Answer:
The net torque is 0.0372 N m.
Explanation:
A rotational body with constant angular acceleration satisfies the kinematic equation:
(1)
with ω the final angular velocity, ωo the initial angular velocity, α the constant angular acceleration and Δθ the angular displacement (the revolutions the sphere does). To find the angular acceleration we solve (1) for α:

Because the sphere stops the final angular velocity is zero, it's important all quantities in the SI so 2.40 rev/s = 15.1 rad/s and 18.2 rev = 114.3 rad, then:

The negative sign indicates the sphere is slowing down as we expected.
Now with the angular acceleration we can use Newton's second law:
(2)
with ∑τ the net torque and I the moment of inertia of the sphere, for a sphere that rotates about an axle through its center its moment of inertia is:
With M the mass of the sphere an R its radius, then:

Then (2) is:

The partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
<u>Explanation:</u>
H₂ + Br₂ ⇒ 2HBr
PH₂ = 0.782atm
PBr₂ = 0.493atm
Kp = (PHBr)²/ (PH₂) (PBr₂) = 1.4 X 10⁻²¹
At equilibrium:
Let 2x = pressure of HBr
PH₂ = 0.782 -x
PBr₂ = 0.493 - x
Kp = (2x)^2 / (0.782-x)(0.493-x)
Now, because Kp is very small, x will be very small compared to 0.782 and 0.493.
Then,
Kp = 1.4X10⁻²¹ = (4x²) / (0.782)(0.493)
x = 1.2X10⁻¹¹
PHBr = 2x = 2.4 X 10⁻¹¹ atm
Therefore, the partial pressures of HBr when the system reaches equilibrium is 2.4 X 10⁻¹¹ atm
Answer:
The electric field strength is 
Solution:
As per the question:
Area of the electrode, 
Charge, q = 50 nC = ![50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m](https://tex.z-dn.net/?f=50%5Ctimes%2010%5E%7B-%209%7D%20C%5B%2Fetx%5D%3C%2Fp%3E%3Cp%3EDistance%2C%20x%20%3D%202%20mm%20%3D%20%5Btex%5D2%5Ctimes%2010%5E%7B-%203%7D%20m)
Now,
To calculate the electric field strength, we first calculate the surface charge density which is given by:

Now, the electric field strength of the electrode is:

where



Heat flows irreversibly from hot to cold