Answer:
Hello, Your answer should be is <u><em>Auditory Nerve</em></u>
Explanation:
Because the "Auditory Nerve" bundle of nerves that carries hearing information, The cochlear nerve, also known as the acoustic nerve, is the sensory nerve that transfers auditory information from the cochlea (auditory area of the inner ear). Its not A, because the <u>"Optic nerve"</u><em> second pairs of cranial transmitting impulses to the brain. Its not really C, Because the </em><u>"Cochlea" </u> is a <em><u>spiral cavity that is inside of the inner ear.</u></em><em> And its not D, because the </em><u><em>"Stirrup" is the stapes or stirrup is on the bone, by the middle ear of humans and mammels too.</em></u><em> So your best answer will be Auditory Nerve.</em>
15 to 35 degrees of equator
Answer:
a) 7200 ft/s²
b) 140 ft
c) 3.7 s
Explanation:
(a) Average acceleration is the change in velocity over change in time.
a_avg = Δv / Δt
We need to find what velocity the puck reached after it was hit by the hockey player.
We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s². Therefore:
v² = v₀² + 2a(x − x₀)
(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)
v₀² = 5200 ft²/s²
v₀ = 20√13 ft/s
So the average acceleration impacted to the puck as it is struck is:
a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)
a_avg = 2000√13 ft/s²
a_avg ≈ 7200 ft/s²
(b) The distance the puck travels before stopping is:
v² = v₀² + 2a(x − x₀)
(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)
x = 140 ft
(c) The time the puck takes to travel 10 ft without friction is:
t = (10 ft) / (20√13 ft/s)
t = (√13)/26 s
The time the puck travels over the rough ice is:
v = at + v₀
(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)
t = √13 s
So the total time is:
t = (√13)/26 s + √13 s
t = (27√13)/26 s
t ≈ 3.7 s
Answer:
4.15 x 10^6 N
Explanation:
Area, A = 1.43 cm^2 = 1.43 x 10^-4 m^2
mass, m = 60.5 kg
Weight, F = m g = 60.5 x 9.8 = 592.9 N
Pressure = Force / Area
P = Weight / Area
P = 592.9 / (1.43 x 10^-4)
P = 4.15 x 10^6 N
