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3 years ago

15

How long will it take, in minutes, for a transformer to transfer 2.3 X 10^6 J of energy from a 120-V circuit to a 345-V circuit

with a current of 1.5 A?

1 answer:

masya89 [10]3 years ago

3 0

Answer:

e

Explanation:

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What is rotational speed

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The number of turns of the object divided by time

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A traffic-safety engineer is designing a deceleration lane. She is basing the length of

Mice21 [21]

The deceleration of the car is observed as 2 m/s^2.

<h3>What is deceleration?</h3>

The term deceleration has to do with the decrease of velocity with time.

Given that;

v = 65 km/h or 18 m/s

u = 126 km/h to or 35 m/s

t = 8.5 s

a = ?

v = u - at

18 - 35 = -8.5a

a = 18 - 35/-8.5

a = 2 m/s^2

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2 years ago

If you have a positive electrical force, and a negative electrical force, what do you expect will happen?

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I expect that they will <em>add</em>, and their effect at every location will be the <em>sum</em> of their individual effects at that location.

For example:

If they're acting at the same point and in opposite directions, the effect will be the same as a single force at that point, with strength equal to their difference, and in the direction corresponding to whichever one is stronger.

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3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

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3 years ago

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HELP ME PLEASE ASAP NO LINKS!!!

Paladinen [302]

Answer:

have you tried number 1, 3, and 5?

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3 years ago