A 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same.
1 answer:
As the temperature of the lead and helium is the same. Thus the average kinetic energy is also the same for lead and helium.
Reason:
It is given that a 5.0-kg bar of lead is placed inside a 12-L chamber filled with helium gas. The temperature of the lead and helium is the same. It is required to compare the average kinetic energy of the lead atoms and helium atoms.
The average kinetic energy is calculated as, .
Here K is the average kinetic energy, R is the gas constant, N is the Avogadro's number, and T is the temperature.
As the temperature is the same for both lead and helium. As a result, the average kinetic energy is also the same for lead and helium.
Learn more about average kinetic energy here,
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Answer:
a) E_total = 6,525 10⁴ N /C ,field direction is radial outgoing
b) E_total = 1.89 10⁶ N / C, field is incoming radial
c) E_total = 0
Explanetion:
For this exercise we can use that the charge in a spherical shell can be considered concentrated at its center and that the electric field inside the shell is zero, since Gauss's law is
Ф = E .dA = q_{int} /ε₀
inside the spherical shell there are no charges
The electric field is a vector quantity, so we calculate the field created by each shell and add it vectorly.
We have two sphere shells with radii 0.050m and 0.15m respectively
a) point where you want to know the electric field d = 0.20 m
shell 1
the point is on the outside,d>ro, therefore we can consider the charge to be concentrated in the center
E₁ = k q₁ / d²
shell 2
the point is on the outside,d>ro
E₂ = k q₂ / d²
the total camp is
E_total = -E₁ + E₂
E_total = k ( )
E_total = 9 10⁹ (-2.1 10⁻⁶+ 5 10⁻⁶ / .2²
E_total = 6,525 10⁵ N /C
The field direction is radial and outgoing ti the shells
b) the calculation point is d = 0.10m
shell 1
point outside the shell d> ro
E₁ = k q₁ / d²
shell 2
the point is inside the shell d <ro
Therefore, according to Gauss's law, since there are no charges in the interior, the electrioc field is zero
E₂ = 0
E_total = E₁
E_total = k q₁ / d²
E_total = 9 10⁹ 2.1 10⁻⁶ / 0.1²
E_total = 1.89 10⁶ N / A
As the charge is negative, this field is incoming radial, that is, it is directed towards the shell 1
c) the point of interest d = 0.025 m
shell 1
point is inside the shell d< ro
as there are no charges inside
E₁ = 0
shell 2
point is inside the radius of the shell d <ro
E₂ = 0
the total field is
E_total = 0
Answer:
Explanation:
The force on the point charge q exerted by the rod can be found by Coulomb's Law.
Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.
In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.
We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.
Applying Coulomb's Law:
The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.
Now, we have to write 'dq' in term of the known quantities.
Now, substitute this into 'dF':
Now we can integrate dF over the rod.
Answer:
5.82812 rad/s
Explanation:
L = Length of meter stick = 1 m = 100 cm
= The center of mass of the stick =
= Angular velocity
Moment of inertia of the system is given by
As the energy in the system is conserved
The maximum angular velocity is 5.82812 rad/s