Answer:
The energy of this particle in the ground state is E₁=1.5 eV.
Explanation:
The energy
of a particle of mass <em>m</em> in the <em>n</em>th energy state of an infinite square well potential with width <em>L </em>is:

In the ground state (n=1). In the first excited state (n=2) we are told the energy is E₂= 6.0 eV. If we replace in the above equation we get that:

So we can rewrite the energy in the ground state as:



Finally

Answer:
a) v = √ g x
, b) W = 2 m g d
, c) a = ½ g
Explanation:
a) For this exercise we use Newton's second law, suppose that the block of mass m moves up
T-W₁ = m a
W₃ - T = M a
w₃ - w₁ = (m + M) a
a = (3m - m) / (m + 3m) g
a = 2/4 g
a = ½ g
the speed of the blocks is
v² = v₀² + 2 ½ g x
v = √ g x
b) Work is a scalar, therefore an additive quantity
light block s
W₁ = -W d = - mg d
3m heavy block
W₂ = W d = 3m g d
the total work is
W = W₁ + W₂
W = 2 m g d
c) in the center of mass all external forces are applied, they relate it is
a = ½ g
The relationship between resistance R and resistivity

is

where L is the length of the wire and A its cross section.
The radius of the wire is half the diameter:

and the cross section is

From the first equation, we can then find the length of the wire when

(copper resistivity:

)
Answer:

Explanation:
According to the statement of the problems, the following identity exists:


After some algebraic handling, the ratio is obtained:


The mass of the object doesn't matter. The change in its momentum is equal to the impulse that changed it ... 15 N-sec.