We assume that horn releases sound of constant frequency. In order for observer to observe different frequency either horn or observer or both must move.
This happens due to Doppler effect. It states that when position of source of sound and observer relative to each other changes, the observed frequency also changes. If the source emits sound of constant frequency than observed frequency will be either higher or lower than original.
When distance between source and observer increases the observed frequency will be lower. This is because same number of sound waves must cover greater distance so they have greater wavelength.
When distance between source and observer decreases the observed frequency will be higher. This is because same number of sound waves must cover smaller distance so they have smaller wavelength.
Wavelength and frequency are inversely proportional meaning when one increases the other drecreases.
From this explanation we can find answer for our question. <span>If we wanted the pitch of a horn to drop relative to an observer we need to move horn away from an observer.</span>
Answer:
Potential energy
Explanation:
Before release, the catapult has potential energy stored in a tension of torsion device in it. Normally a flexible bow like object that could be made of wood or of metal.
Answer:
(A). The speed of the ions is 
(B). The radius of curvature of a singly charged lithium ion is 
Explanation:
Given that,
Electric field = 60000 N/C
Magnetic field = 0.0500 T
(A). We need to calculate the velocity
For no deflection
(B). We need to calculate the radius
Using magnetic force balance by centripetal force
Put the value into the formula
Hence, (A). The speed of the ions is
(B). The radius of curvature of a singly charged lithium ion is
To solve this problem we will begin by finding the pressure through density and average depth. Later we will find the Force, by means of the relation of the pressure and the area.
Here,
h = Depth average
= Density
Moreover,
Replacing,
Finally the force
The inflated balloon shrinks when it is placed in an ice bath with no change in atmospheric pressure.
<u>Explanation:</u>
When the inflated balloon is subjected to an ice bath, it shrinks. This is due to the fact that smaller volume gets occupied by the air/gas inside the balloon as the temperature decreases. Hence, causes the balloon walls to collapse.
An ice bath also lowers the overall air temperature of the balloon inside. As the temperature decreases, the air molecules move more slowly and with lower energy. Because of the particle's lower energy, their collisions with the walls are not enough to keep the inflated balloon.
