Answer:
P-waves travel through liquids and solid while S-waves only travel through solids.
Explanation:
Scientists are able to use the fact that P-waves travel through both solids and liquids and waves travel through only solids to determine what makes the different layers of the Earth.
Answer:
58.8 N
Explanation:
The normal force is calculated as equal to the perpendicular component of the gravitational force.
Thus; N = mg
We are given m = 6 kg
Thus;
N = 6 × 9.8
N = 58.8 N
Thus, magnitude of normal force on the rock = 58.8 N
The three parts of the ear anatomy are the outer ear, the middle ear
and the inner ear. The inner ear is also called the cochlea. (‘Cochlea’
means ‘snail’ in Latin; the cochlea gets its name from its distinctive
coiled up shape.)
The outer ear consists of the pinna, ear canal and eardrum
The middle ear consists of the ossicles (malleus, incus, stapes) and ear drum
The inner ear consists of the cochlea, the auditory (hearing) nerve and the brain
Sound waves enter the ear canal and make the ear drum vibrate. This
action moves the tiny chain of bones (ossicles – malleus, incus, stapes)
in the middle ear. The last bone in this chain ‘knocks’ on the membrane
window of the cochlea and makes the fluid in the cochlea move. The
fluid movement then triggers a response in the hearing nerve.
or
<span>Sound waves enter the ear canal and make the ear drum vibrate. This action moves the tiny chain of bones (ossicles – malleus, incus, stapes) in the middle ear. The last bone in this chain 'knocks' on the membrane window of the cochlea and makes the fluid in the cochlea move.
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</span>
Answer:
The mass of the wheel is 2159.045 kg
Explanation:
Given:
Radius 
m
Force
N
Angular acceleration 
From the formula of torque,
Γ
(1)
Γ
(2)

Find momentum of inertia
from above equation,



Find the momentum inertia of disk,



Kg
Therefore, the mass of the wheel is 2159.045 kg
Answer:
<em>Part A</em><em>:</em>
a) If the wavelength of the light is decreased the fringe spacing Δy will decrease.
<em>Part B</em><em>:</em>
b) If the spacing between the slits is decreased the fringe spacing Δy will increase.
<em>Part C</em><em>:</em>
a) If the distance to the screen is decreased the fringe spacing will decrease.
<em>Part D</em><em>:</em>
The dot in the center of fringe E is
farther from the left slit than from the right slit.
Explanation:
In the double-slit experiment there is a clear contrast between the dark and bright fringes, that indicate destructive and constructive interference respectively, in the central peak and then is less so at either side.
The position of bright fringes in the screen where the pattern is formed can be calculated with


- m is the order number.
is the wavelength of the monochromatic light.- L is the distance between the screen and the two slits.
- d is the distance between the slits.
- Part A: a) In the above equation for the position of bright fringes we can see that if the wavelength of the light
is decreased the overall effect will be that the fringes are going to be closer. That means that the fringe spacing Δy will decrease.
- Part B: b) In the above equation for the position of bright fringes we can see that if the spacing between the slits d is decreased the fringes are going to be wider apart. That means the fringe spacing Δy will increase.
- Part C: a) In the above equation we can see that if the distance to the screen L is decreased the fringes are going to be closer. That means the fringe spacing Δy will decrease.
- Part D: We are told that the central maximum is the fringe C that corresponds with m=0. That means that fringe E corresponds with the order number m=2 if we consider it to be the second maximum at the rigth of the central one. To calculate how much farther from the left slit than from the right slit is a dot located at the center of the fringe E in the screen we use the condition for constructive interference. That says that the path length difference Δr between rays coming from the left and right slit must be
We simply replace the values in that equation :


The dot in the center of fringe E is
farther from the left slit than from the right slit.