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Colt1911 [192]
3 years ago
9

At some instant, a particle traveling in a horizontal circular path of radius 7.90 m has a total acceleration with a magnitude o

f 15.0 m/s2 and a constant tangential acceleration of 12.0 m/s2. Determine the speed of the particle at this instant and (1/8) revolution later.
Physics
1 answer:
DochEvi [55]3 years ago
5 0

Answer:

a) Speed of the particle at this instant

v = 8.43 m/s

b) Speed of the particle at  (1/8) revolution later

v = 14.83 m/s

Explanation:

We apply the equations of circular motion uniformly accelerated :

(a_{T}) ^{2} = (a_{n} )^{2} +(a_{t} )^{2} Formula (1)

a_{n} = \frac{v^{2} }{r} Formula (2)

a_{t} = \alpha *r Formula (3)

v= ω*r Formula (4)

ω² = ω₀² + 2*α*θ  Formula (5)

Where:

a_{T} :  total acceleration, (m/s²)

a_{n} : normal acceleration, (m/s²)

a_{t} :  tangential acceleration, (m/s²)

\alpha : angular acceleration (rad/s²)

r : radius of the circular path (m)

v : tangential velocity (m/s)

ω : angular speed ( rad/s)

ω₀: initial angular speed  ( rad/s)

θ : angle that the particle travels (rad)

Data:

a_{T} = 15 m/s²

a_{t} =  12 m/s²

r=7.90 m  :radius of the circular path

Problem development

In the formula (1) :

a_{n} = \sqrt{(a_{T})^{2} -(a_{t})^{2} }

We replace the data

a_{n} = \sqrt{(15)^{2} -(12)^{2}}

a_{n} = 9 \frac{m}{s^{2} }

We use formula (2)  to calculate v:

9 = \frac{v^{2} }{7.9}  Equation (1)

a)Speed of the particle at this instant

in the equation (1):

v=\sqrt{9*7.9} = 8.43 \frac{m}{s}

b)Speed of the particle at  (1/8) revolution later

We know the following data:

θ =(1/8) revolution=( 1/8) *2π= π/4

a_{t} =  12 m/s²

v₀= 8.43 m/s

r=7.9 m

We use formula (3) to calculate α

12 = \alpha *7.90

\alpha =\frac{12}{7.9} = 1.52  \frac{rad}{s^{2} }

We use formula (4) to calculate ω₀

v₀= ω₀ *r

8.43 =  ω₀*7.9

ω₀ = 8.43/7.9 = 1.067 rad/s

We use formula (5) to calculate ω

ω² = ω₀² +  2*α*θ  

ω²=  (1.067)² + 2*1.52*π/4

ω² =3.526

ω = 1.87 rad/s

We use formula (4) to calculate v

v= 1.87 rad/s * 7.9m

v = 14.83 m/s : speed of the particle at  (1/8) revolution later

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Answer:

The frecuency and wavelength of a photon capable to ionize the nitrogen atom are ν = 3.394×10¹⁵ s⁻¹  and λ = 88.31 nm.

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λ =  (4.136x10⁻¹⁵ eV.s)(3x10¹⁷ nm/s)/14.04 eV

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So the frecuency of a photon, capable to ionize the nitrogen atom, will be 3.394×10¹⁵ s⁻¹ and its wavelength 88.31 nm.

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