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asambeis [7]
3 years ago
14

A transformer changes the 10,000 v power line to 120 v. if the primary coil contains 750 turns, how many turns are on the second

ary?
Physics
1 answer:
inn [45]3 years ago
5 0
For a transformer, the ratio between the number of turns of primary and secondary coil is the same as the ratio between the voltages on the two coils:
\frac{N_p}{N_s}= \frac{V_p}{V_s}
Where N_p and N_s are the number of turns in the primary and secondary coils, while V_p and V_s are the voltages on the two coils.

Using the data of the problem: N_p=750, V_p=10000 V and V_s=120 V, we can find N_s, the number of turns of the secondary coil:
N_s=N_p  \frac{V_s}{V_p}=750  \frac{120 V}{10000 V}=9
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Andreyy89

Answer:

A)T=209.94N

B) F=75.24N

Explanation:

Using the free body diagram and according to Newton's first law, we have:

\sum F_y=Tcos(21^\circ)-mg=0(1)\\\sum F_x=F-Tsin(21^\circ)=0(2)

A) Solving (1) for T:

T=\frac{mg}{cos(21^\circ)}\\T=\frac{20kg(9.8\frac{m}{s^2})}{cos(21^\circ)}\\T=209.94N

B) Solving (2) for F:

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A solid yellow line on your side of the center stripe means pass with care.<br> A. True<br> B. False
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B

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At time t=0, a particle is located at the point (3,6,9). It travels in a straight line to the point (5,2,7), has speed 8 at (3,6
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The particle has constant acceleration according to

\vec a(t)=2\,\vec\imath-4\,\vec\jmath-2\,\vec k

Its velocity at time t is

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\displaystyle\vec r(t)=\vec r(0)+\int_0^t\vec v(u)\,\mathrm du

\vec r(t)=(3+v_{0x}t+t^2)\,\vec\imath+(6+v_{0y}t-2t^2)\,\vec\jmath+(9+v_{0z}t-t^2)\,\vec k

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\|\vec v(0)\|=8\iff{v_{0x}}^2+{v_{0y}}^2+{v_{0z}}^2=64

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\begin{cases}3+v_{0x}T+T^2=5\\6+v_{0y}T-2T^2=2\\9+v_{0z}T-T^2=7\end{cases}\implies\begin{cases}v_{0x}=\dfrac{2-T^2}T\\\\v_{0y}=\dfrac{2T^2-4}T\\\\v_{0z}=\dfrac{T^2-2}T\end{cases}

and in particular we see that

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