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3 years ago

A 18Ω resistance is cut into three equal parts and connected in parallel. Find the equivalent resistance of the combination.

Physics

1 answer:

Andrew [12]3 years ago

5 0

Answer:

2Ω

Explanation:

If a 18Ω resistance is cut into three equal parts each of the resistance will be 18Ω/3 = 6Ω

Equivalent ratio in parallel is expressed as:

1/R = 1/6 + 1/6 + 1/6

1/R = 3/6

Cross multiply

3R = 6

R = 6/3

R = 2Ω

Hence the required equivalent resistance is 2Ω

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What will happen when a light bulb is removed from the circuit below, leaving a gap?

Charra [1.4K]

Answer:

The remaining light bulbs will go out.

Explanation:

The light bulb that was taken out routed power to the other light bulbs, there for, not giving power to the next light bulbs will make them turn off or, "go out". This may be incorrect, as you did not provide a picture of the circuit.

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3 years ago

Read 2 more answers

Consider a monochromatic electromagnetic plane wave propagating in the x direction. At a particular point in space, the magnitud

Scrat [10]

Answer:

a) S = 2.35 10³ J/m²2 ,

b)and the tape recorder must be in the positive Z-axis direction.

the answer is 5

c) the direction of the positive x axis

Explanation:

a) The Poynting vector or intensity of an electromagnetic wave is

S = 1 /μ₀ E x B

if we use that the fields are in phase

B = E / c

we substitute

S = E² /μ₀ c

let's calculate

s = 941 2 / (4π 10⁻⁷ 3 10⁸)

S = 2.35 10³ J/m²2

b) the two fields are perpendicular to each other and in the direction of propagation of the radiation

In this case, the electro field is in the y direction and the wave propagates in the ax direction, so the magnetic cap must be in the y-axis direction, and the tape recorder must be in the positive Z-axis direction.

the answer is 5

C) The poynting electrode has the direction of the electric field, by which or which should be in the direction of the positive x axis

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3 years ago

In a LRC circuit, a second capacitor is connected in parallel with the capacitor previously in the circuit. What is the effect o

likoan [24]

Answer:

Impedance increases for frequencies below resonance and decreases for the frequencies above resonance

Explanation:

See attached file

Explanation:

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3 years ago

A new cat 6 cable run and key stone jack were installed for a workstation that was moved from the sales department to the shippi

Andrei [34K]

Answer:

The cable run exceeds the specifications for Ethernet over twisted pair

Explanation:

The ethernet network's router also serves as a bridge to the Internet. The router connects to the modem, which carries the Internet signal, sending and receiving data packet requests and routing them to the proper computers on the network.

Ethernet is a way of connecting computers together in a local area network or LAN. It has been the most widely used method of linking computers together in LAN s since the 1990 s.

The basic idea of its design is that multiple computers have access to it and can send data at any time.

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3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

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2 years ago