912.
outer ear:
pinna
ear canal
middle ear:
ossicles and ear drum
inner ear:
semcircular canals
cochlea
auditory nerve
13.
frequency = wavespeed ÷ wavelength
14.
if frequency increases you would experience a higher pitch in sound
15.
humans can hear 20Hz to 20kHz
16.
The Doppler effect is the change in frequency or wavelength of a wave for an observer who is moving relative to the wave source. Can be used for machines measuring speed via doppler effect.
17.
Doppler in hospitals can be used for ultrasound to provide images for diagnosis and monitoring.
Covalent bonds. Silicon, carbon, germanium, and a few other elements form covalently bonded solids. In these elements there are four electrons in the outer sp-shell, which is half filled. ... In the covalent bond an atom shares one valence (outer-shell) electron with each of its four nearest neighbour atoms.
Answer:
the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is ![P = 104.04 \hat{i} -314.432 \hat{j}](https://tex.z-dn.net/?f=P%20%3D%20%20104.04%20%5Chat%7Bi%7D%20-314.432%20%5Chat%7Bj%7D)
Explanation:
The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;
- the peripheral velocity that is directed downward
along the y-axis
- the linear velocity
that is directed along the x-axis
Now;
![V_x = \frac{d}{dt}(12t^3+2) = 36 t^2](https://tex.z-dn.net/?f=V_x%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%2812t%5E3%2B2%29%20%3D%2036%20t%5E2)
![V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s](https://tex.z-dn.net/?f=V_x%20%3D%2036%281.7%29%5E2%5C%5C%5C%5CV_x%20%3D%20104.04%5C%20ft%2Fs)
Also,
![-V_y = R* \omega](https://tex.z-dn.net/?f=-V_y%20%3D%20R%2A%20%5Comega)
where
(angular velocity) = ![\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)](https://tex.z-dn.net/?f=%5Cfrac%7Bd%5Ctheta%7D%7Bdt%7D%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%288t%5E4%29)
![-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s](https://tex.z-dn.net/?f=-V_y%20%3D%202%2A32t%5E3%29%5C%5C%5C%5C%5C%5C-V_y%20%3D%202%2A32%281.7%5E3%29%5C%5C%5C%5C-V_y%20%3D%20314.432%20%5C%20ft%2Fs)
∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is ![P = 104.04 \hat{i} -314.432 \hat{j}](https://tex.z-dn.net/?f=P%20%3D%20%20104.04%20%5Chat%7Bi%7D%20-314.432%20%5Chat%7Bj%7D)
Answer: The acceleration of the object is 0.67m/s^2 west.
Explanation: Here we are given the initial velocity and final velocity as well as the time taken. Acceleration is the change in velocity per unit time, thus the equation becomes.
a=dv/t
a=vf-vi/t
a=-2.1-4.7/3.9
a= 0.66m/s^2 west
applied forces would be push for example.
normal forces would seem to be a force such as gravity.
friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.