A 60-Hz single-phase transformer with capacity of 150 kVA has the following parameters: RP = 0.35 Ω RS = 0.002 Ω Rc = 5.2 kΩ XP
1 answer:
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Answer:
e) v₁ = 29.7 m / s
Explanation:
Let's propose the solution of the problem, let's start at the moment
Initial
p₀ = m v₁
Final
= (m + M) v
The moment is preserved
p₀ =
m v₁ = (m + M) v
v = m / (m + M) v₁ (1)
a) energy is conserved
Let's look for kinetic energy
Initial
K₀ = ½ m v₁²
Final
= ½ (m + M) v²
Let's replace v
= ½ (m + M) [m / (m + M) v₁]²
= ½ m² / (m + M) v₁²
Let's look for the relationship of these energies
Ko / = ½ m v₁² / (½ m² / (m + M) v₁²)
Ko / = (m + M) / m = (1 + M / m)²
The kinetic energy changes therefore it is not conserved in the process, the missing energy is converted into potential heat energy during the crash
b) The impulse is conserved because the system is defined as formed by the two bodies and the externals are of action and reaction, so for the complete system the sum is zero and the moment does not change in value
c) in this case the system is already formed by the two bodies and since there is no rubbing the mechanical energy is conserved, transforming from kinetics to potential
d) when the pendulum oscillates the speed changes from v to zero, so the moment is not conserved, this is because there is an external force acting on the system, the force of gravity
e) For this part let's start at the end of the movement
It is system (bullet + block) moves, energy is conserved
Final. Highest point
= U = (m + M) g h
Initial. Lowest point
Em₀ = K = ½ (m + M) v2
Em₀ =
½ (m + M) v² = (m + M) g h
v = √ 2gh
Let's look for the height (h) by trigonometry
Cos 15 = x / L
h = L-x
h = L - L cos 15
h = L (1- cos 15)
We replace
v = √ (2gL (1- cos 15))
Now we use equation (1) of momentum conservation
v = m / (m + M) v1
v₁ = (m + M) / m v
v1 = (0.1 +2.0) /0.1 RA (2 9.8 3 (1- cos 15))
v₁ = 21 √ (2.00)
v₁ = 29.7 m / s