Ask question

Ask question

All categories

3 years ago

12

Are switches connected in series or parallel with the devices they control?

1 answer:

andrezito [222]3 years ago

3 0

In some unusual applications of unusual components, I can think of unusual electric circuits where a switch may be connected in parallel with a device in order to control it.

But I'm sure this is not what's intended in a question on the high-school level.

Until you get in a situation with tricky applications in a tricky circuit, your switches will always be connect <em>in series</em> with the devices they control.

You might be interested in

E20E20: The word “virtual” refers to something that exists in effect but not in actual fact. How does this definition relate to

Lubov Fominskaja [6]

The correct answer (sample response) is:

The image seems to be behind the mirror, but nothing is really there.

Include the following in your response:

The image appears to be behind the mirror.

If someone looks behind the mirror, there is no image there.

$|Huntrw6|$

5 0

3 years ago

Read 2 more answers

Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45 m diamete

Readme [11.4K]

Answer:

In m/s^2:

a=11.3778 m/s^2

In units of g:

a=1.161 g

Explanation:

Since the racing greyhounds are capable of rounding corners at very high speed so we are going use the following formula of acceleration for circular paths.

$a=\frac{v^2}{r}$

where:

v is the speed

r is the radius

Now,

$a=\frac{16^2}{45/2}\\ a=11.3778 m/s^2$

In g units:

$a=\frac{11.3778\ g}{9.8}\\ a=1.161\ g$

7 0

3 years ago

Imagine that the earth and the-moon have positive charges of the same magnitud. How big isäºthe charge necesary to produce an el

lions [1.4K]

Answer:

5.7 x 10^12 C

Explanation:

Let the charge on earth and moon is q.

mass of earth, Me = 5.972 x 10^24 kg

mass of moon, Mm = 7.35 x 10^22 kg

Let d be the distance between earth and moon.

the gravitational force between them is

$F_{g}=G\frac{M_{e} \times M_{m}}{d^{2}}$

The electrostatic force between them is

$F_{e}=\frac{Kq^{2}}{d^{2}}$

According to the question

1 % of Fg = Fe

$0.01 \times 6.67\times10^{-11}\frac{5.97 \times 10^{24}\times7.35 \times 10^{22}}{d^{2}}=9 \times 10^{9}\frac{q^{2}}{d^{2}}$

$2.927 \times 10^{35}=9 \times10^{9}q^{2}$

$3.25 \times 10^{25}=q^{2}$

q = 5.7 x 10^12 C

Thus, the charge on earth and the moon is 5.7 x 10^12 C.

6 0

3 years ago

A coyote can locate a sound source with good accuracy by comparing the arrival times of a sound wave at its two ears. Suppose a

il63 [147K]

Answer:

0.0000109261200583 s

0.0109261200583

Explanation:

$d_2$ = Distance from right ear = 3 m

s = Distance between ears = 15 cm

v = Speed of sound in air = 343 m/s

Distance between the left ear and the bird

$d_1=\sqrt{s^2+d_2^2}\\\Rightarrow d_1=\sqrt{0.15^2+3^2}\\\Rightarrow d_1=3.00374765918\ m=3.004\ m$

Time

$t=\dfrac{Distance}{Speed}$

Time difference would be

$\Delta T=\dfrac{d_1}{v}-\dfrac{d_2}{v}\\\Rightarrow \Delta T=\dfrac{3.00374765918}{343}-\dfrac{3}{343}\\\Rightarrow \Delta T=0.0000109261200583\ s$

The time difference is 0.0000109261200583 s

Time period is given by

$T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{1000}\\\Rightarrow T=10^{-3}\ s$

The ratio is

$\dfrac{\Delta T}{T}=\dfrac{0.0000109261200583}{10^{-3}}\\\Rightarrow \dfrac{\Delta T}{T}=0.0109261200583$

The ratio is 0.0109261200583

7 0

3 years ago

A 500-n crate needs to be lifted 1 meter vertically in order to get it into the back of a pickup truck. what gives the crate a g

worty [1.4K]

The weight an weight of the truck

6 0

3 years ago

Read 2 more answers