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Burka [1]
3 years ago
6

Our balance is maintained, at least in part, by the endolymph fluid in the inner ear. Spinning displaces this fluid, causing diz

ziness. Suppose a dancer (or skater) is spinning at a very fast 3.0 revolutions per second about a vertical axis through the center of the head. Although the distance varies from person to person, the inner ear is approximately 7.0 cm from the axis of spin. What is the radial acceleration (in m/s and in g's) of the endolymph fluid?
Physics
1 answer:
zavuch27 [327]3 years ago
7 0

To solve this problem we will define the given angular velocity, in terms of international units, we will subsequently use the definition of radial acceleration, defined as the product between the square of the angular velocity and the radius. Finally we will convert the units to gravitational terms or units G.

PART A) Our values in SI are,

\omega = 3 rev/s

\omega = 3 \frac{rev}{s} (\frac{2\pi rad}{1 rev})

\omega = 18.85rad/s

Radial acceleration can be described as

a_c = \omega^2 R

a_c = (18.85)^2 (7*10^{-2})

a_c = 24.8726m/s^2

PART A) If we have that 1g is equivalent to 9.8m / s ^ 2 performing the conversion we have to

a_c = 24.8725 m/s^2 (\frac{1g}{9.8m/s^2})

a_c = 2.53g

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Electrons are ejected from a metallic surface with speeds of up to 4.60 3 105 m/s when light with a wavelength of 625 nm is used
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Complete question:

Electrons are ejected from a metallic surface with speeds ranging up to 4.60 x10⁵ m/s when light with a wavelength of 625nm is used. (a) What is the work function of the surface? (b) What is the cutoff frequency for this surface?

Answer:

Part(a) The work function of the surface is 22.177 x 10⁻²⁰ J = 1.384 eV

Part(b) The cutoff frequency for this surface is 3.347 x 10¹⁴ Hz

Explanation:

The kinetic energy (KE) of the emitted photon:

KE = 0.5mv²

m is mass of electron = 9.1 X 10⁻³¹ kg

KE = 0.5 * 9.1 X 10⁻³¹  * (460000)² = 9.628 X 10⁻²⁰ J

in eV = 9.628 X 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 0.601 eV

The photon energy of the incoming radiation:

E = hf = hc/λ

c is speed of light (photon) = 3 x 10⁸

h is Planck's constant = 6.626 × 10⁻³⁴ J.s

E = (6.626 × 10⁻³⁴ *3 x 10⁸) /(625 X 10⁻⁹)

E = 31.805 X 10⁻²⁰ J

in eV = 31.805 X 10⁻²⁰ J x  6.242 X 10¹⁸ ev = 1.985 eV

Part (a) the work function of the surface

KE = hf - W

where;

W is work function

W = hf - KE

W =  31.805 X 10⁻²⁰ J - 9.628 X 10⁻²⁰ J = 22.177 x 10⁻²⁰ J

in eV = 22.177 x 10⁻²⁰ J x 6.242 X 10¹⁸ ev = 1.384 eV

Part(b) the cutoff frequency for this surface

W =hf

f = W/h

f = (22.177 x 10⁻²⁰ J)/(6.626 × 10⁻³⁴ J.s)

f = 3.347 x 10¹⁴ Hz

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3 years ago
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Answer:

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It is the bowl that will have maximum potential energy.

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Unrestrained occupants of a car keep moving during the time the car takes to stop due to ________.
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Answer:

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