I believe it would decrease since the electrical energy is being split up. If it was in a series, the resistors ohms would be added up.
But the resistors put in parallel will be less than the lowest connected resistor. My dad showed me the equation once but I forgot it.
I hope this helps you out!
Answer:
Explanation:
A ) At constant volume :
ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .
For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J
ΔEint = .7 x 12.45 x ( 564 - 300 )
= 2300.76 J .
W₁ = 0 because volume is constant so work done by gas is zero .
Q₁ = ΔEint = 2300.76 J
B )
At constant pressure
Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .
For monoatomic gas ,
Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J
Q₂ = .7 x 20.75 x 264 J
= 3834.6 J
W₂ = work done by gas
= PΔV = nRΔT
= .8 x 8.3 x 264
= 1752.96 J
ΔEint = Q₂ - W₂
= 3834.6 - 1752.96
= 2081.64 J.
Answer:
Remember that the number of protons in the nucleus determines an element's identity. Chemical changes do not affect the nucleus, so chemical changes cannot change one type of atom into another. The number of protons in a nucleus does change sometimes, however. The identity of the atom, therefore, it is true
Answer:
The magnitude of the electric field is 0.1108 N/C
Explanation:
Given;
number of electrons, e = 8.05 x 10⁶
length of the wire, L = 1.03 m
distance of the field from the center of the wire, r = 0.201 m
Charge of the electron;
Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)
Q = 1.2896 x 10⁻¹² C
Linear charge density;
λ = Q / L
λ = (1.2896 x 10⁻¹² C) / (1.03 m)
λ = 1.252 x 10⁻¹² C/m
The magnitude of electric field at r = 0.201 m;
Therefore, the magnitude of the electric field is 0.1108 N/C
The answer is <em>C.) Elevated</em>