It has to be D because the arrow will drop as it moves, if it were a gun, you'd lead the target so fire below it, but due to it being an arrow, you aim high not low. Also, they didnt specify how fast anything is, so you'd probably miss if you actually did it.
Look first for the relation between deBroglie wavelength (λ) and kinetic energy (K):
K = ½mv²
v = √(2K/m)
λ = h/(mv)
= h/(m√(2K/m))
= h/√(2Km)
So λ is proportional to 1/√K.
in the potential well the potential energy is zero, so completely the electron's energy is in the shape of kinetic energy:
K = 6U₀
Outer the potential well the potential energy is U₀, so
K = 5U₀
(because kinetic and potential energies add up to 6U₀)
Therefore, the ratio of the de Broglie wavelength of the electron in the region x>L (outside the well) to the wavelength for 0<x<L (inside the well) is:
1/√(5U₀) : 1/√(6U₀)
= √6 : √5
Answer:
2.43J
Explanation:
Given parameters:
Mass of the arrow = 0.155kg
Velocity = 31.4m /s
Unknown:
Kinetic energy when it leaves the bow = ?
Solution:
The kinetic energy of a body is the energy in motion of the body;
it can be derived using the expression below:
K.E = 
m v²
m is the mass
v is the velocity
Solve for K.E;
K.E = x 0.155 x 31.4 = 2.43J
Explanation:
Because the temperature and the radiation are not correlated, they're not represented as functions of each other, they're represented as independent variables thus using graph 5 you cannot figure out how one affect another
