A lens of focal length 15.0 cm
1 answer:
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Answer:
Explanation:
ω =
k = 2.5 N/m
m = 10 kg
ω = .5 rad /s
x(t) = A cos(ωt + φ₀)
When t = 0 , x(t) = 0
0 = A cos(ωx 0 + φ₀)
cos φ₀ = 0
φ₀ = π /2
x(t) = A cos(ωt +π /2 )
Putting the value of ω
x(t) = A cos(.5 t +π /2 )
Differentiating on both sides
dx(t)/dt = - .5 A sin(.5 t +π /2 )
v(t) = - .5 A sin(.5 t +π /2 )
Given t =0 , v(t) = -5 m/s
-5 = - .5 A sin(.5 x0 +π /2 )
-5 = - .5 A sinπ /2
A = 10 m
x(t) = 10 cos( .5 t +π /2 )
b )
when t = π ( 3.14 s )
x(t) = - 10 m
when t = 2π ( 6.28s )
x(t) = 0
when t = 3π ( 9.42 s )
x(t) = 10 m
and so on
When an electron is accelerated through potential difference then the speed that it attain will be explained by energy conservation
here by energy conservation we can say that
change in kinetic energy of electron = electrostatic potential energy gained through given potential difference
kinetic energy is given as
electrostatic potential energy is given as
now by energy conservation
given that for electron
now by plug in values
So here it is accelerated through potential difference of 2559.4 Volts