Can someone help me with this?

A transformer with X turns in primary coil and Y turns in secondary coil is used to change the magnitude of voltage to 240 V. Ca

Vinil7 [7]

The input voltage is 120 V and the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

<h3>Input voltage </h3>

The input voltage of the transformer is the voltage of the primary coil and it is calculated as follows;

Ns/Np = Es/Ep

where;

Ns is the number of turn in the secondary coil
Np is the number of turn in the primary coil
Es is the secondary voltage
Ep is the primary voltage

2X/X = 240/Ep

2 = 240/Ep

Ep = 240/2

Ep = 120 V

Thus, the transformer is a step up transformer due to increase in the voltage induced in the secondary coil.

Learn more about transformer here: brainly.com/question/25886292

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3 0

2 years ago

What does this equation mean FeDe=FrDr

rjkz [21]

Not exactly the best way to describe it but, it is used to calculate resistance of a lever as in the use of a pry bar or pulley. Technology used to increase output with little input.

3 0

3 years ago

Grop 17 of the periodic table contains the ?

Andrei [34K]

Group 17 is the second column from the right in the periodic table and contains six elements: fluorine (F), chlorine (Cl), bromine (Br), iodine (I), astatine (As), and (Ts). Astatine and are radioactive elements with very short half-lives and thus do not occur naturally.

8 0

3 years ago

A 100-lb child stands on a scale while riding in an elevator. What does the scale read while the elevator slows to stop at the l

Lelechka [254]

Answer: A 100-lb child stands on a scale while riding in an elevator. Then, the scale reading approaches to 100lb, while the elevator slows to stop at the lowest floor

Explanation: To find the correct answer, we need to know more about the apparent weight of a body in a lift.

<h3>What is the apparent weight of a body in a lift?</h3>

Consider a body of mass m kept on a weighing machine in a lift.
The readings on the machine is the force exerted by the body on the machine(action), which is equal to the force exerted by the machine on the body(reaction).
The reaction we get as the weight recorded by the machine, and it is called the apparent weight.

<h3>How to solve the question?</h3>

Here we have given with the actual weight of the body as 100lbs.
This 100lb child is standing on the scale or the weighing machine, when it is riding .
During this condition, the acceleration of the lift is towards downward, and thus, a force of ma .
There is also<em> mg </em>downwards and a normal reaction in the upward direction.
when we equate both the upward force and downward force, we get,

$ma=mg-N\\N=mg-ma$ i.e. during riding the scale reads a weight less than that of actual weight.

When the lift goes slow and stops the lowest floor, then the acceleration will be approaches to zero.

Thus, from the above explanation, it is clear that ,when the elevator moves to the lowest floor slowly and stops, then the apparent weight will become the actual weight.

Learn more about the apparent weight of the body in a lift here:

brainly.com/question/28045397

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7 0

2 years ago

You do 120 j of work while pulling your sister back on a swing, whose chain is 5.10 m long. you start with the swing hanging ver

Goryan [66]

The work done to pull the sister back on the swing is equal to the increase in potential energy of the sister:
$W= \Delta U = mg \Delta h$ (1)

where m is the sister's mass, g is the gravitational acceleration and $\Delta h$ is the increase in altitude of the sister with respect to its initial position.

By calling $\theta$ the angle of the chain with respect to the vertical, the increase in altitude is given by
$\Delta h = L - L \cos \theta = L(1 - \cos \theta)$ (2)
where L is the length of the chain.

Putting (2) inside (1), we find
$W= m g L (1 - \cos \theta)$
from which we can find the mass of the sister:
$m = \frac{W}{g L (1 - \cos \theta)} = \frac{120 J}{(9.81 m/s^2)(5.10 m)(1- \cos 32.0^{\circ})} =15.8 kg$

5 0

3 years ago