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
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
Answer:
Explanation:
a ) No of turns per metre
n = 450 / .35
= 1285.71
Magnetic field inside the solenoid
B = μ₀ n I
Where I is current
B = 4π x 10⁻⁷ x 1285.71 x 1.75
= 28.26 x 10⁻⁴ T
This is the uniform magnetic field inside the solenoid.
b )
Magnetic field around a very long wire at a distance d is given by the expression
B = ( μ₀ /4π ) X 2I / d
= 10⁻⁷ x 2 x ( 1.75 / .01 )
= .35 x 10⁻⁴ T
In the second case magnetic field is much less. It is due to the fact that in the solenoid magnetic field gets multiplied due to increase in the number of turns. In straight coil this does not happen .
Answer:
The following equation can be used.
(32°F − 32) × 5/9=C
