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1 year ago

Sound waves that enter the ear canal are directed to the ____, causing it to vibrate.

1 answer:

4 0

According to the research, the correct option the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

<h3>What is the eardrum?</h3>

It is the membrane found in the middle ear of vertebrate animals, separating this sector from the external auditory canal.

When sound waves enter through the external auditory canal, the eardrum vibrates, transmitting its movement to the middle ear through a series of bones and in this way, the pressure change is transformed into a mechanical movement.

Therefore, we can conclude that according to the research, the correct option is the eardrum. Sound waves that enter the ear canal are directed to the <u>eardrum</u>, causing it to vibrate.

Learn more about the eardrum here: brainly.com/question/12770491

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4. How does nearness to water affect the climate zone of an area?

Afina-wow [57]

Answer:

Choice A.

Nearness to a body of water causes an increase in humidity, due to the higher rate of evaporation.

6 0

2 years ago

Read 2 more answers

A light-rail commuter train accelerates at a rate of 1.35 m/s. D A 33% Part (a) How long does it take to reach its top speed of

Dennis_Churaev [7]

Answer:

a) 17.49 seconds

b) 13.12 seconds

c) 2.99 m/s²

Explanation:

a) Acceleration = a = 1.35 m/s²

Final velocity = v = 85 km/h = $85\frac{1000}{3600}=23.61\ m/s$

Initial velocity = u = 0

Equation of motion

$v=u+at\\\Rightarrow 23.61=0+1.35t\\\Rightarrow t=\frac{23.61}{1.35}=17.49\ s$

Time taken to accelerate to top speed is 17.49 seconds.

b) Acceleration = a = -1.8 m/s²

Initial velocity = u = 23.61\ m/s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61-1.8t\\\Rightarrow t=\frac{23.61}{1.8}=13.12\ s$

Time taken to stop the train from top speed is 13.12 seconds

c) Initial velocity = u = 23.61 m/s

Time taken = t = 7.9 s

Final velocity = v = 0

$v=u+at\\\Rightarrow 0=23.61+a7.9\\\Rightarrow a=\frac{-23.61}{7.9}=-2.99\ m/s^2$

Emergency acceleration is 2.99 m/s² (magnitude)

6 0

3 years ago

Can someone give me an objective and subjective statement example please

shusha [124]

Answer:

Objective: It is raining. Subjective: I love the rain!

Explanation:

Anything objective sticks to the facts, but anything subjective has feelings. Objective and subjective are opposites.

(Hope this helps can I pls have brainlist (crown)☺️)

7 0

2 years ago

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago

A child of mass 40.0 kg is in a roller coaster car that travels in a loop of radius 7.00 m. at point a the speed of the car is 1

pav-90 [236]

I attached the missing picture.
The force of seat acting on the child is a reaction the force of child pressing down on the seat. This is the third Newton's law. The force of a child pressing down the seat and the force of the seat pushing up on the child are the same.
There two forces acting on the child. The first one is the gravitational force and the second one is centrifugal force. In this example, the force of gravity is always pulling down, but centrifugal force always acts away from the center of circular motion.
Part A
For point A we have:
$F_a=F_cf-F_g$
In this case, the forces are aligned, centrifugal is pointing up and gravitational is pulling down.
$F_a=m\frac{v^2}{r}-mg=179 $N$
Part B
At the point, B situation is a bit more complicated. In this case force of gravity and centrifugal force are not aligned. We have to look at y components of this forces, y-axis, in this case, is just pointing upward.
$F=F_{cf}\cos(30)-mg=m\frac{v^2}{r}\cos(30)-mg=153.2$N$
Part C
The child will stay in place at point A when centrifugal force and force of gravity are in balance:
$F_g=F_{cf}\\
mg=m\frac{v^2}{r}\\
gr=v^2\\
v=\sqrt{gr}=8.29\frac{m}{s}$

6 0

3 years ago