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.
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Answer:
10.52 m
Explanation:
The power radiated by a body is given by
P = σεAT⁴ where ε = emissivity = 0.97, T = temperature = 30 C + 273 = 303 K, A = surface area of human body = 1.8 m², σ = 5.67 × 10⁻⁴ W/m²K⁴
P = σεAT⁴ = 5.67 × 10⁻⁸ W/m²K⁴ × 0.97 × 1.8 m² × (303)⁴ = 834.45 W
This is the power radiated by the human body.
The intensity I = P/A where A = 4πr² where r = distance from human body.
I = P/4πr²
r = (√P/πI)/2
If the python is able to detect an intensity of 0.60 W/m², with a power of 834.45 W emitted by the human body, the maximum distance r, is thus
r = (√P/πI)/2 = (√834.45/0.60π)/2 = 21.04/2 = 10.52 m
So, the maximum distance at which a python could detect your presence is 10.52 m.
Check this Light doesn't have mass or gravity right?
So if it doesn't have mass or gravity so light can only affect objects with mass
Does that make sense?
The black hole has gravity and remember light doesn't have gravity so does it affect the light?
To answer that yes, and since light doesn't have gravity it gets "pulled" into the black hole
I hope this helps you
There is approximately 2.54 cm that equals to 1 inch. So your closet answer would be the first choice. :)
Answer:
Hello your question is incomplete below is the complete question
Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000
answer : V = 1.624* 10^-5 m/s
Explanation:
First we have to calculate the value of a
a = 93 * 10^6 mile/m * 1609.344 m
= 149.668 * 10^8 m
next we will express the distance between the earth and the sun
--------- (1)
a = 149.668 * 10^8
E (eccentricity ) = ( 1/60 )^2
= 90°
input the given values into equation 1 above
r = 149.626 * 10^9 m
next calculate the Earths velocity of approach towards the sun using this equation
------ (2)
Note :
Rc = 149.626 * 10^9 m
equation 2 becomes
(
therefore : V = 1.624* 10^-5 m/s