Answer:
Gray-ish looking like as if the atmosphere were full of only pollution.
Answer:
<em>v=14 m/s</em>
Explanation:
<u>Mechanical Energy
</u>
The kinetic energy of a body (K) is the capacity of doing work due to its speed. It can be expressed as
The potential energy (U) is the capacity of doing work due to its height respect to a certain reference level.
The mechanical energy is the sum of both
The principle of conservation of mechanical energy states it must remain the same if no external force is acting on it. The diver drops from the diving board, which means its initial speed is zero (and so its initial kinetic energy). Thus, the mechanical energy at the jumping time is
When the diver is about to get into the water, his height reaches zero and the speed is at maximum. All the potential energy became kinetic energy, so
Rearranging
The final speed of the diver is
Mass of an electron = 9.110 x 10⁻³¹ kg.
Mass of a proton = 1.6727 x 10⁻²⁷ kg
∴ mass of a proton/mass of an electron = 1.6727 x 10⁻²⁷ kg/9.110 x 10⁻³¹ kg.
~1836
∴ mass of a proton = 1836 x mass of an electron.
∴ mass of an electron is insignificant to the mass of an atom.
∴mass of an atom = mass of protons + mass of neutrons
The answer should be A weighlifter lifting 100lb. bell bar from his chest to a certain height.
Answer:
True
Explanation:
If a thin, spherical, conducting shell carries a negative charge, We expect the excess electrons to mutually repel one another, and, thereby, become uniformly distributed over the surface of the shell. The electric field-lines produced outside such a charge distribution point towards the surface of the conductor, and end on the excess electrons. Moreover, the field-lines are normal to the surface of the conductor. This must be the case, otherwise the electric field would have a component parallel to the conducting surface. Since the excess electrons are free to move through the conductor, any parallel component of the field would cause a redistribution of the charges on the shell. This process will only cease when the parallel component has been reduced to zero over the whole surface of the shell
According to Gauss law
∅ = EA =-Q/∈₀
Where ∅ is the electric flux through the gaussian surface and E is the electric field strength
If the gaussian surface encloses no charge, since all of the charge lies on the shell, so it follows from Gauss' law, and symmetry, that the electric field inside the shell is zero. In fact, the electric field inside any closed hollow conductor is zero
