Answer:
Magnitude of the net force on q₁-
Fn₁=1403 N
Magnitude of the net force on q₂+
Fn₂= 810 N
Magnitude of the net force on q₃+
Fn₃= 810 N
Explanation:
Look at the attached graphic:
The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.
Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:
F= (k*q*q)/(d)²
F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N
Magnitude of the net force on q₁-
Fn₁x= 0
Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N
Fn₁=1403 N
Magnitude of the net force on q₃+
Fn₃x= 810- 810 cos 60° = 405 N
Fn₃y= 810*sin 60° = 701.5 N
Fn₃ = 810 N
Magnitude of the net force on q₂+
Fn₂ = Fn₃ = 810 N
Answer:
Hydrated iron(III) oxide, or ferric oxide
Explanation:
the answer would be B. the back emf increases, and the current drawn from the socket increases
more current is needed to make the motor move, like when you try to self crank a motor and the back wires are touching its harder to crank. and the emf increases since more current is being drawn in, strengthening the emf or increasing the emf
Answer:
241 kPa
Explanation:
The ideal gas law states that:
where
p is the gas pressure
V is its volume
n is the number of moles
R is the gas constant
T is the absolute temperature of the gas
We can rewrite the equation as
For a fixed amount of gas, n is constant, so we can write
Therefore, for a gas which undergoes a transformation we have
where the labels 1 and 2 refer to the initial and final conditions of the gas.
For the sample of gas in this problem we have
So we can solve the formula for 
, the final pressure:
