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3 years ago

What is the only bone of the skull that moves?

Physics

2 answers:

NikAS [45]3 years ago

8 0

Answer:

mandible

Explanation:

which is also the lower jaw and it allows the mouth to open and close

hope i helped

Darina [25.2K]3 years ago

3 0

Answer:

Mandible.

Explanation:

The mandible in the bone that allows you to open and close your mouth, otherwise known ad the jawbone.

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(a) Write an equation describing a sinusoidal transverse wave traveling on a cord in the positive of a y axis with an angular wa

Vedmedyk [2.9K]

Missing data: the wave number

$k=60 cm^{-1}$

(a) $z = 0.003 sin (6000y-31.4 t)$

For a transverse wave travelling in the positive y-direction and with vibration along the z-direction, the equation of the wave is

$z = A sin (ky-\omega t)$

where

A is the amplitude of the wave

k is the wave number

$\omega$ is the angular frequency

t is the time

In this situation:

A = 3.0 mm = 0.003 m is the amplitude

$k = 60 cm^{-1} = 6000 m^{-1}$ is the wave number

$T = 0.20 s$ is the period, so the angular frequency is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.20}=31.4 rad/s$

So, the wave equation (in meters) is

$z = 0.003 sin (6000y-31.4 t)$

(b) 0.094 m/s

For a transverse wave, the transverse speed is equal to the derivative of the displacement of the wave, so in this case:

$v_t = z' = -A \omega cos (ky-\omega t)$

So the maximum transverse wave occurs when the cosine term is equal to 1, therefore the maximum transverse speed must be

$v_{t}_{max} =\omega A$

where

$\omega = 31.4 rad/s\\A = 0.003 m$

Substituting,

$v_{t}_{max}=(31.4)(0.003)=0.094 m/s$

(c) 5.24 mm/s

The wave speed is given by

$v=f \lambda$

where

f is the frequency of the wave

$\lambda$ is the wavelength

The frequency can be found from the angular frequency:

$f=\frac{\omega}{2\pi}=\frac{31.4}{2\pi}=5 Hz$

While the wavelength can be found from the wave number:

$\lambda = \frac{2\pi}{k}=\frac{2\pi}{6000}=1.05\cdot 10^{-3} m$

Therefore, the wave speed is

$v=(5)(1.05\cdot 10^{-3} )=5.24 \cdot 10^{-3} m/s = 5.24 mm/s$

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3 years ago

Explain what happens to the energy in a group in a system if one object loses energy according to the Law of Conservation of Ene

konstantin123 [22]

energy never disappears, for example, if you give some kinetic energy to a ball and it stops few seconds later, friction steals this energy to ground which ball was going on. "Law of Conservation of Energy" tell us that energy can't disappear

5 0

3 years ago

A spherical Christmas tree ornament is 8.00 cm in diameter. What is the magnification of an object placed 12.0 cm away from the

LiRa [457]

The magnification of the ornament is 0.25

To calculate the magnification of the ornament, first, we need to find the image distance.

Formula:

1/f = u⁻¹+v⁻¹.................... Equation 1

Where:

f = Focal length of the ornament
u = image distance
v = object distance.

make u the subject of the equation

u = fv/(f+v)................ Equation 2

From the question,

Given:

f = 8/2 = 4 cm
v = 12 cm

Substitute these values into equation 2

u = (12×4)/(12+4)
u = 48/16
u = 3 cm.

Finally, to get the magnification of the ornament, we use the formula below.

M = u/v.................. Equation 3

Where

M = magnification of the ornament.

Substitute these values above into equation 3

M = 3/12
M = 0.25.

Hence, The magnification of the ornament is 0.25

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2 years ago

What is the direction of the normal contact force of the road on the wheels?

Gekata [30.6K]

Answer:

The direction of the contact forces acting on a body is not necessarily perpendicular to the contact surface. The resolution of contact forces in two components i.e. perpendicular to contact surface and along surface. Perpendicular component is normal force and parallel component is friction.

Explanation:

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2 years ago

Someone please help me

schepotkina [342]

The answer is oil and water. If you want I could give you the reason too

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3 years ago

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