The tension in a string from which a 4.0 kg object is suspended in an elevator is equal to 44 N. What is the acceleration of the
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Answer:
a) In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.
b) The new output is 86% of the old output
c) The losses in the new line are 74% the losses in the old line.
Explanation:
a) To relate the turns of primary and secondary to the ratio of voltage we have this expression:
In the old transformer the ratio of voltages was:
In the new transformer the ratio of voltages is:
In the new transformer there are 42 turns in the secondary per turn in the primary, while in the old transformer there were 32 turns per turn in the primary.
b) The new current ratio is
If the old current output was 425 kV, the ratio of current was:
Then, the ratio of the new output over the old output is:
The new output is 86% of the old output (smaller output currents lower the losses on the transmission line).
c) The power loss is expressed as:
Then, the ratio of losses is (R is constant for both power losses):
The losses in the new line are 74% the losses in the old line.
Answer:
a) a = 34.375 m / s², b) v_f = 550 m / s
Explanation:
This problem is the launch of projectiles, they tell us to ignore the effect of the friction force.
a) Let's start with the final part of the movement, which is carried out from t= 16 s with constant speed
v_f =
we substitute the values
v_f =
The initial part of the movement is carried out with acceleration
v_f = v₀ + a t
x₁ = x₀ + v₀ t + ½ a t²
the rocket starts from rest v₀ = 0 with an initial height x₀ = 0
x₁ = ½ a t²
v_f = a t
we substitute the values
x₁ = 1/2 a 16²
x₁ = 128 a
v_f = 16 a
let's write our system of equations
v_f =
x₁ = 128 a
v_f = 16 a
we substitute in the first equation
16 a =
16 4 a = 6600 - 128 a
a (64 + 128) = 6600
a = 6600/192
a = 34.375 m / s²
b) let's find the time to reach this height
x = ½ to t²
t² = 2y / a
t² = 2 5100 / 34.375
t² = 296.72
t = 17.2 s
We can see that for this time the acceleration is zero, so the rocket is in the constant velocity part
v_f = 16 a
v_f = 16 34.375
v_f = 550 m / s