An electric fan is turned off, and its angular velocity decreases uniformly from 500 rev/min to 200 rev/min in 4.00 s.
1 answer:
Answer:
a) -1.25 rev/s² and 23.3 rev
b) 2.67s
Explanation:
a) ω = (500 rev/min)(1min/ 60s) => 8.333 rev/s
ω= (200 rev/min)(1min/ 60s) => 3.333rev/s
time 't'= 4 s
angular acceleration 'α'=?
constant angular acceleration equation is given by,
ω= ω
+ α
t
α= (ω
- ω
)/t => (3.333-8.333)/4
α= -1.25 rev/s²
θ-θ = ω
t + 1/2α
t²
=(8.333)(4) + 1/2 (-1.25)(4)²
=23.3 rev
b) ω=0 (comes to rest)
ω = 3.333 rev/s
α= -1.25 rev/s²
ω= ω
+ α
t
t= (ω - ω
)/α
=> (0- 3.333)/-1.25
t= 2.67s
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Answer:
the light emitting must be of greater wavelength
Explanation:
For this exercise we must use the Planck equation
E = h f
And the speed of light
c = λ f
f = c / λ
We replace
E = h c / λ
The wavelength of the green light is of the order of 500 nm, let's calculate the energy
E = 6.63 10⁻³⁴ 3 10⁸ /λ
E = 1,989 10⁻²⁵ /λ
λ = 500 nm = 500 10⁻⁹ m
E = 1,989 10⁻²⁵ / 500 10⁻⁹
E = 3,978 10⁻¹⁹ J
That is the energy of the transition for a transition is an intermediate state the energy must be less, this implies that the wavelength must increase. For the explicit case of a state with half of this energy
= E / 2
= 3,978 10⁻¹⁹ / 2 = 1,989 10⁻¹⁹
Let's clear and calculate
λ = h c / E
λ = 1,989 10⁻²⁵ / 1,989 10⁻¹⁹
λ = 1 10⁻⁶ m
Let's reduce to nm
λ = 1000 nm
This wavelength is in the infrared region
the light emitting must be of greater wavelength