magnetude is
→ Its direction
→ = 28.6 ,
= -34.3
Substitute this values in the rules above
→
→ Its direction
The negative sign means the direction is below the horizontal
<em>The final velocity is 44.7 m/s in the direction 50.2° below the horizontal</em>