All categories

2 years ago

A(n) ______ can act as an electronic switch that opens or closes the circuits for electrical charges.

Physics

1 answer:

uysha [10]2 years ago

8 0

Answer: diode

Explanation: A diode is an electronic device that has the ability to work in 1 direction of electric current.

At forward bias ( when the positive terminal of the diode is connected to the positive terminal of the source and negative terminal to the negative of the source) the diode allows current to flow thus acting as a close switch.

At reverse bias ( when the positive terminal of the diode is connected to the negative terminal of the source and the negative terminal of the diode to the positive terminal of the source) the diode stops question from flowing thus acting like an open circuit or an open switch

You might be interested in

Which of these equations will you used to find the final velocity if the initial

Svetach [21]

Answer:

2.77 would be the answer for this

5 0

2 years ago

How would you find the resistance of a parallel circuit with n identical resistors?

VMariaS [17]

Answer:

write the resistance of one resistor

Explanation:

This is because when resistors are connected in parallel current flows in different directions.

3 0

2 years ago

Read 2 more answers

to what height will a 250g soccer ball rise to if it is kicked directly upwards at 8 meters per second

Nonamiya [84]

Answer:

3.2 m

Explanation:

we know

Hmax = V²/2g

= 8² / 2*10 = 3.2

6 0

2 years ago

Read 2 more answers

SHOW WORK

Helga [31]

Answer:

Follows are the solution to the given question:

Explanation:

For point a:

$T= 2\pi \sqrt{\frac{m}{k}}\\\\k = \frac{4 \pi^2 m}{T^2}\\\\= \frac{4 \times (3.14)^2 \times 3}{2^2}\\\\=29.578 \ \frac{N}{m}\\\\$

For Point b:

$E=\frac{1}{2} m a^2 w^2\\\\$

$=\frac{1}{2} \times m \times a^2 \times \frac{4\pi^2}{T^2}\\\\=\frac{1}{2} \times 3 \times (0.15)^2 \times \frac{4\times 3.14^2}{2^2}\\\\=0.332 \ J$

For Point C:

$V_{max}= a w$

$= (0.15) \times \frac{2\pi}{T}\\\\= (0.15) \times \frac{2\times 3.14}{2}\\\\=0.471 \frac{m}{s}$

For point D:

$X= a \sin (wt+ \phi)\\\\0.91=0.15 \sin(\frac{2\pi}{T} \times t+\phi)\\\\0.91=0.15 \sin(\frac{2\times 3.14}{2} \times 0.5+\phi)\\\\0.60 = \sin(3.14 \times 0.5+\phi)\\\\0.60 = \sin(1.57+\phi)\\\\1.57 +\phi =\sin^{-1} 60^{\circ}\\\\1.57 +\phi = 36.86^{\circ}\\\\=35.29^{\circ}\\\\So, X=15 \sin(3.14t+35.29^{\circ}) \ cm$