What are formed by the outermost electrons
2 answers:
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Answer:
773.25 Hz
Explanation:
Concept : In an open organ pipe in fundamental mode of vibration
wave length of wave λ = 2L
where L is length of the pipe
frequency = velocity of sound / λ
Given values: fundamental frequency = 288 Hz
fluid is air. velocity of sound = 340 m/s
⇒ 288 = 340/2L
⇒L = 59.02 cm
The point to be noted is if the pipe is filled with helium initially at the same temperature, there would be change in the sound velocity .Then, frequency of note produced will also be changed .
We know that velocity of sound is inversely proportional to square root of molar mass of gas
velocity of sound in air / velocity of sound in helium = Square root of (Molar mass of Helium/ molar mass of air)
Now, frequency = velocity of sound / λ
= 912.75 / (2 x 0.5902)
= 773.25 Hz
Answer:
<h2>a)Explanation:
If a ball dropped from a stationary hot-air balloon that is at an altitude of 576 ft, an expression for the altitude of the ballast after t seconds can be expressed using the equation of motion;
S is the altitude of the ballest
u is the initial velocity
a is the acceleration of the body
t is the time taken to strike the ground
Since the body is dropped from a stationary air balloon, the initial velocity u will be zero i.e u = 0m/s
Also, since the ballast is dropped from a stationary hot-air balloon, the body is under the influence of gravity, the acceleration will become acceleration due to gravity i.e a = +g
Substituting this values into the equation of the motion;
a) An expression for the altitude of the ballast after t seconds is therefore
b) Given S = 576ft and g = 32ft/s², substituting this into the formula in (a);
This means that the ballast strikes the ground after 6secs
c) To get the velocity when it strikes the ground, we will use the equation of motion v = u + gt.
v = 0 + 32(6)
v = 192ft