You pull a solid iron ball with a density of 7.86 g/cm3 and a radius of 1.50 cm upward through a fluid at a constant speed of 9.
00 cm/s. The fluid exerts a drag force that is directly proportional to speed, and the proportionality constant is 0.950 kg/s. What is the magnitude of the force (in N) you exert on the ball
1 answer:
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Answer:
A) d = v² / (2g (μ cos θ + syn θ) B) μ = tan θ
Explanation:
Part A
We can work this part with the work and energy theorem, where the work of the friction forces is equal to the energy change of the system.
The work is
W = fr .d
With the force of friction it opposes the movement
W = - fr d
The energy at the lowest point is
Em₀ = K = ½ m v²
The energy at the highest point
= U = m g y
The height (y) can be found by trigonometry
sin θ = y / d
y = d sin θ
W = –Em₀
-fr d = mg d sin θ - ½ m v²
The equation for the force of friction is
fr = μ N
From Newton's second law
N - W cos Te = 0
We replace
-μ (mg cos θ) d - mg d sin θ = - ½ m v²
d g (μ cos θ + sin θ) = ½ v²
d = v² / (2g (μ cos θ + syn θ)
Part B
The block is stopped, what is the Angle tet, let's use Newton's second law
fr - W sin θ = 0 ⇒ fr = W sin θ
N - W cos θ= 0 ⇒ N = w cos θ
fr = μ N
μ (mg cos θ) = mg syn θ
μ = syn θ / cos θ
μ = tan θ
Answer:
F = 268 Hz
Explanation:
The beat frequency is given as:
|
So, for the first flute and tuning fork:
where,
F = Frequency of tuning fork
F = 248 Hz ± 20 Hz
F = 268 Hz (OR) 228 Hz
Now, for the second flute and tuning fork:
where,
F = Frequency of tuning fork
F = 288 Hz ± 20 Hz
F = 268 Hz (OR) 308 Hz
Since, 268 Hz is common from both calculations. Therefore, it will be the frequency of the tuning fork.
<u>F = 268 Hz</u>