Answer:
A conductor allows free charges to move about within it. The electrical forces around a conductor will cause free charges to move around inside the conductor until static equilibrium is reached. Any excess charge will collect along the surface of a conductor. Conductors with sharp corners or points will collect more charge at those points.
Explanation:
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
Answer:
The value is 
Explanation:
From the question we are told that
The time taken to travel to the planet from earth is 
The time to be spent on the ship is
Generally speed can be obtained using the mathematical relation represented below

The 2 in the equation show that the trip is a round trip i.e going and coming back
=> 
=> 
Answer:

Explanation:
The work function of the metal corresponds to the minimum energy needed to extract a photoelectron from the metal. In this case, it is:

So, the energy of the incoming photon hitting on the metal must be at least equal to this value.
The energy of a photon is given by

where
h is the Planck's constant
c is the speed of light
is the wavelength of the photon
Using
and solving for
, we find the maximum wavelength of the radiation that will eject electrons from the metal:

And since
1 angstrom = 
The wavelength in angstroms is

Answer:
3000 J
Explanation:
Kinetic energy is:
KE = ½ mv²
If m = 15 kg and v = -20 m/s:
KE = ½ (15 kg) (-20 m/s)²
KE = 3000 J