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Reptile [31]
3 years ago
10

You generally need a transformer to operate a doorbell because doorbells are designed to work with much smaller voltages than wh

at you typically get from wiring in a house. One particular doorbell needs a potential difference of 16V to work, while the standard outlet voltage (in the United States) is 120V. You have a transformer with 2250 turns in the primary coil. How many turns do you need in the secondary coil to power the doorbell?
Physics
1 answer:
mars1129 [50]3 years ago
7 0

Answer: 300 turns

Explanation: In order to explain this problem we have to consider the relationship between the voltage for the primiry and secondary and the turns for their coils.

This expression is consequence of the Faraday law because the magnetic flux is the same for both coils then the induced emf are the same for both circuits.

They are reñated as follow:

Vs=Vp*Ns/Np  where Vs and Vp are the voltage for the secondary and primary coil, respectively.

Ns and Np are the turns for the secondary and primary coil, respectively.

Then we have:

Ns=Vs*Np/Vp= 16*2250/120=300

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3 years ago
The action of two forces. One is a forward force of 1157 N provided by traction between the wheels and the road. The other is a
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Complete question

A 2700 kg car accelerates from rest under the action of two forces. one is a forward force of 1157 newtons provided by traction between the wheels and the road. the other is a 902 newton resistive force due to various frictional forces. how far must the car travel for its speed to reach 3.6 meters per second? answer in units of meters.

Answer:

The car must travel 68.94 meters.

Explanation:

First, we are going to find the acceleration of the car using Newton's second Law:

\sum\overrightarrow{F}=m\overrightarrow{a} (1)

with m the mass , a the acceleration and \sum\overrightarrow{F} the net force forces that is:

(F-f) (2)

with F the force provided by traction and f the resistive force:

(2) on (1):

(F-f)=ma

solving for a:

a=\frac{F-f}{m} =\frac{1157N-902N}{2700kg} =0.094\frac{m}{s^{2}}

Now let's use the Galileo’s kinematic equation

Vf^{2}=Vo^{2}+2a\varDelta x (3)

With Vo te initial velocity that's zero because it started from rest, Vf the final velocity (3.6) and \varDelta x the time took to achieve that velocity, solving (3) for \varDelta x:

\varDelta x= \frac{Vf^{2}}{2a} = t= \frac{(3.6\frac{m}{s})^2}{2*0.094\frac{m}{s^{2}}}

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3 years ago
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Check out the picture I drew for a minute before reading this...

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Answer:

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