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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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Is the image formed by a plane mirror always real?<br><br>True or false?​
mafiozo [28]
This statement is false
6 0
2 years ago
Lightweight, vertically suspended spiral spring with a spring constant of 8.6 N / m is fitted with 64 g weight. The weight shall
alexandr1967 [171]

Answer:

Explanation:

Given that

Force constant k=8.6N/m

Weight =64g=64/1000=0.064kg

Extension is 45mm=45/1000= 0.045m

It will have it highest spend when the Potential energy is zero

Therefore energy in spring =change in kinetic energy

Ux=∆K.e

½ke² = ½mVf² — ½mVi²

Initial velocity is 0, Vi=0m/s

½ke² = ½mVf²

½ ×8.6 × 0.045² = ½ ×0.064 ×Vf²

0.0087075 = 0.032 Vf²

Then, Vf² = 0.0087075/0.032

Vf² = 0.2721

Vf=√0.2721

Vf= 0.522m/s

The time it will have this maximum velocity?

Using equation of motion

Vf= Vi + gr

0.522= 0+9.81t

t=0.522/9.81

t= 0.0532sec

t= 53.2 milliseconds

5 0
3 years ago
An airplane has an effective wing surface area of 17.0 m2 that is generating the lift force. In level flight the air speed over
olga_2 [115]

To solve this problem it is necessary to apply the equations given from Bernoulli's principle, which describes the behavior of a liquid moving along a streamline. Mathematically this expression can be given as,

P_1 + \frac{1}{2}\rho*v_1^2 + P_2 + \frac{1}{2}*\rho*v_2^2=0

Where,

P_i = Pressure at each state

\rho= Density

v_i = Velocity

Re-organizing the expression we can get that

P_1 - P_2 = \frac{1}{2}\rho (v_2^2 - v_1^2)

Our values are given as

v_1 = 40m/s

v_2 = 55m/s

\rho_{water} = 1.2kg/m^3 \rightarrow Normal Conditions

Replacing we have,

P_1 -P_2 = \frac{1}{2}*1.2*(55^2-40^2)

P_1 - P_2 = 855Pa

If we consider that there is a balance between the two states, the Force provided by gravity is equivalent to the Support Force, therefore

F_l = F_g

Here the lift force is the product between the pressure difference previously found by the effective area of the aircraft, while the Force of gravity represents the weight. There,

F_g = W

F_l = (P_2-P_1)A

Equating,

(P_1 - P_2)*A = W

W = 855*17

W = 14535 N

Therefore the weight of the plane is 14535N

3 0
3 years ago
A 1.30-kg object is held 1.10 m above a relaxed, massless vertical spring with a force constant of 315 N/m. The object is droppe
pshichka [43]

Answer:

0.345m

Explanation:

Let x (m) be the length that the spring is compress. If we take the point where the spring is compressed as a reference point, then the distance from that point to point where the ball is held is x + 1.1 m.

And so the potential energy of the object at the held point is:

E_p = mgh

where m = 1.3 kg is the object mass, g = 10m/s2 is the gravitational acceleration and h = x + 1.1 m is the height of the object with respect to the reference point

E_p = 1.3 * 10 * (x + 1.1) = 13(x + 1.1) = 13x + 14.3 J

According to the conservation law of energy, this potential energy is converted to spring elastic energy once it's compressed

E_p = E_k = kx^2/2 = 13x + 14.3

where k = 315 is the spring constant and x is the compressed length

315x^2 = 26x + 28.6

315x^2 - 26x - 28.6 = 0

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

x = \frac{26 \pm \sqrt{26^2 - 4*(-28.6)*315}}{2*315}

x = \frac{26 \pm 191.6}{630}

x = 0.345 m or x = -0.263 m

Since x can only be positive we will pick the 0.345m

6 0
3 years ago
Can you please explain and be able to help with further more questions?!
bagirrra123 [75]
The answer is 0m because at point 8s the displacement is at zero m for example at 3s the displacement is at 8m
5 0
3 years ago
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