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3 years ago

Which of the following is a transformer used for?

2 answers:

5 0

A transformer is used to increase or decrease a voltage. BUT ... it has to be an AC voltage, or the transformer doesn't work.

denis23 [38]3 years ago

5 0

Answer: Increase or decrease a voltage

Explanation: Transformers are designed <u>for increasing or decreasin</u>g the voltage in an electrical circuit, using the principle of e<em>lectromagnetic induction</em>.

These only work with alternating currents, because this alternating current drifts in the generation of permanent magnetic flux. The arrangement to make this possible consists of a pair of coils with a number of different spirals isolated from each other, and a ferromagnetic material is added to the core to optimize the magnetic flux. Thanks to the different number of spirals in the coils the increase or decrease of the voltage is achieved.

The voltage transformation follows the following mathematical formula:

$\frac{V_{s} }{V_{p} }=\frac{N_{s} }{N_{p} }$

Where

[tex]\V_{s} = Secondary voltage

[tex]\V_{p} = Primary voltage

[tex]\[N_{s} = Secondary number of spirals

[tex]\N_{p} = Primary number of spirals

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The spring is compressed a total of 3.0 cm, and used to set a 500 gram cart into motion. Find the speed of the cart at the insta

BartSMP [9]

Answer:

1.15 m/s

Explanation:

Part of the question is missing. Found the missing part on google:

"1. A hanging mass of 1500 grams compresses a spring 2.0 cm. Find the spring constant in N/m."

Solution:

First of all, we need to find the spring constant. We can use Hooke's law:

$F=kx$

where

$F=mg=(1.5 kg)(9.8 m/s^2)=14.7 N$ is the force applied to the spring (the weight of the hanging mass)

x = 2.0 cm = 0.02 m is the compression of the spring

Solving for k, we find the spring constant:

$k=\frac{F}{x}=\frac{14.7}{0.02}=735 N/m$

In the second part of the problem, the spring is compressed by

x = 3.0 cm = 0.03 m

So the elastic potential energy of the spring is

$U=\frac{1}{2}kx^2=\frac{1}{2}(735)(0.03)^2=0.33 J$

This energy is entirely converted into kinetic energy of the cart, which is:

$U=K=\frac{1}{2}mv^2$

where

m = 500 g = 0.5 kg is the mass of the cart

v is its speed

Solving for v,

$v=\sqrt{\frac{2K}{m}}=\sqrt{\frac{2(0.33)}{0.5}}=1.15 m/s$

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The correct option is A.
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