The experiments that claimed to demonstrate cold fusion were found
to have been faulty by others who reviewed them. Also, nobody else
was
able to reproduce the finding in other laboratories. In the world
of
Science, this pretty much says that the initial claims were unfounded.
The particles can undergo small oscillations around x₂.
The given parameters;
- <em>initial energy of the particles = E₁</em>
- <em>final energy of the particles, E₂ = 0.33E₁</em>
The movement of the particles depends on the kinetic energy of the particles.
When kinetic energy of the particles is 100%, the particles can oscillate from x₁ to x₅.
However, when the total energy of this particles is reduced to one-third (¹/₃) or 33% of the initial energy of the particle, the oscillation of the particles will be reduced.
- The maximum position the particle can oscillate is x₅
- The half position the particles can oscillate is x₃
Since 33% is less than the half of the energy of the particle, the particle will oscillate between x₁ and x₂.
Thus, we can conclude that the particles can undergo small oscillations around x₂.
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Answer:
A) ( - 200t + 40 ) volts
B) b) anticlockwise , c) anticlockwise , d) clockwise , e) clockwise
Explanation:
Given data:
magnetic flux (Φm) = 5.0t^2 − 2.0t
number of turns = 20
<u>a) determine induced emf </u>
E = - N 
= - N ( 10t - 2 ) = - 20 ( 10t - 2 )
= - 200t + 40 volts
<u>b) Determine direction of induced current </u>
i) at t = 0
E = - 0 + 40 ( anticlockwise direction )
ii) at t = 0.10
E = -20 + 40 = 20 ( anticlockwise direction )
iii) at t = 1
E = - 200 + 40 = - 160 ( clockwise direction)
iv) at t = 2
E = -400 + 40 = - 360 ( clockwise direction )
Answer:
uniform acceleration
Explanation:
The definition for uniform acceleration is:
if an object travels in a straight line and its velocity increases or decreases by equal amounts in equal intervals of time, then the acceleration is said to be uniform.
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Answer:
The peak-to-peak ripple voltage = 2V
Explanation:
120V and 60 Hz is the input of an unfiltered full-wave rectifier
Peak value of output voltage = 15V
load connected = 1.0kV
dc output voltage = 14V
dc value of the output voltage of capacitor-input filter
where
V(dc value of output voltage) represent V₀
V(peak value of output voltage) represent V₁
V₀ = 1 - (
)V₁
make C the subject of formula
V₀/V₁ = 1 - (1 / 2fRC)
1 / 2fRC = 1 - (v₀/V₁)
C = 2fR ((1 - (v₀/V₁))⁻¹
Substitute for,
f = 240Hz , R = 1.0Ω, V₀ = 14V , V₁ = 15V
C = 2 * 240 * 1 (( 1 - (14/15))⁻¹
C = 62.2μf
The peak-to-peak ripple voltage
= (1 / fRC)V₁
= 1 / ( (120 * 1 * 62.2) )15V
= 2V
The peak-to-peak ripple voltage = 2V