Answer:
-10.9 rad/s²
Explanation:
ω² = ω₀² + 2α(θ - θ₀)
Given:
ω = 13.5 rad/s
ω₀ = 22.0 rad/s
θ - θ₀ = 13.8 rad
(13.5)² = (22.0)² + 2α (13.8)
α = -10.9 rad/s²
= Heat released to cold reservoir
= Heat released to hot reservoir
= maximum amount of work
= temperature of cold reservoir
= temperature of hot reservoir
we know that
eq-1
maximum work is given as
= -
using eq-1
= 
-
Answer:
Explanation:
(1.7 m/cycle)(46 cycle/s) = 78.2 m/s
Which amplitude of the following longitudinal waves has the greatest energy?
amplitude = 10 cm; wavelength = 6 cm; period = 4 seconds
Ey = 375 cos [ kx - (92.20x10^14).t ]
<span>the formula of the electric field of an electromagnetic wave : </span>
<span>=> Ey = Emax cos (k x - ω t) </span>
<span>=> Ey = Emax cos [(2π/λ) - (2π f).t ] </span>
<span>then : (2π f).t = (92.20x10^14).t </span>
<span>===> 2π f = 92.20x10^14 </span>
<span>===> f = 92.20x10^14 / 2π </span>
<span>===> f = 1.46 x 10^15 hertz </span>
<span>the speed of electromagnetic wave : c = 3 x10^8 m/s </span>
<span>then : c = f λ </span>
<span>======> λ = c / f </span>
<span>======> λ = 3 x10^8 / 1.46 x 10^15 </span>
<span>======> λ = 2.0547 x 10^-7 meters = 0.205 μm</span>
