The fundamental frequency of the tube is 0.240 m long, by taking air temperature to be 
C is 367.42 Hz.
A standing wave is basically a superposition of two waves propagating opposite to each other having equal amplitude. This is the propagation in a tube.
The fundamental frequency in the tube is given by
where, 
Since, T=37+273 K = 310 K
v = 331 m/s
Using this, we get:
Hence, the fundamental frequency is 367.42 Hz.
To learn more about Attention here:
brainly.com/question/14673613
#SPJ4
Answer:
A) a = 73.304 rad/s²
B) Δθ = 3665.2 rad
Explanation:
A) From Newton's first equation of motion, we can say that;
a = (ω - ω_o)/t. We are given that the centrifuge spins at a maximum rate of 7000rpm.
Let's convert to rad/s = 7000 × 2π/60 = 733.04 rad/s
Thus change in angular velocity = (ω - ω_o) = 733.04 - 0 = 733.04 rad/s
We are given; t = 10 s
Thus;
a = 733.04/10
a = 73.304 rad/s²
B) From Newton's third equation of motion, we can say that;
ω² = ω_o² + 2aΔθ
Where Δθ is angular displacement
Making Δθ the subject;
Δθ = (ω² - ω_o²)/2a
At this point, ω = 0 rad/s while ω_o = 733.04 rad/s
Thus;
Δθ = (0² - 733.04²)/(2 × 73.304)
Δθ = -537347.6416/146.608
Δθ = - 3665.2 rad
We will take the absolute value.
Thus, Δθ = 3665.2 rad
During a total solar eclipse, the moon passes between Earth and the sun. This completely blocks out the sun’s light. However, the moon is about 400 times smaller than the sun. How can it block all of that light?
1 g = 1 ÷ 1000 kg
= 0.001 kg
1 cm³ = 1 ÷ 100 ÷ 100 ÷ 100 m³
= 0.000001 m³
1 g/cm³ = 1 g / 1 cm³
= 0.001 kg / 0.000001 m³
= 1000 kg/m³
The density is 1000 kg/m³.
Answer:
The answer is 3.
Explanation:
The answer to this question can be found by applying the right hand rule for which the pointer finger is in the direction of the electron movement, the thumb is pointing in the direction of the magnetic field, so the effect that this will have on the electrons is the direction that the middle finger points in which is right in this example.
So as a result of the magnetic field directed vertically downwards which is at a right angle with the electron beams, the electrons will move to the right and the spot will be deflected to the right of the screen when looking from the electron source.
I hope this answer helps.
