W=ΔKE , W=-5000j
KEinitial=(1/2)mv² , KEfinal=0j
ΔKE=-(1/2)mv²
-5000=-(1/2)(100kg)v²
v=10 m/s
Answer:
don't know her boi I'm finna glow october
Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
Path 2.
Displacement is the direction and magnitude of an object from its starting point, so path 2 is the direct route you would need to take to find direction and magnitude.
Answer:
b) The star is moving away from us.
Explanation:
If an object moves toward us, the light waves it emits are compressed - the wavelength of the light will be shorter, making the light bluer. On the other hand, if an object moves away from us, the light waves are stretched, making it redder. If from laboratory measurements we know that a specific hydrogen spectral line appears at the wavelength of 121.6 nanometers (nm) and the spectrum of a particular star shows the same hydrogen line appearing at the wavelength of 121.8 nm, we can conclude that the star is moving away from npos, since the wavelength related to that star is more expanded.
