366J. Energy is conserved and is transformed from completely potential energy at the top to completely kinetic at the bottom.
Assuming the same deformation and elastic modulus, we can use the relationship:
(P/D)1 = (P/D)2
Where P is the load, Dis the diameter of the wire, 1 is the first wire, and 2 is the second wire. Using the given values and solving for the diameter of the second wire:
D = (24 / 0.24) (1)
D = 100 mm
Answer:
The speed of the particle is 13767.35 m/s.
Explanation:
given the mass m = 6.67×10^-27 kg of particle and moving with a speed v. the momentum of the particle is given by P = m×v and the planck constant h = 6.63×10^-34 m^2×kg/s. the wavelength is λ = 7.22 pm ≈7.22×10^-12 m
The de Broglie wavelength is given by:
λ = h/P
λ = h/(m×v)
v = h/λ×m
= (6.63×10^-34)/[(7.22×10^-12)×(6.67×10^-27)]
= 13767.35 m/s
Therefore, the speed of the particle is 13767.35 m/s.
Answer:
5.74s
Explanation:
We can first solve for the initial angular velocity using the following formula
Where 
is the final angular velocity, 
is the angular acceleration and 
is the angular displacement
So for the wheel to get from 22.4 to -22.4 with angular acceleration of -7.8 then the time it takes must be
