Given:
Object in circular motion 25 m/s
1 second to go quarter circle
Required:
Centripetal acceleration:
Solution:
Acceleration = v2/r
Where v is the velocity and r is
the radian
Substituting the values into the
equation,
Acceleration = v2/r = (25
m/s)2/(4*pi/180) = 8952.47 m2/s2
de Broglie wavelength (λ) is given by the equation
λ = h/p
where h=Planck’s constant whose value is 6.62 x 10^(−34) joule-seconds and
p = momentum of the particle(here electron)
In terms of kinetic energy(E) momentum(p) can be written as,
p=(2mE)^1/2
where m=mass of the particle.
Hence λ becomes
1 λ = h(2mE)^-1/2
Given here, E = 13.6 eV = 13.6×1.6×10^-19 joule
m(mass of electron)= 9.1×10^-31 kg
Putting these values in equation (1) we get ,
λ =0.332×10^(-9) meter
=3.32×10^(-10) meter
=3.32 Å
Answer:
right after jumping so 0s and as it falls its speed will keep increasing
Explanation:
The best and most correct answers among the choices provided by your question are he second and third choices.
<span>The velocity at any instant the average velocity during some time interval cannot be obtained from the graph alone.</span>
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