Explanation:
The emf is equal to the work done on the charge per unit charge (ϵ=dWdq) when there is no current flowing. Since the unit for work is the joule and the unit for charge is the coulomb, the unit for emf is the volt (1V=1J/C).
Answer:
The force will be zero
Explanation:
Due to the symmetric location of the +2μC charges the forces the excert over the +5μC charge will cancel each other resulting in a net force with a magnitude of zero.However in this case it would be an unstable equilibrium, very vulnerable to a kind of bucking. If the central charge is not perfectly centered on the vertical axis the forces will have components in that axis that will add together instead of canceling each other.
Answer: v = 4.4 m/s
Explanation:
In the absence of friction, the total mechanical energy will be constant
KE₀ + PE₀ = KE₁ + PE₁
0 + mg(6) = ½mv₁² + mg(5)
½mv₁² = mg(6 - 5)
v = √(2g(1)) = 4.4 m/s
Answer:
a) i₈ = 0.5 i₄, b) i₁₀ = 0.3 i₃, i₁₀ = 0.8 i₈
Explanation:
For this exercise we use ohm's law
V = i R
i = V / R
we assume that the applied voltage is the same in all cases
let's find the current for each resistance
R = 4 Ω
i₄ = V / 4
R = 8 Ω
i₈ = V / 8
we look for the relationship between these two currents
i₈ /i₄ = 4/8 = ½
i₈ = 0.5 i₄
R = 3 Ω
i₃ = V3
R = 10 Ω
i₁₀ = V / 10
we look for relationships
i₁₀ / 1₃ = 3/10
i₁₀ = 0.3 i₃
i₁₀ / 1₈ = 8/10
i₁₀ = 0.8 i₈
Answer:
The force is 
Explanation:
The diagram for this question is shown on the first uploaded image
At Equilibrium the summation of the of force on the vertical axis is zero
i.e 
=> 
is the is the speed of water at the nozzle which can be mathematically evaluated as
substituting 
for R and 
for 
is the is the speed of water at the pipe which can be mathematically evaluated as
substituting for R and 
for 
is he density of water with value 
Substituting values into the equation above
At Equilibrium the summation of the of force on the horizontal axis is zero
i.e 
=> 
Since The speed at both A and B nozzle are the same then remains the same
Substituting values
=> 
Hence the force acting on the flange bolts required to hold the nozzle in place is
