All categories

3 years ago

Transformers will not work on ______ electrical systems.

Physics

2 answers:

Andru [333]3 years ago

5 0

Transformers only work with AC.

If you put DC into a transformer, all that comes out is smoke.

(a)

Sphinxa [80]3 years ago

3 0

Transformers will not work on direct current in electrical systems.

Answer: Option A

Explanation:

Transformers work only on the alternate current of electrical system. And they are not suitable to work upon with the direct current. As the direct current has the property to transfer high voltage directly and is unable to direct the current with high and low voltages, causes the transformer to burn out the coil leading to burning of the transformer circuit and causing smoked transformers.

You might be interested in

Plz do all of it i will give brainlest and thanks to best answer plz do it right

AlladinOne [14]

The answer is a rainforest I’m pretty sure

3 0

2 years ago

Read 2 more answers

HELP ASAP PLS! Explain Newton’s 3 laws of motion by using the example of a rollercoaster.

dybincka [34]

For every action there is an equal and opposite reaction." So that applies to a roller coaster, between the ride vehicles and the track. When a ride goes up and down the hill, it creates different forces onto the track.

8 0

3 years ago

I would like help with this physics problem

Darina [25.2K]

(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.

Recall that

$v_f=v_i+at$

We have $v_f=-v_i$ , so that

$-2v_i=at\implies-2\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)=\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)t\implies t=8\,\mathrm s$

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.

(d) The position of the ball $x_f$ at time $t$ is given by

$x_f=x_i+v_it+\dfrac12at^2$

Take the starting position to be the origin, $x_i=0$ . Then after 6 seconds,

$x_f=\left(8\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(-2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2=42\,\mathrm m$

so the ball is 42 m away from where it started.

We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

$x_f-x_i=\dfrac{v_i+v_f}2t\implies42\,\mathrm m=\dfrac{8\,\frac{\mathrm m}{\mathrm s}+v_f}2(6\,\mathrm s)\implies v_f=6\,\dfrac{\mathrm m}{\mathrm s}$

Since the velocity is positive, the ball is still moving up the incline.

8 0

3 years ago

Solve the given parallel circuit by computing the desired quantities.

Alex Ar [27]

Hope this provides some answers

7 0

3 years ago

You are exploring a planet and drop a small rock from the edge of a cliff. In coordinates where the +y direction is downward and

Lelu [443]

Answer:

value of the acceleration of gravity on the planet is 5.00 m/s²

Explanation:

The problem is similar to a free fall exercise, with another gravity value, the expression they give us is the following:

y-yo = ½ gₐ t² (1)

They tell us that they make a squared time graph with the variation of the distance, it is appropriate to clarify this in a method to linearize a curve, which is plotted the nonlinear axis to the power that is raised, specifically, the linearization of a curve The square is plotted against the other variable.

Let's continue our analysis, as we have a linear equation, write the equation of the line.

y1 = m x1 + b (2)

where “y1” the dependent variable, “x1” the independent variable, “m” the slope and “b” the short point

In this case as the stone is released its initial velocity is zero which implies that b = 0,

We plot on the “y” axis the time squared “t²” and on the horizontal axis we place “y-yo”. To better see the relationship we rewrite equation 1 with this form

t² = 2 /gₐ (y-yo)

With the two expressions written in the same way, let's relate the terms one by one

y1 = t²

x1 = (y-yo)

m = 2/gap

b= 0

We substitute and calculate

m = 2/gp

gₐ = 2/m

gₐ = 2/ 0.400

gₐ = 5.00 m / s²

This is the value of the acceleration of gravity on the planet, note that the decimals are to keep the figures significant

6 0

3 years ago