The answer is 'Singularity'. It is the one dimensional point where huge mass with density unmeasured, gravity being infinite lying at the center point of this cosmos entity. It is also described as the point where the laws of physics does not apply.
Even though the object is weightless, it would need inertia, I.e, you pushing it or any form of transportation. So you would still have to push that 500kg to just keep it moving in space. Say if it were a planet with less gravitational force, it would be weightless.
Answer:
To increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.
Explanation:
First, we have to calculate the work function of the element. The maximum kinetic energy as a function of the wavelength is given by:
Here h is the Planck's constant, c is the speed of light, is the wavelength of the light and W the work function of the element:
Now, we calculate the wavelength for the new maximum kinetic energy:
This wavelength corresponds to ultraviolet radiation. So, to increase the maximum kinetic energy of electrons to 1.5 eV, it is necessary that ultraviolet radiation of 354 nm falls on the surface.
The starting angle θθ of a pendulum does not affect its period for θ<<1θ<<1. At higher angles, however, the period TT increases with increasing θθ.
The relation between TT and θθ can be derived by solving the equation of motion of the simple pendulum (from F=ma)
−gsinθ=lθ¨−gainθ=lθ¨
For small angles, θ≪1,θ≪1, and hence sinθ≈θsinθ≈θ. Hence,
θ¨=−glθθ¨=−glθ
This second-order differential equation can be solved to get θ=θ0cos(ωt),ω=gl−−√θ=θ0cos(ωt),ω=gl. The period is thus T=2πω=2πlg−−√T=2πω=2πlg, which is independent of the starting angle θ0θ0.
For large angles, however, the above derivation is invalid. Without going into the derivation, the general expression of the period is T=2πlg−−√(1+θ2016+...)T=2πlg(1+θ0216+...). At large angles, the θ2016θ0216 term starts to grow big and cause
Explanation:
The position vector r:
The velocity vector v:
The acceleration vector a: