Answer:
Assessment zone
Explanation:
It is the assessment zone in various security zones where active and passive security measures are employed to identify, detect, classify and analyze possible threats inside the assessment zones.
Answer:
20cm
Explanation:
A convex lens has a positive focal length and the object placed in front of it produce both virtual and real image <em>(image distance can be negative or positive depending on the nature of the image</em>).
According to the lens equation
where;
f is the focal length of the lens
u is the object distance
v is the image distance
If the magnification is - 0.6
mag = v/u = -0.5
v = -0.5u
since v = 10cm
10 = -0.5u
u = -10/0.5
u =-20 cm
Substitute u = -20cm ( due to negative magnification)and v = 10cm into the lens formula to get the focal length f
Hence the focal length of the convex lens is 20cm
Answer:
i can't see the picture pls can you send a link or no
Explanation:
Answer:
tympanic membrane (eardrum)
Explanation:
The sound waves spread through the air and reach the outer ear, into which they penetrate through the ear canal. In doing so, they stimulate the eardrum, which closes the inner end of the duct. By vibrating this membrane, the vibration of a chain of ossicles located in the middle ear is induced. These ossicles transmit their vibration to the oval window, which is a membranous structure that communicates the middle ear with the cochlea of the inner ear. When the oval membrane moves, it moves the liquid (perilymph) that fills one of the three cavities of the cochlea generating waves in it. These waves mechanically stimulate the sensory cells (hair cells) located in the organ of Corti, within the cochlea in the central cavity, the middle ramp. This cavity is filled with a liquid rich in K +, endolymph. The cells embedded in the endolymph, change their permeability to K + due to the movement of the cilia and respond by releasing a neurotransmitter that excites the nerve terminals, which initiate the auditory sensory pathway.
Answer:
527 Hz
Solution:
As per the question:
Beat frequency of the player, 
Frequency of the tuning fork, f = 523 Hz
Now,
The initial frequency can be calculated as:
when
when
But we know that as the length of the flute increases the frequency decreases
Hence, the initial frequency must be 527 Hz
