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3 years ago

What type of circuit is illustrated?

Physics

2 answers:

Elis [28]3 years ago

8 0

A. Parallel circuit

Short circuits are circuits that is like plugging in a USB into your computer and the other end to your iPhone, so it's not D.

Series circuits can come in all kinds of different shapes, so it's not C.

Open circuits are circuits that are opened, meaning that whenever you turn on the light and it's an open circuit, it won't work cause the two wires are not connected, meaning it's not D.

Parallel circuits are circuits that are in one straight line, just like parallel lines, meaning the answer would be A. (Look at the picture and don't get confused with the other question that's the same question and has the same answers, they both have different pictures, so look at the pictures they give you whenever they ask the question so your getting the right answer)

Hope this helped.

Ber [7]3 years ago

4 0

It will either be <span>A.parallel circuit or C.series circuit it depends on the picture because there are 2 of the same questions with the same picture on the test. ( i know this because i use K12).
Hope This Helps!
~Cupkake </span>

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3 years ago

Michael Jordan, el célebre basquetbolista, ganó el torneo de clavadas de la NBA en 1988. Para lograr la hazaña saltó 1.35 metros

kozerog [31]

(a) 0.40 s

First of all, let's find the initial speed at which Jordan jumps from the ground.

The maximum height is h = 1.35 m. We can use the following equation:

$v^2-u^2=2gh$

where

v = 0 is the velocity at the maximum height

u is the initial velocity

$g=-9.8 m/s^2$ is the acceleration of gravity

Solving for u,

$u=\sqrt{-2gh}=\sqrt{-2(-9.8)(1.35)}=5.14 m/s$

The time needed to reach the maximum height can now be found by using the equation

$v=u+gt$

Solving for t,

$t=\frac{v-u}{g}=\frac{0-5.14}{-9.8}=0.52s$

Now we can find the velocity at which Jordan reaches a point 20 cm below the maximum height, so at a height of

h' = 1.35 - 0.20 = 1.15 m

Using again the equation

$v'^2-u^2=2gh'$

we find

$v'=\sqrt{u^2+2gh}=\sqrt{5.14^2+2(-9.8)(1.15)}=1.97 m/s$

And the corresponding time is

$t'=\frac{v'-u}{g}=\frac{1.97-5.14}{-9.8}=0.32s$

So the time to go from h' to h is

$\Delta t = t-t'=0.52-0.32=0.20 s$

And since we have also to take into account the fall down (after Jordan reached the maximum height), which is symmetrical, we have to multiply this time by 2 to get the total time of permanence in the highest 20 cm of motion:

$\Delta t=2\cdot 0.20 = 0.40 s$