The answer is B frequency. When frequency increases more wave crests pass a fixed point each second. That means the wavelength shortens. So, as frequency increases, wavelength decreases
Answer:
If the temperature of the air in the balloon is less than the temperature of the air surrounding the balloon then the balloon will appear slightly deflated because of the difference in temperature.
As the temperature of the air in the balloon reaches the surrounding air temperature, then the balloon will appear to be fully inflated because the temperature of the air in the balloon is the same as the surrounding air temperature.
The best answer is A) <span>keep moving at a constant velocity until some forces act on them
As the man you're probably tired of hearing about said:
"Every object persists in its state of rest or in uniform motion in a straight line unless a new force acts upon it"
This is Isaac Newton's 1st law of motion, or the law of inertia.
Put more simply, objects in motion tend to stay in motion, and tend the maintain the same velocity (direction and speed) and objects at rest tend to stay at rest. </span>
A) 
The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):
(1)
where k is the spring constant.
The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:
(2)
where x is the displacement, m the mass, and v the speed.
We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

Using (2) we can rewrite this as

And using (1), we find

Substituting
into the last equation, we find the value of x:

B) 
In this case, the kinetic energy is 1/10 of the total energy:

Since we have

we can write

And so we find:

Answer:
a. wavelength of the sound, 
b. observed frequecy, 
Given:
speed of sound source,
= 80 m/s
speed of sound in air or vacuum,
= 343 m/s
speed of sound observed,
= 0 m/s
Solution:
From the relation:
v =
(1)
where
v = velocity of sound
= observed frequency of sound
= wavelength
(a) The wavelength of the sound between source and the listener is given by:
(2)
(b) The observed frequency is given by:


(3)
Using eqn (2) and (3):

