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3 years ago

A student wants to create a 6.0V DC battery from a 1.5V DC battery. Can this be done using a transformer alone

Physics

1 answer:

shusha [124]3 years ago

8 0

Answer:

Therefore, we need an invert, and a rectifier, along with the transformer to do the job.

Explanation:

A transformer, alone, can not be used to convert a DC voltage to another DC voltage. If we apply a DC voltage to the primary coil of the transformer, it will act as short circuit due to low resistance. It will cause overflow of current through winding, resulting in overheating pf the transformer.

Hence, the transformer only take AC voltage as an input, and converts it to another AC voltage. So, the output voltage of a transformer is also AC voltage.

So, in order to convert a 6 V DC to 1.5 V DC we need an inverter to convert 6 V DC to AC, then a step down transformer to convert it to 1.5 V AC, and finally a rectifier to convert 1.5 V AC to 1.5 V DC.

<u>Therefore, we need an invert, and a rectifier, along with the transformer to do the job.</u>

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With the intensity at about half of maximum, how does the current change when you increase the frequency of the light?

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3 years ago

A rock is dropped (from rest) off a bridge over the Merrimack River. The falling rock

rewona [7]

Answer:

31.25 meters or ~31 meters approximately

Explanation:

Let's see which of the 5 variables we are given since this is a constant acceleration problem.

$v_i \ \ \ \ \ \ t \\ v_f \ \ \ \ \ \triangle x \\ a$

We want to find the height of the bridge, aka the vertical displacement of the rock. Let's set the upwards direction to be positive and the downwards direction to be negative.

We are told that the acceleration is 10 m/s² downward, so we have a = -10 m/s².

We are also told that the time it takes the rock to hit the water is 2.5 seconds. Time is the same regardless of the x- or y- direction, so we can say that t = 2.5 seconds.

Now, we aren't told this directly, but we can figure out that the velocity in the y-direction is 0 m/s, since the rock is dropped from rest off the bridge. Therefore, $v_i=0 \frac{m}{s}$ .

We want to find the vertical displacement, the height of the bridge, so we can say that $\triangle x= \ ?$

We have 4 out of 5 variables:

$v_i,\ a, \ t, \ \triangle x$

Look through the constant acceleration equations to see which equation has all 4 of these variables. You should come up with this one (no final velocity):

$x_f=x_i+v_it+\frac{1}{2}at^2$

Subtract $x_i$ from both sides of the equation to get:

$\triangle x=v_it+\frac{1}{2}at^2$

Substitute in our known variables and solve for delta x.

$\triangle x=(0\frac{m}{s})(2.5s) + \frac{1}{2} (-10\frac{m}{s^2})(2.5s)^2$

0 m/s multiplied by 2.5 s is 0, so we have:

$\triangle x =\frac{1}{2} (-10)(2.5)^2$

Evaluate the exponent first and multiply the terms together.

$\triangle x =(-5)(6.25)$
$\triangle x =-31.25$

The vertical displacement is -31.25 meters from the rock's starting position, so we can say that the height of the bridge is 31.25 meters, which is approximately 31 meters tall.

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3 years ago

Read 2 more answers

Please help!! I’m so confused and struggling rn

DIA [1.3K]

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